Problem 1484

### Part (a): Prove that any square matrix has at least a pseudo-inverse.

1. Matrix Decomposition:
Let $A \in \mathcal{M}_n(\mathbb{C})$. We can decompose $A$ using Gaussian elimination into the form:
$$A = PDP^{-1}$$
where $D$ is a block diagonal matrix with $I_k$ (the identity matrix of size $k$) and a zero matrix, and $P$ is an invertible matrix.

2. Constructing the Pseudo-Inverse:
Define $B$ as:
$$B = P^{-1}DP^{-1}$$
We need to verify that $B$ satisfies the conditions $A = ABA$ and $B = BAB$.

3. Verification:
- Compute $ABA$:
$$ABA = A(P^{-1}DP^{-1})A = PDP^{-1}(P^{-1}DP^{-1})PDP^{-1} = PD(P^{-1}P)D(P^{-1}P)DP^{-1} = PDP^{-1} = A$$
- Compute $BAB$:
$$BAB = (P^{-1}DP^{-1})A(P^{-1}DP^{-1}) = P^{-1}D(P^{-1}P)DP^{-1} = P^{-1}DDP^{-1} = P^{-1}DP^{-1} = B$$

Thus, $B$ is a pseudo-inverse of $A$.

### Part (b): For which matrix $A$ is the pseudo-inverse unique?

1. Invertible Matrix:
If $A$ is invertible, then the pseudo-inverse $B$ must be the inverse of $A$, i.e., $B = A^{-1}$. This is because:
$$A = ABA \implies A = A(A^{-1})A = A$$
$$B = BAB \implies A^{-1} = A^{-1}A(A^{-1}) = A^{-1}$$
Hence, the pseudo-inverse is unique and is $A^{-1}$.

2. Non-Invertible Matrix:
If $A$ is not invertible, consider the decomposition $A = PDP^{-1}$ where $D$ contains zero blocks. We can construct different pseudo-inverses by choosing different matrices $L$ such that $LD = L$ and $DL = D$. For example:
$$B_1 = P^{-1}DP^{-1}$$
$$B_2 = P^{-1}(2D - I)P^{-1}$$
These are distinct pseudo-inverses, showing that the pseudo-inverse is not unique in this case.

Therefore, the pseudo-inverse is unique if and only if $A$ is invertible.

$\blacksquare$