Homework 5

True/False?

a. the matrix product $\cA\cB$ is defined for any two matrices $\cA$ and $\cB$.

b. if the matrix product $\cA\cB$ is defined then the matrix product $\cB\cA$ also is defined.

c. if $\cA$ is an $n\times m$ matrix, and if $\cA x=0$ for some $x\in\R^m$ then $\cA=\cO$. ($\cO$ is the $n\times m$ zero matrix.)

d. if $\cA$ and $\cB$ are two $n\times n$ matrices then $(\cA+\cB)^2=\cA^2+2\cA\cB+\cB^2$

a. False $\cA$ and $\cB$ have to have compatible sizes: if $\cA$ is an $n\times m$ matrix then $\cB$ must be an $m\times k$ matrix.

b. False If $\cA$ is $n\times m$ and $\cB$ is $m\times k$, then $\cA\cB$ is defined. But, unless $k=n$, $\cB\cA$ is not defined.

c. False $\cA x=0$ only means that $x\in N(\cA)$.

d. False This is only true if $\cA\cB=\cB\cA$. See problem 6 below.

1. Composition of linear transformations is also linear

If $A:V\to W$ and $B:U\to V$ are linear transformations of vector spaces, then show that the composition $AB:U\to W$ also is linear. The composition is defined by $(AB)(x)=A(B(x)))$

We have to show that $(AB)(sx+ty) = s(AB)x+t(AB)y$ for all $x,y\in V$ and all $s,t\in \F$:

$\begin{aligned} (AB)(sx+ty) &= A\left(B(sx+ty)\right) & (\text{def of }AB)\\ &=A\left(sBx+tBy\right) &(B\text{ is linear}) \\ &=sA(Bx) + tA(By) & (\text{$A$ is linear}) \\ &=s(AB)(x) + t(AB)(y) & (\text{def of $AB$, twice}) \end{aligned}$

Therefore $AB$ is linear.

2. Reflection, projection, and their matrices

Let $\alpha\in(0, \frac\pi2)$ be a given angle, and let $\ell_\alpha$ be the line in $\R^2$ through the origin whose slope is $\tan\alpha$ (its equation is $y=x\tan\alpha$).
Let $S_\alpha:\R^2\to\R^2$ be reflection in the line $\ell_\alpha$. You may assume that $S_\alpha$ is linear.

a. Find the matrix of $S_\alpha$ with respect to the standard basis $\{e_1, e_2\}$

b. Find the matrix of $S_\alpha$ with respect to the basis $\{v, w\}$ where $v=\binom{\cos\alpha}{\sin\alpha}$, $w=\binom{-\sin\alpha}{\cos\alpha}$. Suggestion: first make a drawing of the line $\ell_\alpha$ and the vectors $v,w$.

c. Verify in both cases a and b that the matrix $\mathcal{M}$ you get satisfies $\mathcal{M}^2=I$.

d. Let $P_\alpha:\R^2\to\R^2$ be perpendicular projection on the line $\ell_\alpha$. Find the matrix of $P_\alpha$ with respect to the standard basis $\{e_1, e_2\}$. Hint: use geometry to show that $P_\alpha x = \frac12 (S_\alpha x+x)$ and conclude that $P_\alpha= \tfrac12 (S_\alpha + I)$.

e. Find the matrix of the projection $P_\alpha$ with respect to the basis $v=\binom{\cos\alpha}{\sin\alpha}$, $w=\binom{-\sin\alpha}{\cos\alpha}$.

a. Use a drawing to compute the coordinates of $S_\alpha e_1$ and $S_\alpha e_2$. Finding $S_\alpha e_2$ :

$S_\alpha e_1 = \binom{\cos2\alpha}{\sin2\alpha}$, and $S_\alpha e_2=\binom{\cos(\frac{\pi}{2}-2\alpha)}{-\sin(\frac{\pi}{2}-2\alpha)} = \binom{\sin 2\alpha}{-\cos 2\alpha}$.

