Given $A^2+B^2=\left(\begin{smallmatrix}1402&2022\\2022&1402\end{smallmatrix}\right)$ for which $A,B\in M_2(\mathbb{R})$, show that $AB\neq BA$

Question

$M_2(\mathbb{R})$ is the set of all $2\times2$ matrices that their entries are in $\mathbb{R}$. Now consider $A,B\in M_2(\mathbb{R})$. We have $$A^2+B^2= \begin{bmatrix}1402&&2022\\ 2022 && 1402\\ \end{bmatrix} $$

Show that $AB\neq BA$.

I tried writing $(A+B)^2=A^2+AB+BA+B^2$ and we know if we assume $AB=BA$, then $(A+B)^2=A^2+2AB+B^2$. Then I tried to find the form of a squared matrix and compare these to reach a contradiction but I couldn't. I even don't know if it helps or not. Any help is appreciated!

Answer 1

Hints.

1. $A^2+B^2$ is a real symmetric $2\times2$ matrix with a positive trace and a negative determinant. Therefore, by a similarity transform, it can be diagonalised to some matrix $D=\operatorname{diag}(p,-q)$ with $p,q>0$. Now, let $A$ and $B$ be transformed to $X$ and $Y$ accordingly, so that $X^2+Y^2=D^2$. If $AB=BA$, then $XY=YX$. Hence both $X$ and $Y$ commutes with $D$. However, given that $D=\operatorname{diag}(p,-q)$, can you determine the set of all matrices that it commutes with?
2. Alternatively, note that $\det(A^2+B^2)=\det(A+iB)\det(A-iB)=|\det(A+iB)|^2$ when $AB=BA$.

Answer 2

Slightly different solution, in my opinion more elementary. Suppose that $AB=BA$. Then both of matrices commute with the matrix in the right-hand side. It follows that matrices $A$ and $B$ have the form: $$ A= \left( \begin{array}{cc} x & y \\ y & x \\ \end{array} \right),\ B= \left( \begin{array}{cc} u & v \\ v & u \\ \end{array} \right). $$ Then $$ x^2+y^2+u^2+v^2=1402 $$ and $$ 2xy+2uv=2022. $$ Subtracting the second equality from the first, we get $$ (x-y)^2+(u-v)^2=-620. $$ The obtained contradiction gives the required one.