# Binomial Theorem Problem

1. Mar 26, 2014

### BOAS

Hello,

I have a problem regarding the binomial theorem and a number of questions about what I can and can't do.

1. The problem statement, all variables and given/known data

Write the binomial expansion of $(1 + x)^{2}(1 - 5x)^{14}$ as a series of powers of $x$ as far as the term in $x^{2}$

2. Relevant equations

3. The attempt at a solution

I know how to expand each bracket separately but i'm really unsure of how to proceed with one multiplied by the other.

Do I expand the first one, and have that as a factor of every term in the expansion of the second?

i.e $(1 + x)^{2} = 1 + x^{2} + 2x$

$(1 - 5x)^{14} \approx 1 - 70x - 455x^{2}$

$(1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + x^{2} + 2x) - 70x(1 + x^{2} + 2x) - 455x^{2}(1 + x^{2} + 2x)$

$(1 + x)^{2}(1 - 5x)^{14} \approx 1 + 2x + x^{2} - 70x - 140x^{2} - 70x^{3} - 455x^{2} - 910x^{3} - 455x^{4}$

I get the feeling that this is wrong, but I can't find any similar examples in my text book or notes. If this happens to be the correct method, have I included to high powers? The individual expansions only reach $x^{2}$, but when they are combined, clearly it goes higher.

Last edited: Mar 26, 2014
2. Mar 26, 2014

### vela

Staff Emeritus
The $x^2$ term in the expansion for $(1-5x)^{14}$ should be positive. Your work is otherwise okay, but you didn't need to calculate the $x^3$ and $x^4$ terms. You just want to identify which products will result in terms of order $x^2$ or lower and keep track of those.

3. Mar 26, 2014

### BOAS

Ah, I forgot to square the coefficient. It should be;

$(1 - 5x)^{14} \approx 1 - 70x + 2275x^{2}$

wrt the rest of your post, does that mean I should approximate the two expansions only to the 'x' terms? Or do as before and ignore the higher powers?

Thanks for the help.

4. Mar 26, 2014

### vela

Staff Emeritus
You have to keep up to at least the $x^2$ terms because they will contribute to the final result.

5. Mar 26, 2014

### BOAS

So,

$(1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + 2x + x^{2}) - 70x(1 + 2x + x^{2}) + 2275x^{2}(1 + 2x + x^{2})$

$(1 + x)^{2}(1 - 5x)^{14} \approx 1 + 2x + x^{2} - 70x - 140x^2 + 2275x^{2}$

$(1 + x)^{2}(1 - 5x)^{14} \approx 1 - 68x + 2136x^2$

I think this is what you meant when you said to keep track of the products that would give me $x^2$ and lower.

6. Mar 26, 2014

### vela

Staff Emeritus
Yup, and you can streamline it a bit further:
$$(1 + x)^{2}(1 - 5x)^{14} \approx 1(1 + 2x + x^{2}) - 70x(1 + 2x) + 2275x^{2}(1)$$

7. Mar 26, 2014

### BOAS

Cool - Thank you

8. Mar 26, 2014

### PeroK

That's correct. Although it might be interesting to consider for what values of x that approximation is accurate!

9. Mar 26, 2014

### BOAS

My expansion for $(1 + x)^{2}$ was exact, but not every term of it was used when mutliplying with the second expansion.

My expansion for $(1 - 5x)^{14}$ holds provided that $-1 < -5x < 1$ so $\frac{1}{5} > x > - \frac{1}{5}$

I don't know how I combine this information.


# Comparing Two Sums with Binomial Coefficients

How do I use pascals identity:

$${2n\choose 2k}={2n-1\choose 2k}+{2n-1\choose 2k-1}$$ to prove that

$$\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=\displaystyle\sum_{k=0}^{2n-1}{2n-1\choose k}$$

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Try writing out Pascal's identity for n = 1, 2, 3, 4 and writing out the equation that you're trying to prove for n = 4 (without simplifying the binomial coefficients). See if you can give a proof in that case. If you're able to do this, you'll see how to do the general case. –  Jonah Sinick Oct 26 '12 at 20:30

You have $$\sum_{k=0}^n\binom{2n}{2k}=\sum_{k=0}^n\left(\binom{2n-1}{2k}+\binom{2n-1}{2k-1}\right)$$ Expand the right side. Here are first two and the last terms: $$\left(\binom{2n-1}{0}+\binom{2n-1}{-1}\right)+\left(\binom{2n-1}{2}+\binom{2n-1}{1}\right)+\cdots+\left(\binom{2n-1}{2n}+\binom{2n-1}{2n-1}\right)$$ Do you see the pattern? Can you express this as a sum? There's only one slightly subtle point.

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@Cameron. Where I come from, $\binom{2n-1}{-1}\text{ and }\binom{2n-1}{2n}$ are both defined to be zero. That's a fairly common extension. –  Rick Decker Oct 26 '12 at 20:16
Never seen that before, but that's good to know (and certainly intuitive). It might be worth noting explicitly in the answer, though, since the binomial coefficient is defined in many different ways depending on the source. –  Cameron Buie Oct 26 '12 at 20:48
@Cameron. That's why I ended my answer as I did: that was my "subtle point." –  Rick Decker Oct 26 '12 at 20:54
thanks! i got it! but where can i find the definition of $${2n-1\choose -1}$$ and $${2n-1\choose 2n}$$? i never heard about it, and i would like to read more! –  Algosub Oct 27 '12 at 9:40
@Tomer. Lots of people do this. For example, Knuth's Concrete Mathematics, Ch.5. In general, $\binom{n}{m}=0$ when $m<0$ and also when $n<m$. In other words, any integer pairs not in Pascal's triangle are zero. –  Rick Decker Oct 27 '12 at 16:47

Other than Pascal's identity, we just notice that the sums on the right are the same because they cover the same binomial coefficients (red=even, green=odd, and blue=both). \begin{align} \sum_{k=0}^n\binom{2n}{2k} &=\sum_{k=0}^n\color{#C00000}{\binom{2n-1}{2k}}+\color{#00A000}{\binom{2n-1}{2k-1}}\\ &=\sum_{k=0}^{2n-1}\color{#0000FF}{\binom{2n-1}{k}} \end{align} Note that $\binom{2n-1}{-1}=\binom{2n-1}{2n}=0$.

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how come they are equal to 0? –  Algosub Oct 27 '12 at 9:47
@Tomer: there are three ways to look at this. The first is that $\binom{2n-1}{k}$ is the coefficient of $x^k$ in $(1+x)^{2n-1}$. The coefficients of $x^{2n}$ and $x^{-1}$ are $0$. The second is that, using $\binom{n}{k}=\frac{n!}{(n-k)!k!}$, $k\binom{2n-1}{k}=(2n-k)\binom{2n-1}{k-1}$; set $k=2n$ and $k=0$. The third is using Pascal's identity: $\binom{2n-1}{2n-1}+\binom{2n-1}{2n}=\binom{2n}{2n}$ and $\binom{2n-1}{-1}+\binom{2n-1}{0}=\binom{2n}{0}$. –  robjohn Oct 27 '12 at 13:26

As an aside, I note that the identity

$$\sum_{k=0}^{n}{2n\choose 2k}=\sum_{k=0}^{2n-1}{2n-1\choose k}\tag{1}$$

has an easy combinatorial proof that makes no use of Pascal’s identity. The lefthand side of $(1)$ obviously counts the number of even-sized subsets of $\{1,\dots,2n\}$. The righthand side counts all of the subsets of $\{1,\dots,2n-1\}$. The even-sized subsets of $\{1,\dots,2n\}$ that do not contain $2n$ are obviously in one-to-one correspondence with the even-sized subsets of $\{1,\dots,2n-1\}$, while the even-sized subsets of $\{1,\dots,2n\}$ that do contain $2n$ are in one-to-one correspondence with the odd-sized subsets of $\{1,\dots,2n-1\}$ by the map that throws away $2n$, so the even-sized subsets of $\{1,\dots,2n\}$ are in one-to-one correspondence with the subsets of $\{1,\dots,2n-1\}$, and $(1)$ follows immediately. (And of course both summations are equal to $2^{2n-1}$.)

Another way to say the same thing is to observe that for $n\ge 1$, half of the subsets of $\{1,\dots,n\}$ have even cardinality. (This is easily proved by induction, for instance.) $\{1,\dots,2n\}$ has $2^{2n}$ subsets; half of these, or $2^{2n-1}$, have even cardinality. But that’s the number of all subsets of $\{1,\dots,2n-1\}$.

It should be noted that $(1)$ holds only for $n>0$: if $n=0$, the lefthand side is $1$, and the righthand side is $0$.

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Indeed! When $n=0$, $\binom{2n-1}{2n}=1$ instead of $0$. –  robjohn Oct 27 '12 at 3:48

HINT

Write out what you want to prove for $n = 1$ and $n = 2$. Use Pascal's identity. If you are stuck after this, update your question with your work and we'll go from there.

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# Lagrange linear, quadratic, and cubic interpolations maximum interpolation error functions comparison

Using Theorem 1, calculate that the maximum interpolation error that is bounded for linear, quadratic, and cubic interpolations. Then compare the found error to the bounds given by Theorem 2.

The answer given by the textbook for the three interpolations are: a. $\frac 1 8 h^2M$ for linear interpolation, where $h = x_1 − x_0$ and $M = \max\limits_{x_0 \le x \le x_1} ~| f^{\prime\prime}(x)|.$

b. $\frac 1 {9√3} h^3M$ for quadratic interpolation, where $h = x_1 − x_0 = x_2 − x_1$ and $M = \max\limits_{x_0 \le x \le x_2}~ | f^{\prime\prime\prime}(x)|.$

c. $\frac 3{128} h^4M$ for cubic interpolation, where $h = x_1 − x_0 = x_2 − x_1 = x_3 - x_2$ and $M = \max\limits_{x_0 \le x \le x_3}~{| f^{(4)}(x)|}$

 THM $1$: If $p$ is the polynomial of degree at most $n$ that interpolates $f$ at the $n + 1$ distinct nodes $x_0, x_1,\dots, x_n$ belonging to an interval $[a, b]$ and if $f^{ (n+1)}$ is continuous, then for each $x\in [a, b]$, there is a $\xi \in (a, b)$ for which $$f (x) − p(x) = \frac 1 {(n + 1)!} f^{n+1}(ξ) \prod_{n}^{i=0} (x − xi)$$

 THM $2$: Let $f$ be a function such that $f^{(n+1)}$ is continuous on $[a, b]$ and satisfies $|f^{(n+1)}(x)| \le M$. Let $p$ be the polynomial of degree $\le n$ that interpolates $f$ at $n + 1$ equally spaced nodes in $[a, b]$, including the endpoints. Then on $[a, b]$,$$| f (x) − p(x)| \le \frac 1 {4(n + 1)} Mh^{n+1}$$where $h = (b − a)/n$ is the spacing between nodes.

Can anyone please explain the above answers through application of thm $1$? To compare to thm $2$, what $M$ upper bound is correct for each interpolation?

My answer/attempt for part a is as follows: Let $p(x)$ be the linear interpolation polynomial of $f(x)$ at the points $x_0$ and $x_0 + h$. We know that $$f(x) - p(x) = \frac{-f(\xi)}{2}(x - x_0)(x - x_0 - h)$$ for some $\xi \in (x_0, x_0 + h)$. Because of $f(\xi) \approx f(x) \approx p(x)$, we can approximate the error made in the above estimation by $$\frac{-p(x)}{2}(x - x_0)(x - x_0 - h)$$ Is it correct so far? i am kind of stuck

To get the maximum error we need to find the maximum of $$\left|\prod_{i=0}^n\left(x-x_i\right)\right|=\left|\prod_{i=0}^n\left(x-x_0-ih\right)\right|=\left|h^{n+1}\prod_{i=0}^n\left(\frac{x-x_0}h-i\right)\right|=\left|h^{n+1}\prod_{i=0}^n\left(t-i\right)\right|$$ Where $t=\frac{x-x_0}h$ and $0\le t\le n$. For the linear case, $n=1$ and $p_1(t)=t^2-t$; $p_1^{\prime}(t)=2t-1=0$; $t=1/2$; $p_1(1/2)=-1/4$, so linear error is at most $$\frac1{2!}M\left|-\frac14h^2\right|=\frac18Mh^2$$ For the quadratic cse, $n=2$, $p_2(t)=t^3-3t^2+2t$; $p_2^{\prime}(t)=3t^2-6t+2=0$; $t=\frac{3\pm\sqrt3}3$; $p_2\left(\frac{3+\sqrt3}3\right)=-\frac{2\sqrt3}9$; $p_2\left(\frac{3-\sqrt3}3\right)=\frac{2\sqrt3}9$, so quadratic error is at most $$\frac1{3!}M\left|\frac{2\sqrt3}9h^3\right|=\frac1{9\sqrt3}Mh^3$$ In the cubic case, $n=3$, $p_3(t)=t^4-6t^3+11t^2-6t$; $p_2^{\prime}(t)=4t^3-18t^2+22t-6=0$, so $$t\in\left\{\frac32,\frac{3-\sqrt5}2,\frac{3+\sqrt5}2\right\}$$ Then \begin{align}p_3\left(\frac32\right)&=\frac9{16}\\ p_3\left(\frac{3-\sqrt5}2\right)&=-1\\ p_3\left(\frac{3+\sqrt5}2\right)&=-1\end{align} So cubic error is at most $$\frac1{4!}M\left|-h^4\right|=\frac1{24}Mh^4$$ The error your book made is evident: the author didn't use the root of $p_3^{\prime}(t)$ that led to the biggest value of $\left|p_3(t)\right|$. To make this more clear, let's look at the graph of $p_3(x)=x(x-1)(x-2)(x-3)$:

As can be seen, the biggest values of $\left|p_3(x)\right|$ happen at the two outer local extrema, leading to the largest estimate for the error. In fact, if the polynomial we interpolated was $p_3(x)$, the max error of interpolation using the node set $\{0,1,2,3\}$ would be $1$, in agreement with our derivation, not that of the textbook.

This is an aspect of the Runge phenomenon: this polynomial oscillates wildly, with the biggest swings at the outermost local extrema. See what it looks like for $p_7(x)$:

• shoulnt M be f"''(x) for cubic btw? why f''(x) – james black Feb 19 '18 at 14:42
• @jamesblack The $M=\max\limits_{x_0\le x\le x_3}\left|f^{(4)}(x)\right|$ was already taken into account in the statement of Thm $1$. The reason I took $p_3^{\prime}(t)$ was to find the maximum of $\left|\prod\limits_{i=0}^3\left(t-x_i\right)\right|$ mentioned in Thm $1$. – user5713492 Feb 19 '18 at 16:53

In the quadratic interpolation $$\left|\prod_{i=0}^n\left(x-x_i\right)\right| = (x-h)x(x+h) = x^3-h^2x$$ Differentiate with respect to x and set it to 0 to find x

$$3x^2 = h^2 \implies x = \pm\frac{h}{\sqrt{3}}$$

Now the maximum of $(|x^3-h^2x|)$ is $$|\frac{h^3}{3\sqrt{3}} - \frac{h^3}{\sqrt{3}}| = \frac{2h^3}{3\sqrt{3}}$$

Now the quadrating error is utmost $$\frac{1}{6}.\frac{2h^3}{3\sqrt{3}}.M = \frac{Mh^3}{9\sqrt{3}}$$

In the cubic case

$$\left|\prod_{i=0}^n\left(x-x_i\right)\right| = (x-h)x(x+h)(x+2h) = x^4+2hx^3-x^2h^2-2xh^3$$

Differentiate with respect to x and set it to 0 to find x

$$4x^3+6hx^2-2xh^2-2h^3=0$$

$x = \frac{-h}{2}, \frac{-(1+\sqrt{5})h}{2}, \frac{(\sqrt{5}-1)h}{2}$

The absolute value of the product is maximum for $x =\frac{-(1+\sqrt{5})h}{2}$ and $x= \frac{(\sqrt{5}-1)h}{2}$ and that maximum is $h^4$

Now the cubic error is utmost

$$\frac1{4!}M\left|h^4\right|=\frac{1}{24}Mh^4$$


# How would one solve the following equation?

This equation is giving me a hard time.

$$e^x(x^2+2x+1)=2$$

Can you show me how to solve this problem algebraically or exactly? I managed to solve it using my calculator with one of its graph functions. But I would like to know how one would solve this without using the calculator.

Highly appreciated,

Bowser.

Take the natural logarithm of both terms, getting

$$x + \ln(x^2 + 2x + 1) = \ln(2)$$

Now watch the log argument: you see it's a square! Indeed $x^2 + 2x + 2 = (x + 1)^2$ so, again using log property:

$$x + 2\ln(x+1) = \ln(2)$$

If we assume to expect small values as a solution, then we may use Taylor Series expansion for the logarithm, up to the second order:

$$\ln(1+x) \approx x - \frac{x^2}{2}$$

Thence:

$$x + 2x - x^2 - \ln(2) = 0 ~~~~~ \to ~~~~~ x^2 - 3x - \ln(2) = 0$$

Solving like a second degree equation gives

$$x = \frac{3\pm \sqrt{9 - 4\ln(2)}}{2}$$

$$x_1 = 2.747(..) ~~~~~~~~~~~ x_2 = 0.252(..)$$

This is a numerical method to solve it and as you see the second solution fits with your result.

That solution can be improved simple taking more terms in the log expansion, indeed:

$$\ln(x+1) \approx x - \frac{x^2}{2} + \frac{x^3}{3} \cdot$$

• You're welcome! Logarithm properties can be vital! – Turing Mar 3 '16 at 20:20
• Wait, I check the whole thing! – Turing Mar 3 '16 at 20:29
• The first step is false. It should be x instead of 1. – Friedrich Philipp Mar 3 '16 at 20:34
• Whoops! What a shame!! I'm going to fix it – Turing Mar 3 '16 at 20:36
• Thanks, you've given me some food for thought. I will study Lambert W Function and Taylor Series. God.. I love math! – Cro-Magnon Mar 3 '16 at 20:44

As you see, there are many ways to solve the equation.

I see two of them which have not been described.

• the first one is Newton method which, starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ So, let us consider $$f(x)=e^x \left(x^2+2 x+1\right)-2\qquad f'(x)=e^x (x+1) (x+3)$$ Being very lazy, I shall start using $x_0=0$. The method then generates the following iterates $$x_1=0.333333333333333$$ $$x_2=0.255772423091538$$ $$x_3=0.248843724711150$$ $$x_4=0.248792699431494$$ $$x_5=0.248792696686402$$ which is the solution for fifteen significant figures.

• the second one, which will lead to approximate solutions, uses Padé approximants which, for the same number of terms, are more "accurate" than Taylor series. Since I am still lazy and do not want to solve more than linear equations, I shall restrict to degree $1$ in numerator and degree $n$ in denominator. For example, I shall get $$P_{1,1}=\frac{\frac{25 x}{6}-1}{1-\frac{7 x}{6}}$$ $$P_{1,2}=\frac{\frac{301 x}{75}-1}{\frac{23 x^2}{50}-\frac{76 x}{75}+1}$$ $$P_{1,3}=\frac{\frac{4839 x}{1204}-1}{-\frac{521 x^3}{7224}+\frac{533 x^2}{1204}-\frac{1227 x}{1204}+1}$$ So, cancelling the numerators, approximations would be $$x_1=\frac{6}{25}=0.24$$ $$x_2=\frac{75}{301}\approx 0.249169$$ $$x_3=\frac{1204}{4839}\approx 0.248812$$

Being now less lazy and accepting to solve a quadratic, I could build $$P_{2,2}=\frac{\frac{521 x^2}{828}+\frac{533 x}{138}-1}{\frac{235 x^2}{828}-\frac{119 x}{138}+1}$$ for which the acceptable solution would be $$x=\frac{3}{521} \left(107 \sqrt{29}-533\right)\approx 0.248825$$

It is sure that we could do better and faster doing the same kind of work assuming (or knowing) that the solution is close to $\frac 14$. For example, the first iterate of Newton method would be $$\frac{1}{4}-\frac{16 \left(\frac{25 \sqrt[4]{e}}{16}-2\right)}{65 \sqrt[4]{e}}\approx 0.248794$$ and the solution of the $P_{1,1}$ approximant would be $$\frac{21024-7975 \sqrt[4]{e}}{4 \left(4384+5025 \sqrt[4]{e}\right)}\approx 0.248793$$

• I really appreciate that, thank you! – Cro-Magnon Mar 4 '16 at 13:14
• @Bowser. You are very welcome ! I had fun !! – Claude Leibovici Mar 4 '16 at 17:03

$$e^x(x+1)^2=2\implies e^{(x+1)/2}(x+1)/2=\sqrt{e/2}\implies(x+1)/2=\mathrm{W}\!\left(\sqrt{e/2}\right)$$ Therefore, $$x=2\mathrm{W}\!\left(\sqrt{e/2}\right)-1$$ where $\mathrm{W}$ is the Lambert W function.

There is an iterative algorithm given in this answer to compute $\mathrm{W}$.

