I know that the inverse of a nonsingular hermitian matrix is also hermitian (symmetric if matrix is real). I also know that a singular hermitian matrix has a hermitian generalized inverse.
But are ALL generalized inverses of a hermitian matrix also hermitian? I could not find an answer anywhere, and the following proof suggests so. Please help me find any mistakes if it is not so.
Definition 1: A matrix $A^- \in \mathbb{C}^{n \times m}$ is a generalized inverse of a matrix $A \in \mathbb{C}^{m \times n}$ if $A A^- A = A$.
Lemma 2: Every hermitian matrix $A \in \mathbb{C}^{n \times n}$ can be decomposed into a product of a matrix and its conjugate transpose: $A = X^\dagger X$, where $X \in \mathbb{C}^{r \times n}$ (proof).
Now, I will take this $X$ and construct a perpendicular (orthogonal) projection matrix onto its column space $C(X)$.
Definition 2: $M \in \mathbb{C}^{n \times n}$ is a perpendicular projection matrix onto $X$ if and only if:
- $v \in C(X) \Rightarrow Mv = v$
- $w \perp C(X) \Rightarrow Mw = 0$
Lemma 3: $M$ is a perpendicular projection matrix onto $C(M) = C(X)$ if and only if:
- $MM = M$
- $M^\dagger = M$
The proof for lemma 3 is taken from Plane Answers to Complex Questions Proposition B.32 and Theorem B.33 and slightly modified to account for complex matrices.
Lemma 4: Perpendicular projection matrices are unique (Proposition B.34 from Plane Answers to Complex Questions).
Lemma 5: If $G$ is a generalized inverse for $X^\dagger X$, then $X G X^\dagger X = X$ (Proposition B.43 from Plane Answers to Complex Questions).
Lemma 6: The (unique) perpendicular projection matrix onto $C(X)$ is $X (X^\dagger X)^- X^\dagger$.
Proof: Show that this matrix holds properties in definition 2:
For $v \in C(X)$, $v = Xb$ for some $b \in \mathbb{C}^n$. So, using lemma 5: $$X (X^\dagger X)^- X^\dagger v = X (X^\dagger X)^- X^\dagger Xb = Xb = v$$
For $w \perp C(X)$, $X^\dagger w = 0$. So: $$X (X^\dagger X)^- X^\dagger w = 0$$
Since lemma 4 says that perpendicular projection operators are unique, this is the one and only perpendicular projection operator onto $C(X)$.
Since lemma 3 tells us that a perpendicular projection matrix is hermitian, the matrix from lemma 6 must be hermitian:
$$X (X^\dagger X)^- X^\dagger = (X (X^\dagger X)^- X^\dagger )^\dagger = X ((X^\dagger X)^-)^\dagger X^\dagger $$
Since $X^\dagger X = A$, we have:
$$A^- = (A^-)^\dagger $$
This works for any arbitrary hermitian matrix $A$. Does this prove that any generalized inverse of a non-zero hermitian matrix is hermitian?