Let A be a square matrix so that AB = BA for every square matrix B of the same order. Prove that A is a diagonal matrix.

Question

Let A be a square matrix so that AB = BA for every square matrix B of the same order. Prove that A is a diagonal matrix.

I have no clue as to how to even start.

Answer 1

As a baby example, let us do it in the case $2\times2$. Since $A$ is unknown let us denote: $$A=\left( {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right)$$ And $AB=BA$ for every $2\times 2$ matrix $B$ so let us take some particular easy matrices $B$, to see what they implies on $A$. Let us take these two matrices: $$B_1=\left( {\begin{array}{*{20}{c}} 0&1\\ 0&0 \end{array}} \right)~~B_2=\left( {\begin{array}{*{20}{c}} 0&0\\ 1&0 \end{array}} \right)$$ Then $AB_1=B_1A$ implies: $$\left( {\begin{array}{*{20}{c}} 0&a\\ 0&c \end{array}} \right)=\left( {\begin{array}{*{20}{c}} c&d\\ 0&0 \end{array}} \right)$$ So $c=0$ and $a=d$. Now $AB_2=B_2A$ implies: $$\left( {\begin{array}{*{20}{c}} b&0\\ d&0 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 0&0\\ a&b \end{array}} \right)$$ So $b=0$ and $a=d$. With all this conditions our matrix is: $$A=\left( {\begin{array}{*{20}{c}} a&0\\ 0&a \end{array}} \right)$$ which is a diagonal matrix. As homework you have two things to do:

• Check that this $A$ satisfy $AB=BA$ for every $2\times 2$ matrix $B$, which will finish the exercise.

• Generalize the result to $n\times n$ matrices using the same argument.

Both things are quite easy, and it shouldn't take you many time. Hope that helps you to understand the exercise.