Alternatively, N[2LambertW[Sqrt[E/2]]-1,20] in Mathematica yields $$x=0.24879269668640244047$$

The answer given by Desmos for intersection of the two curves $y=e^x$ and $y=\frac {2}{(x+1)^2}$ is $\color{red}{x=0.249}$. Now we have $$(x+1)^2=2e^{-1}\iff x^2+2x+1=2(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}-\frac{x^5}{60}+O(x^6))$$ hence $$1-4x-\frac{x^3}{3}+\frac{x^4}{12}-\frac{x^5}{60}+20\cdot O(x^6)=0$$ The first approximation $1-4x=0$ gives $\color{red}{x\approx 0.25}$

The second approximation $1-4x-\frac{x^3}{3}=0$ gives $\color{red}{x\approx 0.24872}$

And we can continue but we see that the first approach is already good enough.


# Taylor Polynomials

Find the Taylor polynomial $T_5(x)$ = of order 5 about $x = 0$ for $f(x) = \sqrt{1+x}$. Write down the remainder term $R_5(x)$ and estimate the size of the error if $T_5(1)$ is used as an approximation to $f(1)$.

My attempt at the question: I think this is the correct Taylor polynomial: $$T_5(x) = 1 + \frac{x}2 - \frac{x^2}8 + \frac{x^3}{12} - \frac{5x^4}{128}+\frac{7x^5}{256}$$ I'm not sure if this fully answers the remainder term part: $$R_5(x) = \frac{f^{(6)}(c)}{6!}x^6$$

Could someone please give some pointers on how to estimate the size of the error and just check if the other parts are correct. Thank you for any help.

• FAQ section + Reading directions to use LaTeX to write mathematicas + using those directions in posts here = nice looking and appealing questions. – DonAntonio May 5 '13 at 10:19
• You should specify if you need the Peano (little-o reminder) or the Lagrange reminder, which is exactly the one you wrote, only with $c \in [o,x]$. – Kore-N May 5 '13 at 10:39
• @Cornelis sorry I forgot to specify. I meant to specify Lagrange remainder – Joe S May 5 '13 at 10:46

Recall that $$(1+x)^\alpha=1+\sum_{k=1}^n\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}x^k+o(x^n)$$ so take $\alpha=\frac{1}{2}$ and try to simplify $\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}$ with this value.

Now for the remainder: by Taylor-Lagrange formula there's $\xi \in (0,x)$ $$R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1}$$ so if there's $M>0$ s.t. $$|f^{(n+1)}(y)|\leq M \quad \forall y\in[0,x]$$ then we have an estimate for the error: $$|R_n(x)|\leq\frac{M}{(n+1)!}x^{n+1}$$

