Eigendecomposition with rectangular matrices

Question

I have a matrix $M = ABC$ where:

1. $A$ is an $n \times m$ matrix. ($n$ rows and $m$ columns)
2. $B$ is an $m \times m$ diagonal matrix, and all its diagonal elements are positive.
3. $C$ is an $m \times n$ matrix.
4. $AC = I$ ($A$ is the left inverse of $C$)
5. $n \leq m$
6. $M$ is a symmetric $n \times n$ matrix.
7. All matrices are real matrices.

Can I say that $M$ is positive definite?

In the case where $n = m$, $ABC$ is the eigen decomposition of $M$. Because all diagonal elements of $B$ are positive, the eigen values are positive and $M$ is positive definite. But does this generalize to the case where $n \neq m$?

Edit:
I just found out that left inverses aren't actually unique. I would like to clarify that there only needs to exist a positive definite $M$ for some left inverse $A$, given $B$ and $C$. Meaning it is fine for example to only give a proof for the case where $A$ and $C$ are Moore-Penrose inverses of each other (and $AC=I$).

I also did some numerical tests, which seem to suggest my hypothesis is correct. Moreover, it seems like it isn't even necessary for $M$ to be symmetric.

Answer 1

$M$ is always positive semi-definite, meaning that $M$ is positive definite if and only if it's invertible. Also, $M$ being symmetric is indeed an unnecessary restriction.

Let $V_n$ be a vector space of $m$-dimensional vectors whose last $m-n$ components are all zeros (thus any vector in $V_n$ has at most $n$ non-zero components). Let $P$ be an invertible positive definite $m \times m$ matrix such that: for any vector $v$ in the image of $C$, $Pv \in V_n$. This is possible, because $C$ is of rank $n$. Defining $P$ this way means that the last $m-n$ rows of $PC$ are all zeros.

Let's now rewrite $M$: $$M = ABC\ \ \ \parallel ACA = A$$ $$\Rightarrow M = ACABC$$ $$\Rightarrow M = (AP^{-1})(PCABP^{-1})(PC)$$

Let's give names for these parenthesized expressions: $$\hat A = AP^{-1}$$ $$\hat B = PCABP^{-1}$$ $$\hat C = PC$$

$\hat B$ and $\hat C$ have a lot of zeros due to $PC$, meaning we get the following block matrices:

$$\hat A = \left[\begin{array}{}\hat A_m&\hat A_r\end{array}\right]$$ $$\hat B = \left[\begin{array}{}\hat B_m&\hat B_r\\0&0\end{array}\right]$$ $$\hat C = \left[\begin{array}{}\hat C_m\\0\end{array}\right]$$

Here $\hat A_m$, $\hat B_m$ and $\hat C_m$ are all $n \times n$ matrices. Zeros stand for blocks that are all zeros. $M$ is of course a product of these:

$$M= \left[\begin{array}{}\hat A_m&\hat A_r\end{array}\right] \left[\begin{array}{}\hat B_m&\hat B_r\\0&0\end{array}\right] \left[\begin{array}{}\hat C_m\\0\end{array}\right]$$ $$\Rightarrow M = \hat A_m \hat B_m \hat C_m$$

Some properties of these matrices:

$$\hat A \hat C = AP^{-1}PC = I$$ $$\hat A \hat C = \left[\begin{array}{}\hat A_m&\hat A_r\end{array}\right] \left[\begin{array}{}\hat C_m\\0\end{array}\right] = \hat A_m \hat C_m$$ $$\Rightarrow \hat A_m \hat C_m = I$$ $$\Rightarrow \hat A_m = \hat C_m^{-1}$$ $$$$

$$CA=CA \qquad \parallel AC=I$$ $$\Rightarrow C(AC)A = CA$$ $$\Rightarrow (CA)^2 = CA$$ $$\Rightarrow CA \text{ is idempotent}$$ $$\Rightarrow CA \text{ is diagonalizable, and all eigen values are 0 or 1}$$ $$\Rightarrow CA \text{ is positive semi-definite}$$ $$$$

$\hat B = PCABP^{-1}$, as defined. $P$, $CA$, $B$ and $P^{-1}$ are all positive semi-definite, meaning that $\hat B$ is also positive semi-definite. Thus, for every vector $v$ in $V_n$, $\ v^T \hat B v \geq 0 \Rightarrow v^T \left[\begin{array}{}\hat B_m&0\\0&0\end{array}\right] v \geq 0$. This means that $\hat B_m$ is positive semi-definite. $\hat B_m$ can thus be decomposed as $Q^{-1}DQ$, where $D$ is diagonal with no negative elements. $$$$

$M$ can now be written as: $$\hat C_m^{-1} Q^{-1}DQ \hat C_m = (Q \hat C_m)^{-1} D (Q \hat C_m)$$

This is an ordinary eigen decomposition, where $Q \hat C$ represents the eigen vectors and $D$ represents the eigen values. None of the eigen values are negative, so $M$ must be positive semi-definite.