# Eigenvalues of a product of matrices, involving Moore-Penrose pseudo inverse

I have recently came across the following question in my linear algebra research, which I suspect has a basic straightforward answer:

Suppose we have a real square matrix $A$, and a certain real square matrix $B$, which is not necessarily invertible. If $B$ is invertible, then the matrices $A$ and $BAB^{-1}$ and $B^{-1}AB$ have the same eigenvalues due to similarity. However, if $B$ is not invertible, then what about the matrices $BAB^{\dagger}$ and $B^{\dagger}AB$ ($\dagger$ denotes Moore-Penrose pseudoinverse) Do they share eigenvalues with $A$? Is there an analogy here?

I really have no ability to answer this question, I am not an expert on the pseudo inverse of matrices, so I am hoping someone here can provide the answer, which I suspect exists. I thank all helpers.

Suppose that $A$ in invertible. Then $0$ is not an eigenvalue of $A$. But if $B$ is the null matrix, then $B^\dagger$ is the null matrix too, and therefore $BAB^\dagger$ is the null matrix, which has a single eigenvalue: $0$.
Here's a less radical case. Suppose that$$A=\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\text{ and that }B=\begin{pmatrix}1&2&3\\3&2&1\\2&0&-2\end{pmatrix}.$$Then$$B^\dagger=\frac1{12}\begin{pmatrix}0&2&2\\1&1&0\\2&0&-2\end{pmatrix}$$ and the eigenvalues of $B^\dagger AB$ are $2\pm\frac{\sqrt3}3$, whereas the eigenvalues of $A$ are, of course, $1$, $2$, and $3$.
If $B=USV^T$ is a singular value decomposition, then $BAB^+$ is similar to $S(V^TAV)S$ and $B^+AB$ is similar to $S(U^TAU)S$. While $V^TAV$ is similar to $U^TAU$ (and both are similar to $A$), the left- and right- multiplications of $S$ introduce basis dependence and hence they break similarity. So, in general, we cannot expect nonzero eigenvalues of $BAB^+,\ B^+AB$ and $A$ to have any relations with each other.
To illustrate, suppose $A=\pmatrix{1&2\\ -4&1}$ and $B=\pmatrix{1&0\\ 1&0}$ so that $B^+=\frac12B^T$. The matrix $A$ is not symmetric. Nor does it have any real eigenvalues. But interestingly, both $B^+AB=0$ and $BAB^+=\frac12\pmatrix{1&1\\ 1&1}$ are symmetric positive semidefinite. Also, $B^+AB$ has a zero spectrum but $BAB^+$ has an eigenvalue $1$.