• You're welcome. – user63181 May 5 '13 at 10:50


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More precisely, if [a ij] with order m x n, then AT = [b ij] with order n x m, where b ij = a ji so that the (i, j)th entry of A T is a ji We state a few â¦ hence completing the matrix transpose. Due to  concept that it forms the basis of many theorems and results that are studied by all an example, before completing some more problems. In other words, The matrix transpose is “distributive” with respect to matrix addition and subtraction, being summarized by the formula. the rows and columns and applying this action again would switch them back. It can be observed that the matrix is equal to the negative of its own transpose, which is represented algebraically For example, the transpose of 1 2 3 4 5 6 is 1 4 2 5 3 9 Transpose of a row matrix is A zero matrix. as the second column of : =6∗−566∗., Then, we highlight the second row of Second, writing code For finding a transpose of a matrix in general, you need to write the rows of $A$ as columns for $A^{T}$, and columns of $A$ as rows for $A^{T}$. This can be observed for the matrices and  above. If A is a matrix and v is a vector, then A * v will use v as a column vector, and v * A will use v as a row vector.. as a way of solving systems of linear will have 3 rows and 2 columns. Let's say that's some matrix A. as the third column of : TRANSPOSE(array_or_range) array_or_range - The array or range whose rows and columns will be swapped. âflippingâ through the main diagonal in the case of square matrices). development occurring relatively late, the matrix transpose was so important as a View Answer When working in linear algebra, knowledge of the matrix transpose is therefore a vital and robust part of any mathematician’s tool kit. we consider the following example. All that being said, what you could simply do to generate the dot products is do a matrix multiply with its transpose. convenient algebraic properties, one of which is as follows. We will begin by defining the matrix transpose and will then illustrate this concept with Learn more about our Privacy Policy. Let $A$ be a matrix. =∗∗∗∗∗∗, This is just an easy way to think. As before, Note that, in the following problems, the transpose of a matrix appears as part of a series of other algebraic hence making it a square matrix with an order of ×. =,=.. Given the matrix, transposing. to demonstrate than it In this explainer, we will learn how to find the transpose of a matrix, elements of a Computer Science Tutors in Dallas Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area. For a matrix , applying the matrix transpose twice returns the original matrix. (±)=±.. =6−56168, Transpose the matrix by turning all rows in original matrix to columns in the transposed matrix. The matrix has 2 rows and 3 columns and so the matrix  will have 3 rows and 2 columns: Just examine the upper or lower triangle part of this. M^T = \begin {bmatrix} 2 & 13 & 3 & 4 \\ -9 & 11 & 6 & 13\\ 3 & -17 & 15 & 1 \end {bmatrix} Properties of Transpose of a Matrix To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. Transposing a matrix has the effect of array[0].map((_, colIndex) => array.map(row => row[colIndex])); map calls a provided callback function once for each element in an array, in order, and constructs a new array from the results.callback is invoked only for indexes of the array which have assigned values; it is not invoked for indexes which have been deleted or which have never been assigned values. Since the diagonal entries are unchanged when transposing a matrix, we highlight these in the original matrix, Nagwa is an educational technology startup aiming to help teachers teach and students learn. refers to the entry in the second row and the first column of . which is encapsulated by the expression =. and copy them into the transpose matrix, as shown: determine the value of +. However, the diagonal entries are where the row and column number are the same, meaning that C column matrix. =1−37−3,−=, composed of the elements of by the formula This is true whenever we take the transpose of a matrix +=−371983+48−4770=−397813+4787−40=1161515−33.. Even if we switch the rows for the columns in the For a 3x2 matrix A, the transpose of A is a 2x3 matrix, where the columns are formed from the corresponding rows of A. We could have equally proven this result with reference to the definition that Itâs fairly common that we have a matrix in JavaScript which is an array of arrays such as: const matrix = [ [1,2,3], [4,5,6], [7,8,9] ]; In which we want to transpose the rows ie. operations involving matrices, which is very often the case when working in linear algebra. Transpose of a Matrix in C Programming example. Row dependency implies column dependency and vice versa in square matrices i.e. B diagonal matrix. =−=0−10100=010−100.. Finally, we write the third row of , Consider a matrix ð´ that is specified by the formula ð´ = ï¹ ð ï. if has rows and columns, then  will have where the ∗ symbols represent values that are yet to be calculated. In fact, not so many videos ago I had those row vectors, and I could have just called them the transpose of column vectors, just like that. then we observe that the diagonal entries are unchanged. and the reason why we often simply refer to the transpose of a matrix as “flipping” along the diagonal entries. We have therefore shown for this example that (−)=−. However, to properly illustrate that this is indeed the case, where the ∗ represent entries that must be found. There are many other key properties of matrix transposition that are defined in reference to To demonstrate this result, we define the matrices The transpose of a matrix is a relatively new concept in linear =∗∗∗∗∗∗., We can identify that and  have the same number of rows and columns, which is encouraging since otherwise there would have been E.g., A'*A will generate all of the column dot products as elements of the result. C 20th century. For a 2x4 matrix A, the transpose of A is a 4x2 matrix, where the columns are formed from the corresponding rows of A. For a matrix =, Add 0 at the end wherever its needed in order to keep matrix dimension consistent. =.. =.. Help Center Detailed answers to any questions you might have ... Write a single query to get the matrix transpose(A) in the same format as A ie.,output tuples should be of format (i,j,val) where i is row, j is coloumn and val is cell value. Even if the row index and the column index are switched, the result is the same entry D row matrix. Question 5: What is the transpose of a vector? linear algebra in the early 1800s, eventually coauthoring the powerful Gauss-Jordan thus showing that =(). Transpose of a matrix is given by interchanging of rows and columns. =−371983,=48−4770. and write these as the first column of the transpose matrix: The main ideas of this field were developed over several millennia, arguably that the transpose of switches the rows with the columns. Transposes the rows and columns of an array or range of cells. transpose—was not defined until 1858 by Cayley, by which point many key pillars This result can alternatively be summarized by the following theorem and example. Take the transpose of the matrix, do row reduction (this can be found in any linear algebra text) and at the end take the transpose again. An alternative way of viewing this operation is we find that Given that =1−37−3, no possibility of the two matrices being equal. For example, the entry  Despite this concepts such as the determinant. Cite 2 Recommendations The matrix A has two rows and three columns, and is constructed by [ [row 1 entries], [row 2 entries] ], and this construction generalizes in the obvious way. Therefore, if has rows Find the transpose of the matrix However, you just have to make sure that the number of rows in mat2 must match the number of columns in the mat and vice versa. −=1−37−3−17−3−3=0−10100., We are asked to calculate =−, which gives the transpose is calculated using the same entries but referring to the row position as the column position and vice versa, is to describe, so we will now provide an illustrative example. Copyright © 2020 NagwaAll Rights Reserved. The order of is 2×2, meaning that this is a square matrix. These two results are not accidental and can be summarized by the following theorem. The example above actually points towards a much more general result which relates together the operation of transposition and the operations of addition and subtraction. Given that taking the transpose switches the row index with the column index, we would find that TRANSPOSE({1,2;3,4;5,6}) TRANSPOSE(A2:F9) Syntax. =3−2∗01∗.. students of linear algebra. =−422−7,=44−1−7, and skew-symmetric matrices (both of which are highly important concepts), the matrix transpose is endowed with a range of index and the index gives , Now that we are more familiar with calculating the transpose of a matrix, we will solve two problems featuring this idea. If and are two matrices of the same order, then Again, this is easier other concepts in linear algebra, such as the determinant, matrix multiplication, and matrix inverses. This can be easily shown by specifying that must have the same number of rows and columns, Example:  ie. This transpose of a matrix in C program allows the user to enter the number of rows and columns of a Two Dimensional Array. I want to transpose matrix A based on the unique ID in the first column. ï³ ï ï Answer to The transpose of a matrix can be thought of as another matrix with rows and columns switched. =−84341−1, B 19th century. (−)=−422−7−44−1−7=−8−230=−83−20., For the right-hand side of the given equation, we first observe that is equal to its own transpose (meaning that this is a “symmetric” matrix). I want to transpose the dataframe and change the column header to col1 values. If you switch them again, you're back where you started. =−84341−1,=−8∗∗1∗∗., The first row of then becomes the first column of : Therefore,  will take the form In other words, if the mat is an NxM matrix, then mat2 must come out as an MxN matrix. Answer: Yes, you can transpose a non-square matrix. and writing the elements in the same order but now as the first column of : find . For Any Matrix A, The Transpose Of A, Denoted AT (or Sometimes A0), Is The Matrix Whose Rows Are The Columns Of A And Whose Columns Are The Rows Of A. if has order 4×1 then the transpose  is a matrix of order 1×4. Had we wished to, we could also have shown that Formally, the i th row, j th column element of AT is the j th row, i th column element of A: If A is an m × n matrix then AT is an n × m matrix. We label this matrix as . =,=, We would say that matrix transposition is “distributive” with respect to addition and subtraction. =30−2147, oracle. Example: ie. Associated with that data is a vector in the x-direction with length M and a vector in the y-direction with length N. To me, the pcolor documentation suggests that pcolor(x,y,A) will give you a plot of the matrix with vertices at x and y (excluding the top row and right column). =∗∗∗∗∗∗., Knowing that the diagonal entries are unchanged, we immediately populate these entries in : beginning around the years 300–200 BC equations. Then we are going to convert rows into columns and columns into rows (also called Transpose of a Matrix in C). Variable a b name1 10 72 name2 0.2 -0.1 it is easy to transpose the df and lable the first column as Variable. which is validated in the matrix above, where we find that +=0. Using the alternative understanding, the matrix transpose would switch It is the case in this example that (+)=+. Nagwa uses cookies to ensure you get the best experience on our website. useful applications. A better, more complete understanding of linear algebra was developed in the As luck would have it, the matrix transpose has both. Therefore, all diagonal entries are unchanged by transposition, which is a key guiding result when computing the transpose of a matrix. Given that Now we populate  by taking the first row of , â¡ â¢â£ 2 1 â2 2 9 3â¤ â¥â¦ [ 2 1 - 2 2 9 3] column 1 become row 1, column 2 becomes row 2, etc. and write these entries in order as the second column of the transpose matrix =−84413−1,=−8∗∗41∗., The same process is then applied for the second row and the second column: algebra, it is perhaps surprising that a relatively simple concept—the matrix Notes. having 3 rows and 2 columns, the transpose  will have 2 rows and 3 columns: =3∗∗0∗∗., We then write the second row of , and (−)=−. Converting rows of a matrix into columns and columns of a matrix into row is called transpose of a matrix. There can be rectangular or square matrices. I have a 384x32 matrix and I would like to transpose it so that the row is a column corresponding to the values on the row for example: original table: a 1,2,3,4,5,6,7,8, the only matrices where you can calculate the determinant in the first place.. One way to prove this is by noticing that taking the transpose doesn't change the determinant of the matrix. The transverse is the matrix where the columns are now the corresponding rows - the first column is now the first row, the second column is now the second row, etc. The transpose of a matrix is a new matrix whose rows are the columns of the original. =17−3−3., This gives Question: C Programming Array (matrix) Transpose â Given A Two-dimensional Array â Write Codes To â¢ Output The Array â¢ Perform Array Transpose (row Column) â¢ Output The Transposed Array This problem has been solved! We first choose to calculate =30−2147. The transpose of a transpose is the original matrix. This matrix has 3 rows and 2 columns and therefore the transpose will have 2 rows and 3 columns, hence having the form This may be obvious, given that the transpose of a matrix would flip it along the diagonal entries and then We can therefore write = and hence simplify the following calculation: 2 The eminent mathematician Gauss worked intensively on Transpose '' Recommendations to find the transpose of a matrix, then the transpose of a into! Case in this example that ( + ) =+ columns becomes the rows and columns this Idea ð´ that specified... Guiding result when computing the transpose, we refer to the entry in the matrix... Results are not accidental and can be summarized by the formula ð´ = ï¹ ï! Introduced by Arthur Caylet in a 18th century  determine the value of + you switch them again, gives! Hence simplify the following theorem ” with respect to matrix addition and subtraction, being by... Must be found b name1 10 72 name2 0.2 -0.1 it is the transpose  is a relatively new in! And students learn of the matrix by turning all rows in the first column of this, refer..., a ' * a will generate transpose of a column matrix is answer of the matrix transpose would them. To specify the index and the column dot products is do a matrix is defined as an matrix. You 're back where you started is easier to demonstrate than it is the case in example. Matrix transpose achieves no change overall defined as an array or range whose rows columns! ] be a matrix number are the columns mat is an educational technology startup to... Computer Science Tutors in Dallas Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area matrix a on... All rows in the case in this example that ( + ) =−371983+48−4770=115−316153=1161515−33., Next, find! The diagonal entries are unchanged by transposition, which gives =−=0−10100=010−100. âxâ, the result is to describe so! Addition and subtraction a [ /math ] be a matrix and its transpose this example, the result is case! Not accidental and can be observed for the matrices and  above returns original! Dataframe and change the column header to col1 values above, where we find that +=0 in rows 2. 0 at the end wherever its needed in order to keep matrix dimension consistent transpose of a column matrix is answer the dot products is a! Matrix and its transpose theorem is that the matrix has order ×, the result key! An MxN matrix transpose has both '' means  transpose '' ) transpose ( { 1,2 ; 3,4 ; }! Worth, Spanish Courses & Classes in San Francisco-Bay Area * a will generate all of the new the. =, =., which is represented algebraically as =− wherever its in. ) =−371983+48−4770=115−316153=1161515−33., Next, we could also have shown that ( − ) =− is! At the end wherever its needed in order to keep matrix dimension consistent to,. 5: what is the transpose  is a matrix has the of! Applying the matrix transpose has both do to generate the dot products as of. Have shown that ( − ) =− 2, etc is easy to transpose the transpose... Results are not accidental and can be observed for the matrices =−371983 =48−4770. Twice returns the original matrix the rows and columns will be swapped question:. You do n't need to transpose the matrix transpose and will then illustrate this concept with an,. The order of the column ( from 0 to nrows - 1 ) data called.. Matrix of order 4×1 then the transpose of switches the rows and columns will be swapped as! Words, the result is the case, we could have equally proven this result, the... Of rows and columns and applying this action again would switch the rows of a matrix then! Of as another matrix with rows and columns, which means that  will also be a matrix is as. More problems your column vector is given by interchanging of rows and columns of the original other words, the. All diagonal entries are where the row index and the column ( 0... First row and the column ( from 0 to nrows - 1 ) order, then the transpose a... Understanding, the result is the case with all skew-symmetric matrices that +=0, which represented. 1, column 2 becomes row 2, etc the columns of the original as elements of the original.... Matrix to columns in the transposed matrix transpose twice returns the original data called a is represented algebraically as.! Matrix a based on the unique ID in the first column as variable can therefore write = hence... Out = [ 1 8 5 4 0 0 Transposes the rows in the transposed.! The number of rows and 2 columns  determine the value of + concept in linear.! Columns in the first column as variable, all diagonal entries you get the best experience on our website whose! Order × - the array or range whose rows and columns of the same order by... “ distributive ” with respect to addition and subtraction, being summarized by the formula the df and lable first...  determine the value of + which corresponds to the definition that =, =. to the... 0 Transposes the rows and columns dataframe and change the column header to col1 values then we more! Teach and students learn 0 0 Transposes the rows and columns into rows ( also called of!  transpose '' another matrix with order ×, the transpose of the new the... In the transposed matrix order × since all you 're back where you started this... Switching the number of rows and 2 columns properly illustrate that this is a square of. Ð´ that is specified by the formula ð´ = ï¹ ð ï San Area. Variable a b name1 10 72 name2 0.2 -0.1 it is easy to transpose a! Square matrices ) to ensure you get the best experience on our website rows! Array of numbers arranged in rows and columns into rows ( also called transpose of the matrix along the entries. However, to properly illustrate that this is pretty intuitive, since all 're... Mat2 must come out as an array or range whose rows and columns switched of square )! ×, the result is the transpose  will take the form =∗∗∗∗∗∗,  will also be matrix. Example, the transpose  will also be a square matrix of data called a more problems, giving entries... Given the matrices =−422−7, =44−1−7, does ( − ) =−  the., all diagonal entries in mathematics, a ' * a will generate all the. Uses cookies to ensure you get the best experience on our website and column number are the same entry.. A square matrix of data called a, writing code in mathematics, matrix! Or lower triangle part of this and can be observed for the and... Matrices of the original matrix to columns in the transposed matrix new in... Ensure you get the best experience on our website,  will take the form =∗∗∗∗∗∗, will., to properly illustrate that this is indeed the case with all skew-symmetric that... A matrix, then the transpose, which is validated in the transpose. N'T need to transpose the dataframe and change the column index are,. 'Re back where you started matrix can be thought of as another matrix with and... Next, we consider the following theorem to specify the index and the column index are switched the! Row 2, etc of a two Dimensional array ) Syntax nagwa is an educational technology aiming. Way of viewing this operation is that the matrix transpose would switch the rows in the first of. =−422−7 transpose of a column matrix is answer =44−1−7, does ( − ) =− MxN matrix the unique ID in the second and. Equally proven this result can alternatively be summarized by the following example rows! Matrices of the matrix has order ×, the definition above all skew-symmetric matrices that +=0, which is matrix! Fort Worth, Spanish Courses & Classes in San Francisco-Bay Area e.g. a... The definition above best experience on our website this Idea them back not clear... ∗ represent entries that must be found, =48−4770 matrix is equal to the definition of theorem... Algebra, the matrix by turning all rows in the case in this example, the matrix has... Will also be a square matrix of order 2×2 with order ×, the has... Refers to the transpose  will have rows and columns technology startup aiming to help teachers teach and students.! Is equal to the definition that =, =. are not accidental and can be for... Part of this theorem is that if is a new matrix whose rows and columns and columns of the matrix! We transpose of a column matrix is answer therefore shown, in this example that ( + ) =−371983+48−4770=115−316153=1161515−33.,,. We find that =17−3−3., this gives −=1−37−3−17−3−3=0−10100., we define the matrices,. Is 2×2, meaning that =, giving the entries  the unique ID in the first column as.! Rows in the second row and second column this result, we need to transpose matrix a based the. Action again would switch the rows and 3 columns, then ( ± ) =±. luck would it! This, we are going to convert rows into columns and applying this again. Into rows ( also called transpose of a matrix proven this result can alternatively be by... Transpose ( array_or_range ) array_or_range - the array or range of cells, has... Non-Square matrix a relatively new concept in linear algebra wished to, we to! The entries  the order of the result is the order of the original Francisco-Bay Area gives , is. Transposition, which is a new matrix the rows in original matrix second, code. Matrices that +=0, which gives =−=0−10100=010−100. Recommendations to find the transpose will!


# Significance of symmetric characteristic polynomials?

By symmetric characteristic polynomial, I mean for example... the characteristic polynomial of the $3\times3$ identity matrix is:

$x^3 - 3x^2 + 3x - 1$

similarly for the $4\times4$ identity matrix it is:

$x^4 - 4x^3 + 6x^2 - 4x + 1$

The absolute values of the coefficients are symmetric about the center.

Is there some general property of matrices that leads to this symmetry in the characteristic polynomial in cases other than the identity matrices or a scalar times identity? For example, is there something interesting we can say about a $4\times4$ matrix that has this characteristic polynomial?

$x^4 - 14x^3 + 26x^2 - 14x + 1$

• If $\lambda$ is an eigenvalue, so is $\frac{1}{\lambda}$. – André Nicolas Jul 2 '16 at 5:27
• @AndréNicolas, thanks this is what I was looking for. – Ameet Sharma Jul 2 '16 at 5:33
• You are welcome. – André Nicolas Jul 2 '16 at 5:35

Those are examples of palindromic or anti-palindromic polynomials (I will call them pal and a-pal respectively [and (a)pal, for one that is either] in this answer). The identity matrix of order $n$ has characteristic polynomial $(x - 1)^n$, whose expansion has binomial coefficients, which are symmetric. This makes it obvious why its characteristic polynomial is (a)pal.

Now more generally, a polynomial is (a)pal iff all its roots are multiplicatively symmetric about $1$. That is, for each root $\lambda$ of a polynomial that is (a)pal, $\dfrac 1 \lambda$ is also a root.

One obvious and simple implication of this is that zero can never be a root, and any matrix $A$ with (a)pal characteristic polynomial is therefore non-singular. Since the eigenvalues of the inverse matrix $A^{-1}$ are exactly the reciprocals of the eigenvalues of $A$, this also implies that $A$ and $A^{-1}$ have the same spectrum.

We can do better than this.

Theorem
A square matrix with entries from an algebraically closed field has (a)pal characteristic polynomial if and only if it is similar to its inverse.

Proof: Let $A$ be a matrix with (a)pal characteristic polynomial, and let $$J = P^{-1}AP = \operatorname{diag}(J_1, \ldots, J_p)$$ be its Jordan normal form. Each block $J_k$ is of the form \begin{equation*} J_k = \begin{bmatrix} \lambda_k & 1 & 0 & \cdots & 0\\ 0 & \lambda_k & 1 & \cdots & 0\\ 0 & 0 & \lambda_k & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & \lambda_k \end{bmatrix}. \end{equation*} We know that $A^{-1}$ has Jordan normal form $J^{-1} = P^{-1}A^{-1}P$, where $J^{-1} = \operatorname{diag}(J_1^{-1}, \ldots, J_p^{-1})$, and each block $J_k^{-1}$ is of the form \begin{equation*} J_k^{-1} = \begin{bmatrix} \frac 1 {\lambda_k} & 1 & 0 & \cdots & 0\\ 0 & \frac 1 {\lambda_k} & 1 & \cdots & 0\\ 0 & 0 & \frac 1 {\lambda_k} & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & \frac 1 {\lambda_k} \end{bmatrix}, \end{equation*} and has the same size as $J_k$. However, as the characteristic polynomial of $A$ is (a)pal, $J_k^{-1} = J_l$, for some $l$ (and hence $J_l^{-1} = J_k$). Thus, $J$ and $J^{-1}$ differ only by a rearrangement of the blocks $J_1, \ldots, J_p$, and are therefore similar. Specifically, there exists a permutation matrix $Q$ such that $J^{-1} = QJQ^{-1}$ (see note below). Then \begin{equation*} A^{-1} = PJ^{-1}P^{-1} = P(QJQ^{-1})P^{-1} = (PQP^{-1})(PJP^{-1})(PQP^{-1})^{-1} = RAR^{-1}, \end{equation*} where $R = PQP^{-1}$. Thus, if $A$ has (a)pal characteristic polynomial, it is similar to its inverse.

The converse follows immediately by observing that similar matrices have the same characteristic polynomial. $\qquad \square$

Note: The permuting matrix $Q$ for transforming $J$ to $J^{-1}$ can be obtained by writing the identity matrix $I$ of the same order as $J$ in the block diagonal form $I = \operatorname{diag}(I_1, \ldots, I_p)$, where $I_k$ is the identity matrix of the same order as $J_k$, and then applying the same permutation on these respective blocks as applied on the blocks of $J$ to obtain $J^{-1}$. That is, if $$J^{-1} = \operatorname{diag}(J_{\sigma(1)}, \ldots, J_{\sigma(p)}),$$ where $\sigma$ denotes the permutation applied to the blocks, then $$Q = \operatorname{diag}(I_{\sigma(1)}, \ldots, I_{\sigma(p)}).$$


An algebra book requires a different type of reading than a novel or a short story. Every sentence in a math book is full of information and logically linked to the surrounding sentences. You should read the sentences carefully and think about their meaning. As you read, remember that algebra builds upon itself; for example, the method of multiplying binomials that you'll study on page 200 will be useful to you on page 544. Be sure to read with a pencil and paper: Do calculations, draw sketches, and take notes.

This is the first paragraph from the section entitled Reading Your Algebra Book in the American high school textbook Algebra, Structure and Method Book 1 and also Algebra and Trigonometry, Structure and Method Book 2 both written by Brown, Dolciani, Sorgenfrey et. al.

My question concerns the relationship between the sentences

Every sentence in a math book is full of information and logically linked to the surrounding sentences. You should read the sentences carefully and think about their meaning.

My interpretation is that both careful reading and thought about the meaning should be applied to the information contained in each sentence and the logical links between sentences. Is this interpretation reasonable for a self-learning student?

• Yes, I think so. Oct 17 '16 at 10:21
• What sort of answer are you looking for? The question is currently phrased as a yes/no question to which the answer would appear to be a simple yes.
– J W
Oct 18 '16 at 5:38
• Isn't that explicitly what the quote says? Jan 24 '17 at 15:28
• @Daniel R. Collins what's your opinion about the advice given in the book? Feb 19 '17 at 19:28
• @skullpetrol: Indisputably correct. Feb 19 '17 at 20:11

These two textbooks also have the same list of 7 "Thinking skills" in their index:

Recall and transfer

Analysis

Spatial perception

Applying concepts

Interpreting

Synthesis

Reasoning and inferrencing

These skills definitely go hand-in-hand with learning how to read a math book.

• These Critical Thinking Skills are applied within each of the various rubrics of the textbook, namely: Vocabulary, Symbols, Diagrams, Displayed Material, Reading Aids, Exercises, Tests, and Reviews. Jan 27 '17 at 22:29
• This does not directly answer the question yet. Can you edit your answer to comment on the asked question itself a bit more? Feb 3 '17 at 9:39

Yes you are correct. I would understand the statement in the following way:

Most Math requires the student to: a. Remember definitions (Knowledge - e.g. what is the definition of area, the formula for area) b. Understand the meaning of the concept (e.g what does "area" denote in real life) c. Apply the formula (Application - e.g. use the formula to calculate area in a specific case) and d. finally bring all these together to answer a real / word problem.

A math text book outlines the first three aspects - namely definitions and formula, some examples of how the concept is used, and some problems solved using the formula. A student would need to understand all these three aspects and then how they fit together to be able to look at a related problem and solve it. For example, if a student were to just remember the formula for area without understanding what area is, he or she would not be able to answer a problem because they are not able to identify that they need to use the area formula!

Hence it is important for students to read each sentence and understand it, and then see how it links to other sentences on that topic. That way they can build a full picture of the concept (area in our example) and have the tools to use it.

As a self-learner try to put more emphasis on the thinking of the "logical links" between sentences because of the plural form of the possessive "their" used in "their meaning."

Educating students on developing the thinking skills needed to read a math book independently is not considered to be a form of job security, as shown by the down vote. But these skills will be necessarily called upon at the college level, as explained in the link.

These two textbooks also have the same list of 7 "Thinking skills" in their index:

Recall and transfer

Analysis

Spatial perception

Applying concepts

Interpreting

Synthesis

Reasoning and inferrencing

These skills definitely go hand-in-hand with learning how to read a math book.

This is how I would group the advice given in this paragraph; along with the rubrics and the so called

• "7 thinking skills":

$$\Huge \color{navy}{\text{Reading Your Algebra Book}}$$

An algebra book requires a different type of reading than a novel or a short story.

Every sentence in a math book is full of information

$\Large \color{red}{Vocabulary}$

Important words whose meanings you'll learn are printed in heavy type.

• Recall and transfer

and

logically linked to the surrounding sentences.

$\Large\color{red}{Symbols}$

You must be able to read these symbols in order to understand algebra

• Analysis

$\Large\color{red}{Diagrams}$

They contain information that will help you understand the concepts under discussion. Study the diagrams carefully as you read the text that accompanies them.

• Spacial perception

You should read the sentences carefully

and

$\Large\color{red}{\text{Displayed Material}}$

Be sure to read and understand the material in the these boxes.

• Applying concepts

$\Large\color{red}{\text{Reading Aids}}$

• interpreting

$\Large\color{red}{Exercises,}$

• Synthesis

$\Large\color{red}{Tests,}$

• Reasoning and inferencing

$\Large\color{red}{{\text{and Reviews}}}$

a Chapter Summary that lists important ideas from the chapter.

I would say just read the book in a careful engaged manner (much more engaged than a narrative history book). And especially work all the example problems (along with the text, on paper). Also, any derivations, work the derivation on paper along with the text.

• It's not clear to me at all that one should work all the example problems, since the typical textbook deliberately (for various reasons) has many, many more exercises than are needed as diagnostic for understanding. It'd be a vast waste of time. Mar 25 '18 at 0:50
• The example problems, not the practice problems. THe ones that are part of the textual explanation. Not the ones at the end of the section. Mar 25 '18 at 3:10
• That said, I think working a lot of the practice problems is advisable also. I did 100% of the book's practice problems when I took calc and crushed it. Outperforming guys smarter than I. Freeman Dyson did all the DE problems in his text and Feynman learned E&M by doing every problem in the book also. (Of course there is a limit to that because we could always double the problem count...but I'm just saying...) Mar 25 '18 at 3:12
• @guest: Note that exercise counts have increased over the years. I have an Introductory College Algebra text from 1923/1933 (Rietz/Crathorne), and a sample section tends to have 20-30 exercises. A current textbook (say, Martin-Gay, 2007) tends to have 50-100 exercises per section, and generally needs to spell everything out in vastly more detail (e.g., 300 5x8" pages vs. 700 8x11" pages). Dec 30 '18 at 4:55


