Does this matrix have negative eigenvalues? Suppose I have the following square block-matrix
$A= \begin{pmatrix} 
M M^\dagger & F \\
F^\dagger & M^\dagger M 
\end{pmatrix}$
where $\det(M M^\dagger)=0$.
1) Does the matrix A have a negative or zero eigenvalue? 
2) Is there an upper bound on the smallest eigenvalue?
3) If F is non-zero is the smallest eigenvalue negative?

I think I can answer this myself. Since the matrix A is hermitian, the smallest eigenvalue is bounded above the smallest diagonal element. Since $M M^\dagger$ and $M M^\dagger$ are hermitian, I can diagonalize them with unitary matrices, $U$ and $V$. Then I can transform $A$ by
$ A^\prime = \begin{pmatrix} U^\dagger & 0 \\ 0 & V^\dagger \end{pmatrix} \cdot A \cdot \begin{pmatrix} U & 0 \\ 0 & V \end{pmatrix}$. 
Then $A^\prime$ has a $0$ diagonal entry, since $M M^\dagger$ has a $0$ eigenvalue. Thus, lightest eigenvalue is bounded above by $0$. 
Now if $F\ne 0$, does the eigenvalue have to be negative? What are the conditions on $F$ that insure that the lightest eigenvalue is strictly negative?