Hence the matrix of $S_\alpha$ with respect to the standard basis is

$[S_\alpha]_{e_1, e_2} = \mat\cos2\alpha & \sin 2\alpha \\ \sin 2\alpha & - \cos 2\alpha\rix$

b. The vector $v$ lies on the line of reflection, so $S_\alpha v=v$. The vector $w$ is perpendicular to the line of reflection, so $S_\alpha w=-w$. Therefore the matrix of $S_\alpha$ with respect to the basis $\{v, w\}$ is

$[S_\alpha]_{v,w} = \mat 1 & 0 \\ 0 & -1\rix .$

c. The first matrix we found was $\cM = [S_\alpha]_{e_1, e_2}= \tmat\cos2\alpha & \sin 2\alpha \\ \sin 2\alpha & - \cos 2\alpha\trix$, so we find

$\begin{aligned} \cM^2 &= \mat\cos2\alpha & \sin 2\alpha \\ \sin 2\alpha & - \cos 2\alpha\rix \mat\cos2\alpha & \sin 2\alpha \\ \sin 2\alpha & - \cos 2\alpha\rix \\ &=\mat\cos^22\alpha+\sin^22\alpha & \cos2\alpha \sin2\alpha - \sin2\alpha\cos2\alpha\\ \sin2\alpha\cos2\alpha-\cos2\alpha\sin2\alpha & \sin^22\alpha + \cos^22\alpha \rix\\ &=\mat 1 & 0 \\ 0 & 1\rix = I \end{aligned}$

For the other matrix $\cM=[S_\alpha]_{v,w} = \tmat1 & 0 \\ 0 & -1\trix$ the computation is easier and also gives $\cM^2=I$.

d. $P_\alpha x = \frac12 (x+S_\alpha x)$, so we have $P_\alpha e_1 = \binom{\frac12(1+\cos2\alpha)}{\frac12 \sin2\alpha}$ and $P_\alpha e_2 = \tmat \frac12 \sin2\alpha \\ \frac12 (1-\cos 2\alpha)\trix$. Thus

$[P_\alpha]_{e_1, e_2} = \mat \frac12(1+\cos2\alpha) & \frac12 \sin2\alpha \\ \frac12 \sin2\alpha & \frac12 (1-\cos 2\alpha) \rix$

e. $[P_\alpha]_{v, w}=\tmat 1 & 0 \\ 0 & 0 \trix$

3. Some matrix products

Compute $\mathcal A \mathcal B$ and $\mathcal{B}\mathcal{A}$ for the following matrices

a. $\mathcal{A}=\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$, $\mathcal{B}=\begin{pmatrix} 1 & 1 \\ 4 & 1 \end{pmatrix}$

$\cA\cB= \tmat 9 & 3 \\ 6 & 3 \trix$, $\cB\cA = \tmat3 & 3 \\ 6 & 9\trix$

b. $\mathcal{A}=\begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}$, $\mathcal{B}=\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix}$

$\cA\cB=\tmat 1 & a+b \\ 0 & 1\trix$, and $\cB\cA=\tmat 1 & a+b \\ 0 & 1\trix$, so for these matrices $\cA\cB=\cB\cA$

c. $\mathcal{A}=\begin{pmatrix} 1 & a & 0 \\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$, $\mathcal{B}=\begin{pmatrix} 1 & c & 0 \\ 0 & 1 & d \\ 0 & 0 & 1 \end{pmatrix}$.

$\cA\cB = \tmat1 & a+c & ad \\ 0 & 1 & b+d \\ 0 & 0 & 1\trix$, $\cB\cA = \tmat1 & a+c & bc \\ 0 & 1 & b+d \\ 0 & 0 & 1\trix$

d. $\cA=\mat a_1 & a_2\rix$, $\cB=\mat b_1 \\ b_2\rix$.

$\cA\cB = (a_1b_1+a_2b_2)$, $\cB\cA = \tmat a_1b_1 & a_2b_1 \\a_1b_2 & a_2 b_2 \trix$. In this example $\cA\cB$ and $\cB\cA$ both are defined, but they don't even have the same size.

