Given that $A+B$ is invertible, prove or disprove $A(A+B)B = B(A+B)A$ as well as show that $A(A + B)^{-1}B = B(A + B)^{-1}A$

Question

I have tried coming up with a counterexample for the first one, but it has worked each time so my intuition is that the first statement is true. I've tried using the fact that $\det(A+B)$ does not equal $0$ and I've tried using properties of matrix multiplication to no avail. I can't seem to figure out this problem if anyone has any suggestions that would be greatly appreciated!

Answer 1

A random choice of $2 \times 2$ matrices has a very good chance of giving you a counterexample for the first one. For example, you might try

$$ A = \pmatrix{0 & 1\cr 1 & 1\cr},\ B = \pmatrix{0 & 1\cr 1 & 2} $$

Answer 2

Suppose it were true.

$A(A+B)B = B(A+B)A$

If we add $B(A+B)B$ to both sides (or $A(A+B)A$)

$(A+B)^2B = B(A+B)^2$

$(A+B)^2$ and $B$ commute if they have the same eigenspace, which the generally won't.

So we should be able to find a counter example fairly easily. Let $A = \begin{bmatrix} 1&1\\0&0 \end{bmatrix}, B = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix}$

$A(A+B)B = \begin{bmatrix} 1&1\\0&0 \end{bmatrix}\begin{bmatrix} 1&1\\0&1 \end{bmatrix}\begin{bmatrix} 0&0\\0&1 \end{bmatrix} = \begin{bmatrix} 0&2\\0&0 \end{bmatrix}$

$B(A+B)A = \begin{bmatrix} 0&0\\0&1 \end{bmatrix}\begin{bmatrix} 1&1\\0&1 \end{bmatrix}\begin{bmatrix} 1&1\\0&0 \end{bmatrix} = \begin{bmatrix} 0&0\\0&0 \end{bmatrix}$