Eigenvalues of real diagonal matrix times orthogonal projection In the analysis of the stability of a dynamical system I came across
the following Jacobi matrix $J$ (the eigenvalues of which determine stability):
\begin{equation}
J=KP
\end{equation}
where $K$ is an $m\times m$ real diagonal matrix and 
\begin{equation}
P=A\left(A^{T}A\right)^{-1}A^{T}
\end{equation}
is an orthogonal projection. The non-square matrix $A$ is an $m\times n$ real
matrix, $m\geq n$. 
The eigenvalues of $P$ are 1 with algebraic multiplicity $n$ and
0 with algebraic multiplicity $m-n$ (see this proof exploiting the
idempotence of $P$ and this proof arguging the multiplicity). 
However, when you consider the whole matrix $J=KP$, I am not sure
about its eigenvalues. 

I am particularly interested if anything
  can be said about the sign of the the largest nonzero eigenvalue of $J$, as this is what matters for stability analysis. 

 A: Let $\lambda_{\max}(A)$ be the largest singular value of $A$;  when $A$ is symmetric this is the largest absolute value of the eigenvalues of $A$.
Since $P$ is a projection, $\lambda_{max}(J)\le \lambda_{\max}(K)$.  Equality can be attained:  take $P$ to be the identity, or assume that the range of $P$ contains the largest eigenvector of $K$.
A similar inequality holds for the the maximal eigenvalue $\sigma_{\max}$, provided that $\sigma_{\max}(K)\ge0$.  To see this, notice that any eigenvalue of $J=KP$ is also an eigenvalue of the symmetric matrix $PKP$.  Since $P$ is a projection, $\|Pu\|\le\|u\|$ and therefore
$$
\sigma_{\max}(J)
\le\sigma_{\max}(PKP)
=\sup_{\|u\|\le1}\langle PKPu,u\rangle
=\sup_{\|u\|\le1}\langle KPu,Pu\rangle
\le\sup_{\|v\|\le1}\langle Kv,v\rangle
=\sigma_{\max}(K).
$$
For the eigenvalue statement between $KP$ and $PKP$, notice that if $KPv=\lambda v$, then $P^{1/2}KP^{1/2}(P^{1/2}v)=\lambda (P^{1/2}v)$.  So all the eigenvalues of $KP$ are also eigenvalues of $P^{1/2}KP^{1/2}=PKP$, because $P$ is a projection.  $A^{1/2}$ denotes the matrix square root of a nonnegative definite matrix $A$ (and $P$ is such, as a projection).
You can show that any \emph{nonzero} eigenvalue of $PKP$ is also an eigenvalue of $J$.  It could be that $J$ is not diagonalisable, but this will happen only if there is a nonzero $v$ such that $Pv=v$ and $PKv=0$ yet $Kv\ne0$ ($v$ is the range of $P$ but $Kv$ is in the orthogonal complement of the range of $P$).
We will have equality in the second inequality if and only if the maximising $v$ (or \emph{a} such maximiser if there are many) is in the range of $P$.  How far off we are depends on how close is the range of $P$ to a maximiser, and also on how close are the other eigenvalues of $K$ to the maximal one.
If $\sigma_{\max}(K)<0$, then taking $P=0$ shows that $\sigma_{\max}(J)$ can be larger than $\sigma_{\max}(K)$.  But the same idea shows that $\sigma_{\min}(J)\ge \sigma_{\min}(K)$ if the latter is not positive.
So:  
$K$ negative definite  ---> $\sigma_{\max}(J)\le0$, and it will be zero unless $P$ is the identity.
$K$ not negative definite ---> $0\le\sigma_{\max}(J) \le \sigma_{\max}(K)$.  Note that it doesn't mean that $K$ is nonnegative definite, all it means is that it has at least one nonnegative eigenvalue.
There analogous inequalities for $\sigma_{\min}$ also hold.
A: This is an extended comment on @YZS answer. 
I appreciate that the largest singular values $\lambda_\text{max}$ satisfy $\lambda_{\text{max}}\left(KP\right)\leq\lambda_{\text{max}}\left(K\right)$. Let's say that we call 
\begin{equation}
\sigma_{\text{max}}
\end{equation}
the largest non-zero eigenvalue. From running numerical tests, I strongly suspect that the eigenvalues are bounded in a very similar way to the singular values, perhaps even the same way, $\sigma_{\text{max}}\left(KP\right)\leq\sigma_{\text{max}}\left(K\right)$.
Consider this simulation result, which I found to be representative:

Here $\lambda_\text{max}$ are largest singular values and $\sigma_\text{max}$ largest non-zero eigenvalues. Every dot represents a randomly generated matrix $A$ of dimensions $6\times 10$, the entries of which were integers between $-100$ and $100$. For all simulations, $K$ was always the same diagonal matrix
\begin{equation}
K=\text{diag}(-9,-9,-8,-8,-5,-4,-4,-1,1,3).
\end{equation}
So clearly, 
\begin{equation}
\sigma_{\text{max}}\left(K\right)=3,\quad\lambda_{\text{max}}\left(K\right)=9.
\end{equation}
I have repeated such simulations for various $K$, as well as different dimensions of $A$ (starting with a square and then increasing column size, growing $K$ accordingly). I have found that the stated inequalities for both singular and non-zero eigenvalues were always true in my simulations. 

Of course such tests are not an exhaustive search, but can anyone prove or provide a plausibility argument why $\sigma_{\text{max}}\left(KP\right)\leq\sigma_{\text{max}}\left(K\right)$ might be correct?
Also, given that $KP$ is not symmetric, is it somehow obvious why I never find any imaginary eigenvalues in numerical simulations? 

As an aside, as you change the dimensions of $A$ from a square to increasingly elongated rectangle, it appears that the largest nonzero eigenvalues $\sigma_\text{max}(K P)$ move further and further away from the (supposed) bound $\sigma_\text{max}(K)$. This trend also holds for the max. singular values. In the plot below, $\sigma_\text{max}(K)=-2$ and the dimension of $A$ changes from $6\times 6$ to $6 \times 40$, as the eigenvalues move further and further below the (supposed) bound $\sigma_\text{max}(K)$:

