## Linear Algebra and Its Applications, exercise 1.4.14

Exercise 1.4.14. Show example 2×2 matrices having the following properties:

1. A matrix A with real entries such that $A^2 = -I$
2. A nonzero matrix B such that $B^2 = 0$
3. Two matrices C and D with nonzero product such that CD = -DC
4. Two matrices E and F with all nonzero entries such that EF = 0

Answer: (a) If $A^2 = -I$ we have

$\begin{bmatrix} a_{11}&a_{12} \\ a_{21}&a_{22} \end{bmatrix} \begin{bmatrix} a_{11}&a_{12} \\ a_{21}&a_{22} \end{bmatrix} = \begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix}$

We then have the following:

$a_{11}^2 + a_{12}a_{21} = -1$
$a_{11}a_{12} + a_{12}a_{22} = 0 \Rightarrow a_{12}(a_{11} + a_{22}) = 0 \Rightarrow a_{11} = -a_{22} \: \rm if \: a_{12} \ne 0$
$a_{21}a_{11} + a_{22}a_{21} = 0 \Rightarrow a_{21}(a_{11} + a_{22}) = 0 \Rightarrow a_{11} = -a_{22} \: \rm if \: a_{21} \ne 0$
$a_{21}a_{12} + a_{22}^2 = -1$

Assume that $a_{11} = -a_{22} = 0$. From the first and fourth equations above we then have

$a_{11}^2 + a_{12}a_{21} = a_{12}a_{21} = -1$
$a_{21}a_{12} + a_{22}^2 = a_{21}a_{12} = -1$

which reduces to the single equation $a_{12}a_{21} = -1$. Assume that $a_{12} = a$ for some real nonzero a. We then have $a_{21} = -1/a$.

So a matrix A meeting the above criterion is

$A = \begin{bmatrix} 0&a \\ -1/a&0 \end{bmatrix}$

where a is nonzero, for which

$A^2 = \begin{bmatrix} 0&a \\ -1/a&0 \end{bmatrix} \begin{bmatrix} 0&a \\ -1/a&0 \end{bmatrix} = \begin{bmatrix} a(-1/a)&0 \\ 0&(-1/a)a \end{bmatrix} = \begin{bmatrix} -1&0 \\ 0&-1 \end{bmatrix} = -I$

We can obtain a specific example of A by setting a = 1, in which case

$A = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix}$

(b) If $B^2 = 0$ we have

$\begin{bmatrix} b_{11}&b_{12} \\ b_{21}&b_{22} \end{bmatrix} \begin{bmatrix} b_{11}&b_{12} \\ b_{21}&b_{22} \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$

We then have the following:

$b_{11}^2 + b_{12}b_{21} = 0 \Rightarrow b_{12}b_{21} = -b_{11}^2$
$b_{11}b_{12} + b_{12}b_{22} = 0 \Rightarrow b_{12}(b_{11} + b_{22}) = 0 \Rightarrow b_{11} = -b_{22} \: \rm if \: b_{12} \ne 0$
$b_{21}b_{11} + b_{22}b_{21} = 0 \Rightarrow b_{21}(b_{11} + b_{22}) = 0 \Rightarrow b_{11} = -b_{22} \: \rm if \: b_{21} \ne 0$
$b_{21}b_{12} + b_{22}^2 = 0 \Rightarrow b_{12}b_{21} = -b_{22}^2 \Rightarrow b_{22}^2 = b_{11}^2$

Assume that $b_{12}$ and $b_{21}$ are nonzero, and choose $b_{11} = b$ where b is a nonzero real number. Then from the second and third equations we have $b_{22} = -b_{11} = -b$. From the first and fourth equations we should have $b_{11}^2 = b_{22}^2$ and this is indeed the case, since $b^2 = (-b)^2$.

Substituting into the first equation we then have

$b_{11}^2 + b_{12}b_{21} = 0 \Rightarrow b_{12}b_{21} = -b_{11}^2 \Rightarrow b_{12}b_{21} = -b^2$

(We could have used the fourth equation just as well for this.)

