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## Solution to Linear Algebra Hoffman & Kunze Chapter 1.5

#### Exercise 1.5.1

We have $$AB=\left[\begin{array}{c} 4\\4\end{array}\right],$$so
$$ABC= \left[\begin{array}{c} 4\\4\end{array}\right]\cdot [1\ \ -1] =\left[\begin{array}{cc} 4 & -4\\ 4 & -4 \end{array}\right].$$and
$$CBA=[1\ \ -1] \cdot \left[\begin{array}{c} 4\\4\end{array}\right] = .$$

#### Exercise 1.5.2

We have$$A^2= \left[\begin{array}{ccc}1 & -1 & 1\\2 & 0 & 1\\3 & 0 & 1\end{array}\right] \cdot \left[\begin{array}{ccc}1 & -1 & 1\\2 & 0 & 1\\3 & 0 & 1\end{array}\right]$$$$=\left[\begin{array}{ccc}2 & -1 & 1\\5 & -2 & 3\\6 & -3 & 4\end{array}\right].$$And
$$AB= \left[\begin{array}{ccc}1 & -1 & 1\\2 & 0 & 1\\3 & 0 & 1\end{array}\right] \cdot \left[\begin{array}{cc}2 & -2\\1 & 3\\4 & 4\end{array}\right]$$$$=\left[\begin{array}{cc} 5 & -1 \\ 8 & 0\\ 10 & -2 \end{array}\right].$$Thus
$$A^2B= \left[\begin{array}{ccc}2 & -1 & 1\\5 & -2 & 3\\6 & -3 & 4\end{array}\right] \cdot \left[\begin{array}{cc}2 & -2\\1 & 3\\4 & 4\end{array}\right]$$\begin{equation}
=\left[\begin{array}{cc}
7 & -3 \\
20 & -4\\
25 & -5
\end{array}\right].
\label{x1}
\end{equation}And
$$A(AB)= \left[\begin{array}{ccc}1 & -1 & 1\\2 & 0 & 1\\3 & 0 & 1\end{array}\right] \cdot \left[\begin{array}{cc} 5 & -1 \\ 8 & 0\\ 10 & -2 \end{array}\right]$$\begin{equation}
=\left[\begin{array}{cc}
7 & -3\\
20 & -4\\
25 & -5\end{array}\right].
\label{x2}
\end{equation}Comparing (\ref{x1}) and (\ref{x2}) we see both calculations result in the same matrix.

#### Exercise 1.5.3

For example, $$\left[\begin{array}{cc} 0 & 1\\0 & 0\end{array}\right],\quad \left[\begin{array}{cc} 0 & 0\\1 & 0\end{array}\right]$$are two such matrices.

#### Exercise 1.5.4

$$A=\left[\begin{array}{ccc}1 & -1 & 1\\2 & 0 & 1\\3 & 0 & 1\end{array}\right]$$$$E_1A=\left[\begin{array}{ccc}1 & 0 & 0\\-2 & 1 & 0\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1\\2 & 0 & 1\\3 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 1\\0& 2 & -1\\3 & 0 & 1\end{array}\right]$$$$E_2(E_1A)=\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\-3 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1\\0& 2 & -1\\3 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 1\\0& 2 & -1\\0 & 3 & 0\end{array}\right]$$$$E_3(E_2E_1A)=\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1/2 & 0\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1\\0& 2 & -1\\0 & 3 & 0\end{array}\right]=\left[\begin{array}{ccc}1 & -1 & 1\\0& 1 & -1/2\\0 & 3 & 0\end{array}\right]$$$$E_4(E_3E_2E_1A)=\left[\begin{array}{ccc}1 & 1 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & -1 & 1\\0& 1 & -1/2\\0 & 3 & 0\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 1/2\\0& 1 & -1/2\\0 & 3 & 0\end{array}\right]$$$$E_5(E_4E_3E_2E_1A)=\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & -3 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1/2\\0& 1 & -1/2\\0 & 3 & 0\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 1/2\\0& 1 & -1/2\\0 & 0 & 3/2\end{array}\right]$$$$E_6(E_5E_4E_3E_2E_1A)=\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 2/3\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1/2\\0& 1 & -1/2\\0 & 0 & 3/2\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 1/2\\0& 1 & -1/2\\0 & 0 & 1\end{array}\right]$$$$E_7(E_6E_5E_4E_3E_2E_1A)=\left[\begin{array}{ccc}1 & 0 & -1/2\\0 & 1 & 0\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 1/2\\0& 1 & -1/2\\0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0\\0& 1 & -1/2\\0 & 0 & 1\end{array}\right]$$$$E_8(E_7E_6E_5E_4E_3E_2E_1A)=\left[\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 1/2\\0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0\\0& 1 & -1/2\\0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0\\0& 1 & 0\\0 & 0 & 1\end{array}\right].$$

