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?
 A: 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?
A: Consider the quadratic form $u^\dagger Au$. Partition $u$ into two subvectors $x$ and $y$ of equal lengths, i.e. $u^T=(x^T,y^T)$. Then $u^\dagger Au = \|M^\dagger x\|^2 + \|My\|^2 + 2\operatorname{Re}(x^\dagger Fy)$. Therefore $A$ is positive seimidefinite only if $\ker(M)\subseteq\ker(F)$ and $\ker(M^\dagger)\subseteq\ker(F^\dagger)$. In particular,


*

*$A$ is singular and hence it can never be definite, although it can be semidefinite;

*if $F=0$, $A$ is positive semidefinite;

*if $M\ne0$, $A$ has a positive eigenvalue;

*if $M$ is square and $F$ is invertible, $A$ has a negative eigenvalue.


In general, $A$ is sometimes positive semidefinite and sometimes not. In fact, by pulling out the unitary matrices from the singular value decomposition of $M$, we may transform the problem into one such that $M$ is a nonnegative digaonal matrix. In this case, $A$ is positive semidefinite if and only if the submatrix of $A$ corresponding to the indices of the nonzero diagonal entries of $M$ is positive semidefinite.
In other words, the eigenvalue problem for $A$ is not very different from an ordinary eigenproblem for a Hermitian matrix. I don't think there is any good upper bound to the smallest-sized nonzero eigenvalue of $A$, but anyway, you may apply Gershgorin disc theorem or the like to obtain some (very rough) bounds.
