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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