Group inverse of $\textbf{A}\ast\textbf{B}:=\textbf{B}\textbf{A}+\textbf{A}+\textbf{B}$

Question

This question came up on a recent linear algebra exam of mine, and it's been bothering me ever since. The group is defined such that every element plus the identity matrix is invertible:

$$(G,\ast):=\{\textbf{A}\in G | \textbf{A}+\textbf{E}_{n} \text{ is invertible}\}$$

The neutral element is quite obviously the null matrix, yet finding the inverse seemed to be beyond my best abilities.

I'd greatly appreciate any hints.

Answer 1

Hint.

Let the identity matrix be $I.$ Let the "zero matrix" be $0.$ We have $$A*B=(B+I)(A+I)-I.$$ And $A*0=0*A=A$ so $0$ is the 2-sided identity for the operation $*.$

When $A\in G,$ to find $B$ such that $A*B=0\, \;$ : If $A*B=0$ then $(B+I)(A+I)-I=0$ so $$(B+I)(A+I)=I$$ and $(A+I)$ is invertible with respect to standard matrix multiplication, so .....

Similarly, to find $B$ such that $B*A=0$ ....

Remark. We may verify by direct calculation that $*$ is associative.

Remark. I said $0$ is "the" identity for $*.$ If $A*J=A$ for all $A \in G$ then with $A=0$ we have $0=A=A*J=0*J=J.$ Similarly if $J*A=A$ for all $A\in G$ then $0=J.$

Answer 2

You may first prove that the neutral element $E$ is $0$ by solving the equation $A\ast E=A$. Then you may find $B=A^{-1}$ by solving $A\ast B=E(=0)$.

Alternatively, you may determine $B=A^{-1}$ first by solving $A\ast B\ast A=A$, and then find the neutral element by calculating $A\ast B$.