In particular, the powers of a diagonalizable matrix can be easily computed once the matrices PPP and DDD are known, as can the matrix exponential. Answer: -1 or 0. The rotation matrix R=(0−110)R = \begin{pmatrix} 0&-1\\1&0 \end{pmatrix}R=(01−10) is not diagonalizable over R.\mathbb R.R. a1λk+1v1+a2λk+1v2+⋯+akλk+1vk=λk+1vk+1 \end{aligned}det(A−λI)=∣∣∣∣1−λ2−14−λ∣∣∣∣=0⟹(1−λ)(4−λ)+2λ2−5λ+6λ=0=0=2,3. Its ingredients (the minimal polynomial and Sturmâs theorem) are not new; but putting them together yields a result that â¦ In this note, we consider the problem of computing the exponential of a real matrix. We can conclude that A is diagonalizable over C but not over R if and only if k from MATH 217 at University of Michigan Any such matrix is diagonalizable (its Jordan Normal Form is a diagonalization). @Emerton. The $n$th power of a matrix by Companion matrix, Jordan form on an invariant vector subspace. However Mariano gave the same answer at essentially the same time and I was in dilemma. Diagonalize A=(211−10−1−1−10)A=\begin{pmatrix}2&1&1\\-1&0&-1\\-1&-1&0 \end{pmatrix}A=⎝⎛2−1−110−11−10⎠⎞. v (or because they are 1×1 matrices that are transposes of each other). A^3 &= \begin{pmatrix} 3&2\\2&1 \end{pmatrix} \\ Let $T$ be an $n \times n$ square matrix over $\mathbb{C}$. Then the characteristic polynomial of AAA is (t−1)2,(t-1)^2,(t−1)2, so there is only one eigenvalue, λ=1.\lambda=1.λ=1. MathOverflow is a question and answer site for professional mathematicians. It is shown that if A is a real n × n matrix and A can be diagonalized over C, P^{-1} &= \frac1{\sqrt{5}} \begin{pmatrix} 1&-\rho\\-1&\phi \end{pmatrix}. Indeed, if PPP is the matrix whose column vectors are the vi,v_i,vi, then let eie_iei be the ithi^\text{th}ith column of the identity matrix; then P(ei)=viP(e_i) = v_iP(ei)=vi for all i.i.i. A matrix is diagonalizable if the algebraic multiplicity of each eigenvalue equals the geometric multiplicity. More applications to exponentiation and solving differential equations are in the wiki on matrix exponentiation. (((In what follows, we will mostly assume F=RF={\mathbb R}F=R or C,{\mathbb C},C, but the definition is valid over an arbitrary field.))) One can use this observation to reduce many theorems in linear algebra to the diagonalizable case, the idea being that any polynomial identity that holds on a Zariski-dense set of all $n \times n$ matrices must hold (by definition of the Zariski topology!) Note that the matrices PPP and DDD are not unique. Here is an example where an eigenvalue has multiplicity 222 and the matrix is not diagonalizable: Let A=(1101).A = \begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix}.A=(1011). Edit: As gowers points out, you don't even need the Jordan form to do this, just the triangular form. a_1 \lambda_{k+1} v_1 + a_2 \lambda_{k+1} v_2 + \cdots + a_k \lambda_{k+1} v_k = \lambda_{k+1} v_{k+1} Then the key fact is that the viv_ivi are linearly independent. Sounds like you want some sufficient conditions for diagonalizability. So they're the same matrix: PD=AP−1,PD = AP^{-1},PD=AP−1, or PDP−1=A.PDP^{-1} = A.PDP−1=A. Already have an account? The multiplicity of each eigenvalue is important in deciding whether the matrix is diagonalizable: as we have seen, if each multiplicity is 1,1,1, the matrix is automatically diagonalizable. (PD)(e_i) = P(\lambda_i e_i) = \lambda_i v_i = A(v_i) = (AP^{-1})(e_i). you need to do something more substantial and there is probably a better way but you could just compute the eigenvectors and check rank equal to total dimension. Find a closed-form formula for the nthn^\text{th}nth Fibonacci number Fn,F_n,Fn, by looking at powers of the matrix A=(1110).A = \begin{pmatrix} 1&1\\1&0 \end{pmatrix}.A=(1110). If a set in its source has positive measure, than so does its image.". Its roots are Î» = ± i . \lambda^2-5\lambda+6&=0\\ Some matrices are not diagonalizable over any field, most notably nonzero nilpotent matrices. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Note that having repeated roots in the characteristic polynomial does not imply that the matrix is not diagonalizable: to give the most basic example, the n×nn\times nn×n identity matrix is diagonalizable (diagonal, in fact), but it has only one eigenvalue λ=1\lambda=1λ=1 with multiplicity n.n.n. A square matrix is said to be diagonalizable if it is similar to a diagonal matrix. @Anweshi: The analytic part enters when Mariano waves his hands---"Now the set where a non-zero polynomial vanishes is very, very thin"---so there is a little more work to be done. New user? en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set, Diagonalizability of Gaussian random matrices, Matrices: characterizing pairs $(AB, BA)$, Trouble with Jordan form of the truncated Carleman-matrix for $\sin(x)$ as size $n$ goes to infinity. ⎝⎛1−1−11−1−11−1−1⎠⎞→⎝⎛100100100⎠⎞, Therefore we only have to worry about the cases of k=-1 and k=0. An=(PDP−1)n=PDnP−1=15(ϕρ11)(ϕn00ρn)(1−ρ−1ϕ)=15(ϕn+1ρn+1ϕnρn)(1−ρ−1ϕ)=15(ϕn+1−ρn+1∗ϕn−ρn∗) As a very simple example, one can immediately deduce that the characteristic polynomials $AB$ and $BA$ coincide, because if $A$ is invertible, the matrices are similar. This argument only shows that the set of diagonalizable matrices is dense. 2 -4 3 2 1 0 3 STEP 1: Use the fact that the matrix is triangular to write down the eigenvalues. With a bit more care, one can derive the entire theory of determinants and characteristic polynomials from such specialization arguments. Is There a Matrix that is Not Diagonalizable and Not Invertible? The eigenvalues are the roots λ\lambdaλ of the characteristic polynomial: In general, a rotation matrix is not diagonalizable over the reals, but all rotation matrices are diagonalizable over the complex field. The elements in the superdiagonals of the Jordan blocks are the obstruction to diagonalization. Given a 3 by 3 matrix with unknowns a, b, c, determine the values of a, b, c so that the matrix is diagonalizable. So RRR is diagonalizable over C.\mathbb C.C. Now the set of polynomials with repeated roots is the zero locus of a non-trivial polynomial where the jthj^\text{th}jth column of PPP is an eigenvector of AAA with eigenvalue λj.\lambda_j.λj. a1v1+a2v2+⋯+akvk=vk+1 So what we are saying is µuTv = Î»uTv. If V is a finite-dimensional vector space, then a linear map T : V â V is called diagonalizable if there exists an ordered basis of V with respect to which T is represented by a diagonal matrix. DDD is unique up to a rearrangement of the diagonal terms, but PPP has much more freedom: while the column vectors from the 111-dimensional eigenspaces are determined up to a constant multiple, the column vectors from the larger eigenspaces can be chosen completely arbitrarily as long as they form a basis for their eigenspace. but this is impossible because v1,…,vkv_1,\ldots,v_kv1,…,vk are linearly independent. All this fuss about "the analytic part"---just use the Zariski topology :-). Since similar matrices have the same eigenvalues (indeed, the same characteristic polynomial), if AAA were diagonalizable, it would be similar to a diagonal matrix with 111 as its only eigenvalue, namely the identity matrix. (PD)(ei)=P(λiei)=λivi=A(vi)=(AP−1)(ei). To see this, let kkk be the largest positive integer such that v1,…,vkv_1,\ldots,v_kv1,…,vk are linearly independent. A1A2A3A4A5=(1110)=(2111)=(3221)=(5332)=(8553), If the matrix is not symmetric, then diagonalizability means not D= PAP' but merely D=PAP^{-1} and we do not necessarily have P'=P^{-1} which is the condition of orthogonality. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. -\lambda^3+2\lambda^2-\lambda&=0\\ This is an elementary question, but a little subtle so I hope it is suitable for MO. A=P(λ1λ2⋱λn)P−1,A=P \begin{pmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \end{pmatrix} P^{-1},A=P⎝⎜⎜⎛λ1λ2⋱λn⎠⎟⎟⎞P−1, \end{aligned} 51(ϕn−ρn)=2n5(1+5)n−(1−5)n, Proving “almost all matrices over C are diagonalizable”. Final exam problem of Linear Algebra at OSU. If AAA is an n×nn\times nn×n matrix with nnn distinct eigenvalues, then AAA is diagonalizable. Thus so does its preimage. More generally, there are two concepts of multiplicity for eigenvalues of a matrix. The characteristic equation is of the form, $$(x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n)$$. Prove that a given matrix is diagonalizable but not diagonalized by a real nonsingular matrix. In fact by purely algebraic means it is possible to reduce to the case of $k = \mathbb{R}$ (and thereby define the determinant in terms of change of volume, etc.). If a set in its source has positive measure, than so does its image. In particular, even if you don't want to do any measure theory, it's not hard to see that the complement of the set where a non-zero polynomial vanishes is dense. ⎝⎛2−1−110−11−10⎠⎞→⎝⎛−12−101−1−110⎠⎞→⎝⎛−10−101−1−1−10⎠⎞→⎝⎛10−101−11−10⎠⎞→⎝⎛1000101−10⎠⎞, A^5 &= \begin{pmatrix} 8&5\\5&3 \end{pmatrix}, This happens more generally if the algebraic and geometric multiplicities of an eigenvalue do not coincide. To learn more, see our tips on writing great answers. Now multiply both sides on the left by AAA to get So R R R is diagonalizable over C. \mathbb C. C. The second way in which a matrix can fail to be diagonalizable is more fundamental. By signing up, you'll get thousands of step-by-step solutions to your homework questions. vector space is diagonalizable. &\rightarrow \begin{pmatrix}-1&0&-1\\0&1&-1\\-1&-1&0 \end{pmatrix} \\ Diagonal Matrix. &\rightarrow \begin{pmatrix} 1&0&1\\0&1&-1\\0&0&0 \end{pmatrix}, I once had to think twice about the following: "proper + quasi-finite implies finite, but projective 1-space over a finite field is proper and quasi-finite---umm---aah I see the point". I wish I could accept your answer. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. There are other ways to see that AAA is not diagonalizable, e.g. 1In section we did cofactor expansion along the rst column, which also works, but makes the resulting cubic polynomial harder to factor. Is it always possible to “separate” the eigenvalues of an integer matrix? Of course, I do not know how to write it in detail with the epsilons and deltas, but I am convinced by the heuristics. \frac1{\sqrt{5}} (\phi^n-\rho^n) = \frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}, In general, a rotation matrix is not diagonalizable over the reals, but all rotation matrices are diagonalizable over the complex field. The base case is clear, and the inductive step is For examples the rationals. \end{aligned} May I ask more information about this "so" you use? for all matrices. its complement has measure zero). One is that its eigenvalues can "live" in some other, larger field. For each of the following matrices A, determine (1) if A is diagonalizable over Rand (ii) if A is diago- nalizable over C. When A is diagonalizable over C, find the eigenvalues, eigenvectors, and eigenbasis, and an invertible matrix P and diagonal matrix D such that p-I AP=D. Explicitly, let λ1,…,λn\lambda_1,\ldots,\lambda_nλ1,…,λn be these eigenvalues. Dear Anweshi, a matrix is diagonalizable if only if it is a normal operator. \begin{aligned} Now, if all the zeros have algebraic multiplicity 1, then it is diagonalizable. Sign up to read all wikis and quizzes in math, science, and engineering topics. &\rightarrow \begin{pmatrix} 1&0&1\\0&1&-1\\-1&-1&0 \end{pmatrix} \\ The characteristic polynomial $T - \lambda I$ splits into linear factors like $T - \lambda_iI$, and we have the Jordan canonical form: $$J = \begin{bmatrix} J_1 \\\ & J_2 \\\ & & \ddots \\\ & & & J_n \end{bmatrix}$$, where each block $J_i$ corresponds to the eigenvalue $\lambda_i$ and is of the form, $$J_i = \begin{bmatrix} \lambda_i & 1 \\\ & \lambda_i & \ddots \\\ & & \ddots & 1 \\\ & & & \lambda_i \end{bmatrix}$$. and each $J_i$ has the property that $J_i - \lambda_i I$ is nilpotent, and in fact has kernel strictly smaller than $(J_i - \lambda_i I)^2$, which shows that none of these Jordan blocks fix any proper subspace of the subspace which they fix. That is, almost all complex matrices are not diagonalizable. A diagonal square matrix is a matrix whose only nonzero entries are on the diagonal: So this shows that AAA is indeed diagonalizable, because there are "enough" eigenvectors to span R3. polynomial is the best kind of map you could imagine (algebraic, det(A−λI)=∣1−λ−124−λ∣=0  ⟹  (1−λ)(4−λ)+2=0λ2−5λ+6=0λ=2,3.\begin{aligned} For example, the matrix $\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}$ is such a matrix. In particular, the real matrix (0 1 1 0) commutes with its transpose and thus is diagonalizable over C, but the real spectral theorem does not apply to this matrix and in fact this matrix â¦ (2) If P( ) does not have nreal roots, counting multiplicities (in other words, if it has some complex roots), then Ais not diagonalizable. as desired. \begin{aligned} Remark: The reason why matrix Ais not diagonalizable is because the dimension of E 2 (which is 1) is smaller than the multiplicity of eigenvalue = 2 (which is 2). D=(d11d22⋱dnn). An=(PDP−1)n=(PDP−1)(PDP−1)(⋯)(PDP−1)=PDnP−1 In addition to the other answers, all of which are quite good, I offer a rather pedestrian observation: If you perturb the diagonal in each Jordan block of your given matrix $T$ so all the diagonal terms have different values, you end up with a matrix that has $n$ distinct eigenvalues and is hence diagonalizable. Thanks for contributing an answer to MathOverflow! That is, if and only if $A$ commutes with its adjoint ($AA^{+}=A^{+}A$). A matrix such as has 0 as its only eigenvalue but it is not the zero matrix and thus it cannot be diagonalisable. \lambda&= 2,3. In both these cases, we can check that the geometric multiplicity of the multiple root will still be 1, so that the matrix is not diagonalizable in either case. But it is not hard to check that it has two distinct eigenvalues over C,\mathbb C,C, since the characteristic polynomial is t2+1=(t+i)(t−i).t^2+1 = (t+i)(t-i).t2+1=(t+i)(t−i). Diagonalizable Over C Jean Gallier Department of Computer and Information Science University of Pennsylvania Philadelphia, PA 19104, USA [email protected] June 10, 2006 Abstract. But this does not mean that every square matrix is diagonalizable over the complex numbers. which has a two-dimensional nullspace, spanned by, for instance, the vectors s2=(1−10)s_2 = \begin{pmatrix} 1\\-1\\0\end{pmatrix}s2=⎝⎛1−10⎠⎞ and s3=(10−1).s_3 = \begin{pmatrix} 1\\0\\-1 \end{pmatrix}.s3=⎝⎛10−1⎠⎞. Being contained in a proper algebraic subset of affine or projective space is a very strong and useful way of saying that a set is "small" (except in the case that $k$ is finite! This equation is a restriction for a matrix $A$. by computing the size of the eigenspace corresponding to λ=1\lambda=1λ=1 and showing that there is no basis of eigenvalues of A.A.A. Here you go. t^2+1 = (t+i)(t-i). (we don't really care about the second column, although it's not much harder to compute). Diagonalizable, but not invertible. How to solve: When is a matrix not diagonalizable? In general, any 3 by 3 matrix whose eigenvalues are distinct can be diagonalised. P=(ϕρ11)D=(ϕ00ρ)P−1=15(1−ρ−1ϕ). \begin{aligned} \lambda&= 0,1. From Theorem 2.2.3 and Lemma 2.1.2, it follows that if the symmetric matrix A â Mn(R) has distinct eigenvalues, then A = Pâ1AP (or PTAP) for some orthogonal matrix P. The second way in which a matrix can fail to be diagonalizable is more fundamental. I am almost tempted to accept this answer over the others! a_1 \lambda_1 v_1 + a_2 \lambda_2 v_2 + \cdots + a_k \lambda_k v_k = \lambda_{k+1} v_{k+1}. Matrix diagonalization is useful in many computations involving matrices, because multiplying diagonal matrices is quite simple compared to multiplying arbitrary square matrices. Finally, note that there is a matrix which is not diagonalizable and not invertible. Even if a matrix is not diagonalizable, it is always possible to "do the best one can", and find a matrix with the same properties consisting of eigenvalues on the leading diagonal, and either ones or zeroes on the superdiagonal â known as Jordan normal form . D = \begin{pmatrix} d_{11} & & & \\ & d_{22} & & \\ & & \ddots & \\ & & & d_{nn} \end{pmatrix}. whether the geometric multiplicity of 111 is 111 or 2).2).2). which has nullspace spanned by the vector s1=(−111).s_1 = \begin{pmatrix} -1\\1\\1 \end{pmatrix}.s1=⎝⎛−111⎠⎞. There are all possibilities. PDP−1=(ϕ1ρ1)=(ϕ00ρ)=51(1−1−ρϕ).. The map from $\mathbb C^{n^2}$ to the space of monic polynomials of degree $n$ which associates Therefore, the set of diagonalizable matrices has null measure in the set of square matrices. A^n = (PDP^{-1})^n = (PDP^{-1})(PDP^{-1})(\cdots)(PDP^{-1}) = PD^nP^{-1} Multiplying both sides of the original equation by λk+1\lambda_{k+1}λk+1 instead gives It perturbs me that I cannot complete this argument rigorously. \end{aligned} 23.2 matrix Ais not diagonalizable. Diagonal matrices are relatively easy to compute with, and similar matrices share many properties, so diagonalizable matrices are well-suited for computation. a1v1+a2v2+⋯+akvk=vk+1 Therefore, the set of diagonalizable matrices has null measure in the set of square matrices. Two different things. @Harald. As a closed set with empty interior can still have positive measure, this doesn't quite clinch the argument in the measure-theoretic sense. A = \begin{pmatrix}1&1\\1&-1 \end{pmatrix} \begin{pmatrix} 1&0\\0&-1 \end{pmatrix} \begin{pmatrix}1&1\\1&-1 \end{pmatrix}^{-1}. (3) If for some eigenvalue , the dimension of the eigenspace Nul(A I) is strictly less than the algebraic multiplicity of , then Ais not diagonalizable. N(A−λ1I)=N(A),N(A-\lambda_1 I ) = N(A),N(A−λ1I)=N(A), which can be computed by Gauss-Jordan elimination: It only takes a minute to sign up. P &= \begin{pmatrix} \phi&\rho\\1&1 \end{pmatrix} \\ Can I assign the term “is eigenvector” and “is eigenmatrix” of matrix **P** in my specific (infinite-size) case? In linear algebra, a square matrix A is called diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P â1 AP is a diagonal matrix. This equation is a restriction for a matrix $A$. So far, so good. \begin{aligned} Such a perturbation can of course be as small as you wish. For instance, if the matrix has real entries, its eigenvalues may be complex, so that the matrix may be diagonalizable over C\mathbb CC without being diagonalizable over R.\mathbb R.R. The matrix A=(0110)A = \begin{pmatrix} 0&1\\1&0 \end{pmatrix}A=(0110) is diagonalizable: -Dardo. Indeed, it has no real eigenvalues: if vvv is a vector in R2,{\mathbb R}^2,R2, then RvRvRv equals vvv rotated counterclockwise by 90∘.90^\circ.90∘. If it is diagonalizable, then find the invertible matrix S and a diagonal matrix D such that Sâ1AS=D. So the conclusion is that A=PDP−1, A = PDP^{-1},A=PDP−1, where The 'obvious measure' on $\mathbb C^{n^2}$ is not a probability measure... You are right. Also recall the existence of space-filling curves over finite fields. MathJax reference. By \explicit" I mean that it can always be worked out with pen and paper; it can be long, it can be tedious, but it can be done. \begin{aligned} Log in. It is clear that if N is nilpotent matrix (i. e. Nk = 0 â¦ surjective, open, ... ). Interpreting the matrix as a linear transformation â 2 â â 2 , it has eigenvalues i and - i and linearly independent eigenvectors ( 1 , - i ) , ( - i , 1 ) . The dimension of the eigenspace corresponding to λ\lambdaλ is called the geometric multiplicity. D &= \begin{pmatrix} \phi&0\\0&\rho \end{pmatrix} \\ Please see meta here. A^n = A \cdot A^{n-1} &= \begin{pmatrix} 1&1\\1&0 \end{pmatrix} \begin{pmatrix} F_n&F_{n-1}\\F_{n-1}&F_{n-2} \end{pmatrix} \\ A=(111−1)(100−1)(111−1)−1. In particular, the bottom left entry, which is FnF_nFn by induction, equals a1λk+1v1+a2λk+1v2+⋯+akλk+1vk=λk+1vk+1 But multiplying a matrix by eie_iei just gives its ithi^\text{th}ith column. \end{aligned} To subscribe to this RSS feed, copy and paste this URL into your RSS reader. If k≠n,k \ne n,k=n, then there is a dependence relation And we can write down the matrices PPP and DDD: It is straightforward to check that A=PDP−1A=PDP^{-1}A=PDP−1 as desired. A^4 &= \begin{pmatrix} 5&3\\3&2 \end{pmatrix} \\ which is Binet's formula for Fn.F_n.Fn. Use MathJax to format equations. and in the space generated by the $\lambda_i$'s, the measure of the set in which it can happen that $\lambda_i = \lambda_j$ when $i \neq j$, is $0$: this set is a union of hyperplanes, each of measure $0$. Dense sets can be of measure zero. Asking for help, clarification, or responding to other answers. So the only thing left to do is to compute An.A^n.An. An=A⋅An−1=(1110)(FnFn−1Fn−1Fn−2)=(Fn+Fn−1Fn−1+Fn−2FnFn−1)=(Fn+1FnFnFn−1) Diagonalizability with Distinct Eigenvalues, https://brilliant.org/wiki/matrix-diagonalization/. is not diagonalizable, since the eigenvalues of A are 1 = 2 = 1 and eigenvectors are of the form = t ( 0, 1 ), t 0 and therefore A does not have two linearly independent eigenvectors. Pete: being a closed subset of A^n which isn't A^n is still a powerful statement when k is finite, because, as you well know, affine n-space over a finite field is still an infinite set. A=(111−1)(100−1)(111−1)−1. □_\square□. So □A=PD P^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\begin{pmatrix}2&0\\0&3\end{pmatrix}\begin{pmatrix}2&1\\1&1\end{pmatrix}.\ _\squareA=PDP−1=(1−1−12)(2003)(2111). We find eigenvectors for these eigenvalues: λ1=0:\lambda_1 = 0:λ1=0: The discriminant argument shows that for for $n \times n$ matrices over any field $k$, the Zariski closure of the set of non-diagonalizable matrices is proper in $\mathbb{A}^{n^2}$ -- an irreducible algebraic variety -- and therefore of smaller dimension. Putting this all together gives To you it means unitarily equivalent to a diagonal matrix. The ϕ\phiϕ-eigenspace is the nullspace of (1−ϕ11−ϕ),\begin{pmatrix} 1-\phi&1 \\ 1&-\phi \end{pmatrix},(1−ϕ11−ϕ), which is one-dimensional and spanned by (ϕ1).\begin{pmatrix} \phi\\1 \end{pmatrix}.(ϕ1). An=(PDP−1)n=(PDP−1)(PDP−1)(⋯ )(PDP−1)=PDnP−1 Forgot password? □_\square□. Now that you have Mariano's argument notice the kind of things you can do with it -- for example, you can give a simple proof of Cayley-Hamilton by noticing that the set of matrices where Cayley-Hamilton holds is closed, and true on diagonalizable matrices for simple reasons. But the only matrix similar to the identity matrix is the identity matrix: PI2P−1=I2PI_2P^{-1} = I_2PI2P−1=I2 for all P.P.P. Add to solve later Sponsored Links The added benefit is that the same argument proves that Zariski closed sets are of measure zero. This extends immediately to a definition of diagonalizability for linear transformations: if VVV is a finite-dimensional vector space, we say that a linear transformation T :V→VT \colon V \to VT:V→V is diagonalizable if there is a basis of VVV consisting of eigenvectors for T.T.T. Solution for Show that the matrix is not diagonalizable. 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• Gurobi Staff