If we choose $b_{12} = -b$ we then have $b_{21} = b$. This gives us the following matrix

$B = \begin{bmatrix} b&-b \\ b&-b \end{bmatrix}$

for which

$B^2 = \begin{bmatrix} b&-b \\ b&-b \end{bmatrix} \begin{bmatrix} b&-b \\ b&-b \end{bmatrix} = \begin{bmatrix} b^2 -b^2&-b^2 + b^2 \\ b^2 - b^2&-b^2 + b^2 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix}$

If we set b = 1 then we obtain the specific example

$B = \begin{bmatrix} 1&-1 \\ 1&-1 \end{bmatrix}$

(c) If $CD = -DC$ we have

$\begin{bmatrix} c_{11}&c_{12} \\ c_{21}&c_{22} \end{bmatrix} \begin{bmatrix} d_{11}&d_{12} \\ d_{21}&d_{22} \end{bmatrix} = - \begin{bmatrix} d_{11}&d_{12} \\ d_{21}&d_{22} \end{bmatrix} \begin{bmatrix} c_{11}&c_{12} \\ c_{21}&c_{22} \end{bmatrix}$

By the rules of matrix multiplication we then have

$c_{11}d_{11} + c_{12}d_{21} = -(d_{11}c_{11} + d_{12}c_{21})$
$c_{11}d_{12} + c_{12}d_{22} = -(d_{11}c_{12} + d_{12}c_{22} )$
$c_{21}d_{11} + c_{22}d_{21} = -(d_{21}c_{11} + d_{22}c_{21})$
$c_{21}d_{12} + c_{22}d_{22} = -(d_{21}c_{12} + d_{22}c_{22})$

Taking the second and third equations above and rearranging terms we have

$c_{12}d_{22} + c_{12}d_{11} = -d_{12}c_{11} - d_{12}c_{22} \Rightarrow c_{12}(d_{11} + d_{22}) = -d_{12}(c_{11} + c_{22})$
$c_{21}d_{11} + c_{21}d_{22} = -d_{21}c_{11} - d_{21}c_{22} \Rightarrow c_{21}(d_{11} + d_{22}) = -d_{21}(c_{11} + c_{22})$

The easiest way to satisfy the resulting equations is to set

$c_{11} = c_{22} = d_{11} = d_{22} = 0 \Rightarrow c_{11} + c_{22} = d_{11} + d_{22} = 0$

We then have

$\begin{bmatrix} 0&c_{12} \\ c_{21}&0 \end{bmatrix} \begin{bmatrix} 0&d_{12} \\ d_{21}&0 \end{bmatrix} = - \begin{bmatrix} 0&d_{12} \\ d_{21}&0 \end{bmatrix} \begin{bmatrix} 0&c_{12} \\ c_{21}&0 \end{bmatrix}$

which gives us the following equations:

$c_{12}d_{21} = -d_{12}c_{21}$
$c_{21}d_{12} = -d_{21}c_{12}$

which reduce to the single equation $c_{12}d_{21} = -c_{21}d_{12}$. If we set $c_{12} = c_{21} = c$ where $c \ne 0$ then we have $d_{21} = -d_{12}$. If we set $d_{12} = d$ where $d \ne 0$ then $d_{21} = -d$. We then have the following matrices C and D:

$C = \begin{bmatrix} 0&c \\ c&0 \end{bmatrix} \quad D = \begin{bmatrix} 0&d \\ -d&0 \end{bmatrix}$

with

$CD = \begin{bmatrix} 0&c \\ c&0 \end{bmatrix} \begin{bmatrix} 0&d \\ -d&0 \end{bmatrix} = \begin{bmatrix} -cd&0 \\ 0&cd \end{bmatrix} \ne 0 \: \rm if \: c \ne 0, d \ne 0$

and

$DC = \begin{bmatrix} 0&d \\ -d&0 \end{bmatrix} \begin{bmatrix} 0&c \\ c&0 \end{bmatrix} = \begin{bmatrix} cd&0 \\ 0&-cd \end{bmatrix} = - \begin{bmatrix} -cd&0 \\ 0&cd \end{bmatrix} = -CD$

One example of C and D can be found by setting $c = d = 1$:

$C = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} \quad D = \begin{bmatrix} 0&1 \\ -1&0 \end{bmatrix}$

(d) Using the result of (b) above, if we set

$E = F = \begin{bmatrix} 1&-1 \\ 1&-1 \end{bmatrix}$

then we will have EF = 0 with both E and F having all nonzero entries.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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