Question: 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.

Answer: 
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$$A=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}\text{ and that }B=\begin{bmatrix}1&2&3\\3&2&1\\2&0&-2\end{bmatrix}.$$Then$$B^\dagger=\frac1{12}\begin{bmatrix}0&2&2\\1&1&0\\2&0&-2\end{bmatrix}$$ and the eigenvalues of $B^\dagger AB$ are $0$ and $2\pm\frac{\sqrt3}3$, 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 $1$.