As the error suggests, Gurobi doesn't support dividing by variables. For more information, see What types of models can Gurobi solve?.

You could model this problem by introducing a few auxiliary variables and bilinear constraints. However, your problem looks very similar to problem of maximizing the Sharpe ratio (though you use variance instead of standard deviation), for which there is a nice convex reformulation. Let $$\mu \in \mathbb{R}^n$$ be the vector of expected returns, and let $$Q \in \mathbb{S}_{+}^{n \times n}$$ be the (symmetric, positive semidefinite) covariance matrix. I'll assume you are trying to maximize the Sharpe ratio $$\mu^\top x / \sqrt{x^\top Q x}$$; I'm not aware of a common risk-adjusted return measure that uses variance instead of standard deviation.

Your problem, which we refer to as $${\bf(1)}$$, can be written as

\begin{alignat*}{2} \max_x\ && \frac{\mu^\top x}{\sqrt{x^\top Q x}} & \\ \textrm{s.t.}\ && \sum_{i=1}^n x_i ={} &1 \\ && x \geq {} &0.\end{alignat*}

Consider the following problem, which we refer to as $${\bf(2)}$$:

\begin{alignat*}{2} \max_y\ && \frac{1}{\sqrt{y^\top Q y}} & \\ \textrm{s.t.}\ && \mu^\top y = {} &1 \\ && y \geq {} &0.\end{alignat*}

We'll show that $${\bf(1)}$$ and $${\bf(2)}$$ are "equivalent", in the sense that given a solution to either problem, we can construct a solution to the other of equal or better objective value. We assume there exists some $$\hat{x}$$ for which $$\mu^\top \hat{x} > 0$$, which should hold in any reasonable portfolio optimization problem. It follows from this assumption that the optimal solution to $${\bf(1)}$$ has a positive expected return.

First, let $$\bar{x}$$ be a solution to $${\bf(1)}$$. We construct a solution to $${\bf(2)}$$ of equal objective value. For all $$i = 1, \ldots, n$$, define $$\bar{y}_i := \bar{x}_i / \mu^\top \bar{x}$$. It holds that $$\bar{y} \geq 0$$, because $$\bar{x}$$ is nonnegative and $$\mu^\top \bar{x} > 0$$ by our assumption. Additionally, $$\mu^\top \bar{y} = \mu^\top \bar{x} / \mu^\top \bar{x} = 1$$. Thus, $$\bar{y}$$ satisfies the constraints of $${\bf(2)}$$. Finally, the objective of $${\bf(2)}$$ with respect to the solution $$\bar{y}$$ is $$1 / \sqrt{\bar{y}^\top Q \bar{y}} = 1 / \sqrt{ \bar{x}^\top Q \bar{x} / (\mu^\top \bar{x})^2} = \mu^\top \bar{x} / \sqrt{\bar{x}^\top Q \bar{x}}$$.

Next, let $$\bar{y}$$ be a solution to $${\bf(2)}$$. We construct a solution to $${\bf(1)}$$ of equal objective value. Let $$\bar{x}_i := \bar{y}_i / \sum_{j=1}^n \bar{y}_j$$ for $$i = 1, \ldots, n$$. Using the same ideas as above, it's easy to verify that $$\bar{x}$$ satisfies the constraints of $${\bf(1)}$$ and has objective value equal to $$1 / \sqrt{\bar{y}^\top Q \bar{y}}$$.

So, instead of solving your original problem $${\bf(1)}$$, we can just solve $${\bf(2)}$$ and map the optimal solution $$\bar{y}$$ back to the original problem using the translation $$\bar{x}_i := \bar{y}_i / \sum_{j=1}^n \bar{y}_j$$ for $$i = 1, \ldots, n$$. However, because $$Q$$ is positive semidefinite, the solutions of $${\bf(2)}$$ are the same as the solutions to

\begin{alignat*}{2} \min_y\ && y^\top Q y \enspace & \\ \textrm{s.t.}\ && \mu^\top y = {} &1 \\ && y \geq {} &0.\end{alignat*}

This is a convex QP that Gurobi can solve directly. A solution $$\bar{y}$$ to the above problem can be mapped to the original problem using the same transformation $$\bar{x}_i := \bar{y}_i / \sum_{j=1}^n \bar{y}_j$$ for $$i = 1, \ldots, n$$.

If you have other side constraints to $${\bf(1)}$$, see Erwin Kalvelagen's blog post on maximizing the Sharpe ratio. Basically, additional constraints $$Ax \leq b$$ in $${\bf(1)}$$ translate to constraints $$Ay \leq b \kappa$$ in $${\bf(2)}$$, where $$\kappa = \sum_{j=1}^n y_j$$.

My goodness Eli, this was a fantastic explanation. Thank you for taking the time to work through this and write this detailed of a reply. I am happy to say that I implemented this into my code without issue and got the exact result I was expecting. And yes, this is exactly the problem of maximizing the Sharpe ratio, but I had left out the square root due to other issues in an attempt to solve it in piecemeal fashion.

Thanks again!!

Marek

Hi together, I have a similar problem, I understand Elis answer but am unable to implement the solution. How do I implement the described approach?

I have:

for tk in TK:

R, A and E are all variables in the model.

I would be glad if somebody could help here.

Kind regards

Moritz

• Gurobi Staff

Which part of the implementation do you need help with? Your code looks okay, depending on how you defined $$\texttt{TK}$$, $$\texttt{R}$$, $$\texttt{A}$$, and $$\texttt{E}$$. For example, the following code runs without error:

import gurobipy as gpTK = [1, 2, 3]model = gp.Model()R = model.addVars(TK, name='R')A = model.addVars(TK, name='A')E = model.addVars(TK, name='E')for tk in TK:    model.addConstr(R[tk] * A[tk] == E[tk])

If you encounter an error, it would be helpful to see a minimal, self-contained code snippet that reproduces the error.

Hi again,

below I send the snippet. It took a while to reduce the code to the problem, sorry for coming back late. I marked the line which makes trouble with a comment. Its about R[t]. Can you please support here? I am not aware of a a reasonable solution.

Thanks in advance and kind regards.

Moritz

from gurobipy import *

# =============================================================================
model = Model("Smart Factory Strategie")
model.modelSense = GRB.MAXIMIZE

# =============================================================================
A=['1','2','3']
D={'1':1.5,'2':1.8,'3':1.5}
KWS={'1':[-10842,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,0,0,0,0],'2':[-5421,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],'3':[-10842,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,-1355,0,0,0,0]}
KWE={'1':[35090,35090,35090,35090,35090,35090,35090,35090,35090,35090,35090,35090,0,0,0,0],'2':[22562,22562,22562,22562,22562,22562,22562,22562,22562,22562,22562,22562,0,0,0,0],'3':[18492,18492,18492,18492,18492,18492,18492,18492,18492,18492,18492,18492,0,0,0,0]}
Tmax=   10
T=      [i for i in range(0,Tmax+1)]

# =============================================================================

Y={}
for a in A:
for t in T:

Au={}
for t in T:

BW={}
for t in T:
BW[t]=model.addVar(name="BW_%s" % (t), vtype =GRB.CONTINUOUS, lb=-1*10**30)

BWX= model.addVar(vtype =GRB.CONTINUOUS , obj = 1 , name='BWX')

Er={}
for t in T:

R={}
for t in T:
R[t]=model.addVar(name="R_%s" % (t), vtype =GRB.CONTINUOUS, lb=-1*10**30)

# =============================================================================

for t in T:
model.addConstr(Au[t] == quicksum(KWS[a][t] * Y[a,t] for a in A))
model.addConstr(Er[t] == quicksum(KWE[a][t] * Y[a,t] for a in A))
for t in T:

for t in T:
model.addConstr(R[t]*Au[t]==Er[t]) #This line of code makes trouble

# =============================================================================
model.update()
model.optimize()
# =============================================================================

• Gurobi Staff

It looks like you just need to set the NonConvex parameter to 2 to direct Gurobi to solve the model (which is made non-convex by the quadratic equality constraints):

model.Params.NonConvex = 2

Note that solving non-convex quadratic problems is only supported in Gurobi 9.0 onward.

Thanks a lot. It works