## 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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