## Linear Algebra and Its Applications, exercise 1.4.17

Exercise 1.4.17. Assume A is a 2×2 matrix

$A = \begin{bmatrix} a&b \\ c&d \end{bmatrix}$

and further assume that AB = BA for any 2×2 matrix B, including the matrices

$B_1 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \quad B_2 = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$

Show that a = d and that b and c are zero.

$AB_1 = \begin{bmatrix} a&b \\ c&d \end{bmatrix} \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} = \begin{bmatrix} a&0 \\ c&0 \end{bmatrix}$

and

$B_1A = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \begin{bmatrix} a&b \\ c&d \end{bmatrix} = \begin{bmatrix} a&b \\ 0&0 \end{bmatrix}$

Since AB = BA for all B we have

$AB_1 = B_1A \Rightarrow \begin{bmatrix} a&0 \\ c&0 \end{bmatrix} = \begin{bmatrix} a&b \\ 0&0 \end{bmatrix} \Rightarrow b = c = 0$

Similarly we have

$AB_2 = \begin{bmatrix} a&b \\ c&d \end{bmatrix} \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} = \begin{bmatrix} 0&a \\ 0&c \end{bmatrix}$

and

$B_2A = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} \begin{bmatrix} a&b \\ c&d \end{bmatrix} = \begin{bmatrix} c&d \\ 0&0 \end{bmatrix}$

Since AB = BA for all B we have

$AB_2 = B_2A \Rightarrow \begin{bmatrix} 0&a \\ 0&c \end{bmatrix} = \begin{bmatrix} c&d \\ 0&0 \end{bmatrix} \Rightarrow a = d$

The result is that we have

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

for some a (which could be zero); in other words, A is a multiple of the identity matrix I.

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.

This entry was posted in linear algebra. Bookmark the permalink.