#### Exercise 1.5.5

To find such a $C=\left[\begin{array}{ccc}a & b & c\\d&e&f\end{array}\right]$ we must solve the equation
$$\left[\begin{array}{ccc}a & b & c\\d&e&f\end{array}\right]\left[\begin{array}{cc}1 & -1\\2 & 2\\1 & 0\end{array}\right]=\left[\begin{array}{cc}3 & 1\\-4 & 4\end{array}\right].$$This gives a system of equations
\begin{alignat*}{1}
a+2b+c &=3\\
-a+2b&=1\\
d+2e+f&=-4\\
-d+2e&=4
\end{alignat*}We row-reduce the augmented coefficient matrix
$$\left[\begin{array}{cccccc|c} 1 & 2 & 1 & 0 & 0 & 0 & 3\\ -1 & 2 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 2 & 1 & -4\\ 0 & 0 & 0 & -1 & 2 & 0 & 4 \end{array}\right]$$$$\rightarrow \left[\begin{array}{cccccc|c} 1 & 0 & 1/2 & 0 & 0 & 0 & 1\\ 0& 1 & 1/4 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 1 & 0 & 1/2 & -4\\ 0 & 0 & 0 & 0 & 1 &1/4 & 0 \end{array}\right].$$Setting $c=f=4$ gives the solution
$$C=\left[\begin{array}{ccc}-1 & 0 & 4\\-6 & -1 & 4\end{array}\right].$$Checking:
$$\left[\begin{array}{ccc}-1 & 0 & 4\\-6 & -1 & 4\end{array}\right] \left[\begin{array}{cc}1 & -1\\2 & 2\\1 & 0\end{array}\right]=\left[\begin{array}{cc}3 & 1\\-4 & 4\end{array}\right].$$

#### Exercise 1.5.6

The $ij$-th entry of $AB$ is $\sum_{r=1}^kA_{ir}B_{rj}$. Since the term $B_{rj}$ is independent of $i$, we can view the sum independent of $i$ as $\sum_{r=1}^nB_{rj}\alpha_r$ where $\alpha_r$ is the $r$-th column of $A$. I'm not sure what more to say, this is pretty immediately obvious from the definition of matrix multiplication.

Exercise 1.5.7

Suppose
$$A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right],\quad B=\left[\begin{array}{cc}x & y\\z & w\end{array}\right].$$$$AB=\left[\begin{array}{cc}ax+bz & ay+bw\\cx+dz & cy+dw\end{array}\right].$$Then $AB=I$ implies the following system in $u,r,s,t$ has a solution
\begin{alignat*}{1}
au+bs &= 1\\
cu+ds &= 0\\
ar+bt &= 0\\
cr + dt &= 1
\end{alignat*}because $(x,y,z,w)$ is one such solution. The augmented coefficient matrix of this system is
\begin{equation}
\left[\begin{array}{cccc|c}
a & 0 & b & 0 & 1\\
c & 0 & d & 0 & 0\\
0 & a & 0 & b & 0\\
0 & c & 0 & d & 1
\end{array}\right].
\label{fewf}
\end{equation}As long as $ad-bc\not=0$ this system row-reduces to the following row-reduced echelon form
$$\left[\begin{array}{cccc|c} 1 & 0 &0 & 0 & d/(ad-bc)\\ 0 & 1 & 0 & 0 & -b/(ad-bc)\\ 0 & 0 & 1 & 0 & -c/(ad-bc)\\ 0 & 0 & 0 & 1 & a/(ad-bc) \end{array}\right]$$Thus we see that necessarily $x=d/(ad-bc)$, $y=-b/(ad-bc)$, $z=-c/(ad-bc)$, $w=a/(ad-bc)$. Thus
$$B=\left[\begin{array}{cc} d/(ad-bc) & -b/(ad-bc)\\ -c/(ad-bc) & a/(ad-bc) \end{array}\right].$$Now it's a simple matter to check that
$$\left[\begin{array}{cc} d/(ad-bc) & -b/(ad-bc)\\ -c/(ad-bc) & a/(ad-bc) \end{array}\right] \cdot \left[\begin{array}{cc}a & b\\c & d\end{array}\right] = \left[\begin{array}{cc}1 & 0\\ 0 & 1\end{array}\right].$$

