## 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 .

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