4. Matrix powers

Consider the matrix $\cA=\left(\begin{smallmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{smallmatrix}\right)$

a. Compute $\cA^3$. Then compute $\cA^{2020}$.

$\cA^2 = \tmat0&0&1\\1&0&0\\0&1&0\trix$, $\cA^3=\tmat1&0&0\\0&1&0\\0&0&1\trix$, i.e. $\cA^3=I$. Once you know $\cA^3=I$ you get $\cA^4=\cA\cA^3=\cA I=\cA$, $\cA^5=\cA^2\cA^3=\cA^2$, $\cA^6=\cA^3=I$, ... etc.

Since $2020=673\cdot 3+1$, you get $\cA^{2020} = \cA^{3\cdot673+1}=\cA^1=\cA$.

b. Find the inverse of $\cA$.

$\cA\cA^2=\cA^3=I$ and $\cA^2\cA=\cA^3=I$, so $\cA^2$ is the inverse of $\cA$.

c. Find a $2\times2$ matrix $\cB$ for which $\cB^5=I$. (hint: rotations).

$\cB=I=\tmat 1&0\\0&1\trix$ is a valid answer. To find other answers consider the rotation matrices $\cR(\theta) = \tmat\cos\theta & -\sin\theta \\\sin\theta&\cos\theta\trix$, where $\theta\in\R$.
We know from lecture that $\cR(\theta)\cR(\phi)=\cR(\theta+\phi)$. Therefore

$\cR(\phi)^5 = \cR(\phi)\cR(\phi)\cR(\phi)\cR(\phi)\cR(\phi)=\cR(5\phi)$

Choose $\phi$ so that $5\phi=2\pi$, i.e. choose $\phi=\frac25\pi = 72^\circ$, then

$\cB=\cR(\frac25\pi)=\mat\cos\frac25\pi&-\sin\frac25\pi\\ \sin\frac25\pi&\cos\frac25\pi\rix$

satisfies $\cB^5=I$.

5. Derivatives and matrices

Let $V=\cP_n(\R)$ be the space of polynomials of degree at most $n$, and let $\beta=\{1, x, x^2, \dots, x^n\}$ be the standard basis on $V$

a. Find the matrix of the linear transformation $D:V\to V$ given by $(Df)(x)=f'(x)$.

(When $n=3$) To find the matrix of $D$ with respect to the basis $\{1, x, x^2, x^3\}$ compute $D(1), D(x), D(x^2), D(x^3)$ and express them as linear combinations of $\{1,x,x^2, x^3\}$:

$\begin{aligned} D(1)=0&=0\cdot 1+0\cdot x+0\cdot x^2+0\cdot x^3\\ D(x)=1&=1\cdot 1+0\cdot x+0\cdot x^2+0\cdot x^3\\ D(x^2)=2x&=0\cdot 1+2\cdot x+0\cdot x^2+0\cdot x^3\\ D(x^3)=3x^2&=0\cdot 1+0\cdot x+3\cdot x^2+0\cdot x^3\\ \end{aligned}$

The coefficients of $D(1)$ go in the first column of the matrix of $D$, etc. We get

$\cD = \mat 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \rix$

For any value of $n$ the matrix $\cD$ is an $(n+1)\times(n+1)$ matrix, given by

$\cD = \mat 0 & 1 & 0 & 0& & 0 \\ 0 & 0 & 2 & 0& & 0 \\ 0 & 0 & 0 & 3& & 0 \\ & & & & \ddots& \\ 0 & 0 & 0 & 0& 0& n \\ 0 & 0 & 0 & 0& 0& 0 \rix$

b. Assume that $n=3$ and let $\cD$ be the matrix of $D$ you found in a. Compute $\cD^2$, $\cD^3$, ... and $\cD^k$ for each integer $k\in\N$.

$\cD^2 = \tmat0&0&2&0\\0&0&0&6\\0&0&0&0\\0&0&0&0\trix$, reflecting the fact that $D(D(1))=0$, $D(D(x))=0$, $D(D(x^2))=D(2x)=2$, and $D(D(x^3))=D(3x^2)=6x$.

