# Matrix A where AB=BA

Find all $2 \times 2$ matrices $\mathbf{A}$ that have the property that for any $2 \times 2$ matrix $\mathbf{B}$, $$\mathbf{A} \mathbf{B} = \mathbf{B} \mathbf{A}.$$

I'm having difficulty trying to do this problem. Letting $\mathbf{A}$ equal $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $\mathbf{B}$ equal various cases involving $1$s and $0$s like $\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right), I, \left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right), \left(\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right), \left(\begin{smallmatrix}0&0\\1&1\end{smallmatrix}\right), \left(\begin{smallmatrix}1&0\\1&0\end{smallmatrix}\right)$ etc. And I got conditions like $a=d$ and $b=c$, and under other matrix $\mathbf{B}$ $b=0$ and $c+d=a$, etc, but I don't know how I can be sure that I've found all solutions, and actually if my approach of trying only 1's and 0's is right (I think I can express any 2x2 matrix using 0's and 1's, am I right?).

• For typing math on this site, the backslash character you are using is the wrong one. Some weird unicode backslash? Use this  \  backslash instead. Jan 12, 2018 at 17:32
• You have $a=d$, $b=c$ and $b=0$. That already implies $A=\pmatrix{a&0\\0&a}$. Jan 12, 2018 at 17:33
• I forgot to make clear that a=d and b=c are from the same matrix B I tried, and b=0 is from a different matrix. It is now updated to hopefully make that more clear. Jan 12, 2018 at 17:38
• @Helena but, if these conditions should work for all matrices $B$, that would mean that those conditions you got must although found from separate matrices must in fact be true simultaneously for all matrices. You will arrive at what LordShark mentions then and you can prove that this is sufficient. Jan 12, 2018 at 17:43
• As for typesetting, you are using ∖ instead of \ Jan 12, 2018 at 17:44

Suppose that $A$ is some matrix such that $AB=BA$ for all $2\times 2$ matrices $B$.

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

Since $A\left(\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right)=\left(\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right)A$ should be true we learn that $\begin{pmatrix}a&a\\c&c\end{pmatrix}=\begin{pmatrix}a+c&b+d\\0&0\end{pmatrix}$

This implies that $c=0$ and that $b+d=a$

Similarly, since $A\left(\begin{smallmatrix}0&0\\1&1\end{smallmatrix}\right)=\left(\begin{smallmatrix}0&0\\1&1\end{smallmatrix}\right)A$ should be true we learn that $\begin{pmatrix}b&b\\d&d\end{pmatrix} = \begin{pmatrix}0&0\\a+c&b+d\end{pmatrix}$

This implies $b=0$

Combined with the earlier observations, we learn that $a=d$ and that $b=c=0$, and this in turn implies that is $A$ must be a diagonal matrix with identical entries. In other words, $A$ must be a scalar multiple of the identity matrix.

Indeed, every scalar multiple of an identity matrix satisfies the desired property so there cannot be any additional restrictions for our matrix $A$. Remember that scalar multiplication of matrices can be factored outside, that is to say $BA=B(aI)=a(BI)=aB=(aI)B=AB$