The loose end is that we assumed $ad-bc\not=0$. To tie up this loose end we must show that if $AB=I$ then necessarily $ad-bc\not=0$. Suppose that $ad-bc=0$. We will show there is no solution to (\ref{fewf}), which contradicts the fact that $(x,y,z,w)$ is a solution. If $a=b=c=d=0$ then obviously $AB\not=I$. So suppose WOLOG that $a\not=0$ (because by elementary row operations we can move any of the four elements to be the top left entry). Subtracting $\frac ca$ times the 3rd row from the 4th row of (\ref{fewf}) gives
$$\left[\begin{array}{cccc|c} a & 0 & b & 0 & 1\\ c & 0 & d & 0 & 0\\ 0 & a & 0 & b & 0\\ 0 & c-\frac caa & 0 & d-\frac cab & 1 \end{array}\right].$$Now $c-\frac caa=0$ and since $ad-bc=0$ also $d-\frac cab=0$. Thus we get
$$\left[\begin{array}{cccc|c} a & 0 & b & 0 & 1\\ c & 0 & d & 0 & 0\\ 0 & a & 0 & b & 0\\ 0 & 0 & 0 & 0 & 1 \end{array}\right].$$and it follows that (\ref{fewf}) has no solution.

#### Exercise 1.5.8

We want to know when we can solve for $a,b,c,d,x,y,z,w$ such that
$$\left[\begin{array}{cc}c_{11}&c_{12}\\c_{21}&c_{22}\end{array}\right]=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\left[\begin{array}{cc}x&y\\z&w\end{array}\right]-\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\left[\begin{array}{cc}x&y\\z&w\end{array}\right]$$The right hand side is equal to
$$\left[\begin{array}{cc} bz-cy & ay+bw-bx-dy\\ cx+dz-az-cw & cy-bz\end{array}\right]$$Thus the question is equivalent to asking: when can we choose $a,b,c,d$ so that the following system has a solution for $x,y,z,w$
\begin{equation}
\begin{alignedat}{1}
bz-cy &= c_{11}\\
ay+bw-bx-dy &= c_{12}\\
cx+dz-az-cw &= c_{21}\\
cy-bz &= c_{22}
\end{alignedat}
\label{q9}
\end{equation}The augmented coefficient matrix for this system is
$$\left[\begin{array}{cccc|c} 0 & -c & b & 0 & c_{11}\\ -b & a-d & 0 & b & c_{12}\\ c & 0 & d-a & -c & c_{21}\\ 0 & c & -b & 0 & c_{22} \end{array}\right]$$This matrix is row-equivalent to
$$\left[\begin{array}{cccc|c} 0 & -c & b & 0 & c_{11}\\ -b & a-d & 0 & b & c_{12}\\ c & 0 & d-a & -c & c_{21}\\ 0 & 0 & 0 & 0 & c_{11}+c_{22} \end{array}\right]$$from which we see that necessarily $c_{11}+c_{22}=0$.

Suppose conversely that $c_{11}+c_{22}=0$. We want to show $\exists$ $A,B$ such that $C=AB-BA$.