$\cD^3=\tmat0&0&0&6\\0&0&0&0\\0&0&0&0\\0&0&0&0\trix$, $\cD^4=\tmat0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\trix=O$

$\cD^5=\cD^6=\cdots=O$

c. Compute the matrix $\cT$ of the linear transformation $T:V\to V$ defined by $Tf(x) = f(x+1)$ (assume again $n=3$).

Compute $T(1), T(x), T(x^2), T(x^3)$ and express them as linear combinations of $\{1,x,x^2, x^3\}$:

$\begin{aligned} T(1)=1&=1\cdot 1+0\cdot x+0\cdot x^2+0\cdot x^3\\ T(x)=x+1&=1\cdot 1+1\cdot x+0\cdot x^2+0\cdot x^3\\ T(x^2)=(x+1)^2=x^2+2x+1&=1\cdot 1+2\cdot x+1\cdot x^2+0\cdot x^3\\ T(x^3)=(x+1)^3=x^3+3x^2+3x+1&=1\cdot 1+3\cdot x+3\cdot x^2+1\cdot x^3\\ \end{aligned}$

Therefore the matrix $\cT$ of $T$ is

$\cT=\mat 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \\ \rix$

d. Show that $\cT=I+\cD+\frac1{2!}\cD^2 + \frac1{3!}\cD^3$.

Use the values of $\cD^k$ that we computed in b above:

$\begin{aligned} I+&\cD+\frac1{2!}\cD^2 + \frac1{3!}\cD^3 =\\ \\ &=\tmat 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \trix + \tmat 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \trix +\tfrac12\tmat0&0&2&0\\0&0&0&6\\0&0&0&0\\0&0&0&0\trix +\tfrac16\tmat0&0&0&6\\0&0&0&0\\0&0&0&0\\0&0&0&0\trix \\ &=\tmat 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \trix + \tmat 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \trix +\tmat0&0&1&0\\0&0&0&3\\0&0&0&0\\0&0&0&0\trix +\tmat0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\trix \\ &=\tmat 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \\ \trix\\ &=\cT \end{aligned}$

6. Commutators

The commutator of two square matrices $\cA$ and $\cB$ is defined to be $[\mathcal{A},\mathcal{B}] \stackrel{\rm def}{=}\mathcal{A}\mathcal{B}-\mathcal{B}\mathcal{A}$. It is called the commutator because $\cA\cB = \cB\cA$ (“$\cA$ and $\cB$ commute”) holds if and only if $[\cA,\cB]=0$.

a. Show that $[\cA, \cB]=-[\cB,\cA]$ for all $n\times n$ matrices $\cA,\cB$.

$[\cB, \cA] = \cB\cA-\cA\cB = -\cA\cB+ \cB\cA = -\left(\cA\cB-\cB\cA\right)=-[\cA,\cB]$.

b. Compute $[I, \cA]$ for any $n\times n$ matrix $\cA$ ($I$ is the identity matrix).

For any $n\times n$ matrix $\cA$ one has $\cA I=I\cA=\cA $.
Therefore $[\cA,I]=\cA I-I\cA=\cA-\cA=O$.

c. Show that $[t\cA+s\cB, \cC] = t[\cA,\cC]+s[\cB,\cC]$ for all $n\times n$ matrices $\cA,\cB,\cC$.

$\begin{aligned} [t\cA+s\cB, \cC] &= (t\cA+s\cB)\cC - \cC(t\cA+s\cB) \\ &= t\cA\cC+s\cB\cC - t\cC\cA- s\cC\cB \\ &= t\cA\cC- t\cC\cA+s\cB\cC - s\cC\cB \\ &= t\bigl(\cA\cC- \cC\cA\bigr)+s\bigl(\cB\cC - \cC\cB\bigr) \\ &= t[\cA,\cC]+s[\cB,\cC] \end{aligned}$

d. Let $\cC$ be some $n\times n$ matrix, and compute $[\cC^k,\cC^l]$ where $k,l\in\N$.
Then compute $[\cA,\cB]$ where $\cA= I+\cC^2$ and $\cB=\cC-2\cC^2$.

