Let $L$ and $M$ be matrices over a commutative ring $R$ equipped with an involution "$*$". Define $L \oplus M$ (the "direct sum" of $L$ and $M$) to be $\begin{bmatrix}L & 0 \\ 0 & M \end{bmatrix}$. If $K$ is a matrix over $(R, *)$, define $K^*$ to be the "conjugate-transpose" of $K$, i.e. $(K^*)_{ij} = (K_{ji})^*$. If $L \oplus M$ has a Moore-Penrose pseudoinverse, then do $L$ and $M$ have one?
In attempting to answer the above question, I've defined $X^+$ to mean the pseudoinverse of a matrix $X$. Recall the axioms defining the pseudoinverse $X^+$ of a matrix $X$:
- $XX^+X = X$,
- $X^+XX^+ = X^+$,
- $\left(XX^+\right)^* = XX^+$,
- $\left(X^+X\right)^* = X^+X$.
Write $(L \oplus M)^+$ as a block matrix: $(L \oplus M)^+ = \begin{bmatrix}A & B \\ C & D \end{bmatrix}$. Notice that the aim is to show that $A = L^+$ and $D = M^+$ (or equivalently, that $B$ and $C$ are zero matrices).
Plugging the block matrix representations of $L \oplus M$ and $(L \oplus M)^+$ into the above 4 axioms gives:
$\left[\begin{matrix}L A L & L B M\\M C L & M D M\end{matrix}\right] = \left[\begin{matrix}L & 0\\0 & M\end{matrix}\right]$
$\left[\begin{matrix}A L A + B M C & A L B + B M D\\C L A + D M C & C L B + D M D\end{matrix}\right] = \left[\begin{matrix}A & B\\C & D\end{matrix}\right]$
$\begin{bmatrix}(LA)^* &(MC)^* \\ (LB)^* & (MD)^*\end{bmatrix} = \begin{bmatrix} LA & LB \\ MC & MD\end{bmatrix}$
$\begin{bmatrix} (AL)^* & (CL)^* \\ (BM)^* & (DM)^* \end{bmatrix} = \begin{bmatrix} AL & BM \\ CL & DM\end{bmatrix}$
This shows that $A$ satisfies all of the equations to be the pseudoinverse of $L$, except for axiom 2 ($ALA = A$). Likewise, the same can be said for the relationship between $D$ and $M$ (all the conditions follow except for axiom 2: $DMD = D$). How do I prove that $ALA = A$ and $DMD = D$?
Alternatively, are there any counterexamples to the above claim?
