If matrix $A$ has a left inverse $C$, is $AC$ orthogonally diagonalizable with all eigenvalues $0$ or $1$?

Question

Let $A,C$ be matrices such that $CA = I$. Then $AC$ is defined. I conjecture the following:

1. $AC$ is orthogonally diagonalizable.
2. Its eigenvalues are either $1$ or $0$.
3. It is Hermitian

Informally, if $A$ has a left inverse $C$, then switching their order can either preserve information as is (eigenvalue $1$) or lose it (eigenvalue $0$), but can't do anything else with it (transform it otherwise).

Is the above correct? Is this part of a theory?

Answer 1

Only 2 is correct, because $AC$ is idempotent: $(AC)^2=ACAC=A(CA)C=AC$. For a counterexample to statements 1 and 3, consider $C=\pmatrix{1&1}$ and $A=\pmatrix{1\\ 0}$.

Answer 2

Here is the theory.

Theorem 1:

1. A linear transformation $R : V \to W$ has a left inverse $L : W \to V$ (so that $LR = I$) iff it is injective. $L$ is surjective.
2. $P = RL : W \to W$ is an idempotent ($P^2 = P$), so is diagonalizable with eigenvalues $0$ and $1$.
3. The $1$-eigenspace of $P$ is $$\text{im}(P) = \{ v \in V : Pv = v \} = \text{im}(R)$$ and the $0$-eigenspace is $$\text{ker}(P) = \{ v \in V : Pv = 0 \} = \ker(L)$$ so $P$ breaks $W$ up into a direct sum $$W = \text{im}(P) \oplus \ker(P) = \text{im}(R) \oplus \ker(L).$$
4. The assignment $L \mapsto \ker(L)$ is a bijection between left inverses of $R$ and complements of $\text{im}(R).$

Theorem 2: If in addition $V$ and $W$ are Hilbert spaces (and $L, R$ are bounded), then $\text{im}(R)$ is closed, and the following conditions are equivalent:

1. $P = RL$ is an orthogonal projection.
2. $P$ is Hermitian / self-adjoint.
3. $\ker(L) = \text{im}(R)^{\perp}$ is the orthogonal complement of $\text{im}(R)$.

It follows that a left inverse $L$ satisfying any of these conditions is unique if it exists, and is the Moore-Penrose inverse of $R$.

None of this is particularly difficult to prove, it's mainly a matter of knowing what statements to even look for, so I'll leave these as exercises. There is a dual characterization of right inverses which I'll also leave as an exercise.

Combining these two theorems, 1 and 3 are equivalent and you get counterexamples to them by taking non-orthogonal complements to $\text{im}(R)$.

For a very general discussion of what left inverses look like the keywords are section, retraction, split monomorphism, split epimorphism. See also the section of this blog post titled "the facts of life about idempotents and retracts."