For all $k,l$ one has $\cC^k\cC^l = \cC^{k+l}$, so $[\cC^k,\cC^l]=\cC^k\cC^l-\cC^l\cC^k = \cC^{k+l}-\cC^{l+k}=O$. First computation of $[\cA,\cB]$ where $\cA= I+\cC^2$ and $\cB=\cC-2\cC^2$, directly from the definition of the commutator:

$\begin{aligned} \left[I+\cC^2, \cC-2\cC^2\right] &= (I+\cC^2)(\cC-2\cC^2)-(\cC-2\cC^2)(I+\cC^2) \\ &= \cC-2\cC^2+\cC^3-2\cC^4- \cC+2\cC^2-\cC^3+2\cC^4 \\ &=O \quad\text{(all terms cancel)} \end{aligned}$

Second, alternative, computation using the properties in b and c of this problem:

$\begin{aligned} \left[I+\cC^2, \cC-2\cC^2\right] &=[I, \cC-2\cC^2] + [\cC^2, \cC-2\cC^2] \\ &=O + [\cC^2,\cC] - 2[\cC^2, \cC^2]\\ &=O+O-2O\\ &=O \end{aligned}$

e. Compute the commutator of

$\mathcal{A} = \begin{pmatrix} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \\ \end{pmatrix} \qquad \mathcal{B} = \begin{pmatrix} 0 & -b_3 & b_2 \\ b_3 & 0 & -b_1 \\ -b_2 & b_1 & 0 \\ \end{pmatrix}$

Compare the result with the cross product of the vectors $\tmat a_1\\[1pt] a_2\\[1pt] a_3\trix$ and
$\tmat b_1\\b_2\\b_3\trix$ (for cross product see here or the second minute of this)

$\begin{aligned} \cA\cB &= \tmat -a_2b_2-a_3b_3 & a_2b_1 & a_3b_1\\ a_1b_2 &-a_1b_1-a_3b_3 & a_3b_2\\ a_1b_3 & a_2b_3 & -a_1b_1-a_2b_2 \trix\\ \cB\cA &= \tmat -b_2a_2-b_3a_3 & b_2a_1 & b_3a_1\\ b_1a_2 &-b_1a_1-b_3a_3 & b_3a_2\\ b_1a_3 & b_2a_3 & -b_1a_1-b_2a_2 \trix\\ [\cA,\cB]=\cA\cB-\cB\cA&=\tmat 0&a_2b_1-a_1b_2 & a_3b_1-a_1b_3\\ a_1b_2-a_2b_1& 0 & a_3b_2-a_2b_3\\ a_1b_3-a_3b_1&a_2b_3-a_3b_2 & 0 \trix \end{aligned}$

If we call $\cC=[\cA,\cB]$ then our computation shows that $\cC$ is of the form $\cC=\tmat0&-c_3&c_2\\c_3&0&-c_1\\-c_2&c_1&0\trix$ where

$\tmat c_1\\c_2\\c_3\trix =\tmat a_2b_3-a_3b_2 \\ a_3b_1-a_1b_3 \\ a_1b_2-a_2b_1 \trix =\tmat a_1\\ a_2\\ a_3\trix \times \tmat b_1\\b_2\\b_3\trix$

f. Suppose $\cA$ and $\cB$ are two $n\times n$ matrices and suppose they satisfy

$(\cA+\cB)^2=\cA^2+2\cA\cB+\cB^2.$

Show that $\cA\cB=\cB\cA$.

Expand the square:

$\begin{aligned} (\cA+\cB)^2 &=(\cA+\cB)(\cA+\cB) \\ &=\cA(\cA+\cB) + \cB(\cA+\cB)\\ &=\cA^2 + \cA\cB + \cB\cA+\cB^2 \end{aligned}$

So $(\cA+\cB)^2=\cA^2+2\cA\cB+\cB^2$ only holds if $\cA\cB=\cB\cA$.