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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
$$CBA=[1\ \ -1] \cdot \left[\begin{array}{c}
4\\4\end{array}\right] = [0].$$


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:
$$\left[\begin{array}{cc}0&1\\1&2\end{array}\right]\left[\begin{array}{cc}3&-2\\0&0\end{array}\right]-\left[\begin{array}{cc}3&-2\\0&0\end{array}\right]\left[\begin{array}{cc}0&1\\1&2\end{array}\right] = \left[\begin{array}{cc}2&1\\3&-2\end{array}\right].$$

From http://greggrant.org

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has 2 Comments

  1. Hello!
    I wanna ask about Exercise1.5.8
    Can the conclusion be applied to more generalized situations? Exercise 1.5.8 says for 2by2 matrix C,there exist matrices A and B such that C=AB-BA if and only if traceC=0.
    I guess this conclusion is still right for higher dimension matrix, i.e.for any squared matrix C ,there exist A and B such that C=AB-BA if and only if traceC=0.
    The proof of one direction is easy,if C=AB-BA,then trC=tr(AB)-tr(BA)=0,but I don't have any idea about the other direction,What information can I get from trC=0 to prove the conclusion,can you give me some suggestion?Thanks!
    Looking forward to your reply!

    1. Update: the following is not exactly the one you want. For the desired one, the idea is to show C is similar to a matrix which has zeros on diagonal. I need to use for example either Jordan form or rational form. If you are interested, you may look at this paper On matrices of trace zeros.

      One possible way:
      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.
      You already know that W is a subspace of V. We would like to show that V=W.
      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.

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