We first handle the case when $c_{11}=0$. We know $c_{11}+c_{22}=0$ so also $c_{22}=0$. So $C$ is in the form
$$\left[\begin{array}{cc}0&c_{12}\\c_{21}&0\end{array}\right].$$In this case let$$A=\left[\begin{array}{cc}0 & c_{12}\\-c_{21} & 0\end{array}\right],\quad B=\left[\begin{array}{cc}-1 & 0 \\ 0 & 0\end{array}\right].$$Then$$AB-BA$$$$=\left[\begin{array}{cc}0 & c_{12}\\-c_{21} & 0\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 0 & 0\end{array}\right]-\left[\begin{array}{cc}-1 & 0 \\ 0 & 0\end{array}\right]\left[\begin{array}{cc}0 & c_{12}\\-c_{21} & 0\end{array}\right]$$$$=\left[\begin{array}{cc}0 & 0\\c_{21} & 0\end{array}\right]-\left[\begin{array}{cc}0 & -c_{12}\\0 & 0\end{array}\right]=\left[\begin{array}{cc}0 & c_{12}\\c_{21} & 0\end{array}\right]=C.$$So we can assume going forward that $c_{11}\not=0$. We want to show the system (\ref{q9}) can be solved. In other words we have to find $a,b,c,d$ such that the system has a solution in $x,y,z,w$. If we assume $b\not=0$ and $c\not=0$ then this matrix row-reduces to the following row-reduced echelon form
$$\left[\begin{array}{cccc|c} 1 & 0 & (d-a)/c & -1 & \frac{d-a}{bc}c_{11}-\frac{c_{12}}{b}\\ 0 &1 & -d/c & 0 & -c_{11}/c\\ 0 & 0 & 0 & 0 & -\frac{c_{11}}{b}(d-a)+c_{21}+\frac{c_{12}c}{b}\\ 0 & 0 & 0 & 0 & c_{11}+ c_{22} \end{array}\right]$$We see that necessarily
$$-\frac{c_{11}}{b}(d-a)+c_{21}+\frac{c_{12}c}{b}=0.$$Since $c_{11}\not=0$, we can set $a=0$, $b=c=1$ and $d=\frac{c_{12}+c_{21}}{c_{11}}$. Then the system can be solved and we get a solution for any choice of $z$ and $w$. Setting $z=w=0$ we get $x=c_{21}$ and $y=c_{11}$.

Summarizing, if $c_{11}\not=0$ then:
$$a=0,b=1,c=1$$$$d=({c_{12}+c_{21}})/{c_{11}}$$$$x=c_{21},y=c_{11},z=0,w=0$$For example if $C=\left[\begin{array}{cc}2&1\\3&-2\end{array}\right]$ then $A=\left[\begin{array}{cc}0&1\\1&2\end{array}\right]$ and $B=\left[\begin{array}{cc}3&-2\\0&0\end{array}\right]$. Checking:
$$\left[\begin{array}{cc}0&1\\1&2\end{array}\right]\left[\begin{array}{cc}3&-2\\0&0\end{array}\right]-\left[\begin{array}{cc}3&-2\\0&0\end{array}\right]\left[\begin{array}{cc}0&1\\1&2\end{array}\right] = \left[\begin{array}{cc}2&1\\3&-2\end{array}\right].$$

### This Post Has 2 Comments

1. Hello!
1. Let V be the space of matrices (size of $n\times n$) with trace zero and W the subspace spanned by AB-BA for all A and B.
First show that the dimension of V is $n^2-1$. Then show that the dimension of W is also $n^2-1$ by finding $n^2-1$ linearly independent matrices. To find it, please try $A=E_{ij}$ and $B=E_{kl}$. Then you should be able to find all elements $E_{ij}$ for $i\ne j$ in W and also $E_{ii}-E_{i+1,i+1}=E_{i,i+1}E_{i+1,i}-E_{i+1,i}E_{i,i+1}$in W for $i=1,\dots,n-1$. Hence the number of such vectors will be $$n^2-n+(n-1)=n^2-1.$$Now show they are linearly independent.