Linear Algebra: Is det({{M,F},{F, M})<0 when det(M)=0? Suppose that $M$ and $F$ are real matrices. Let $A$ be the block-matrix
$$
A=
\begin{pmatrix}
 M  & F \\
 F  & M
\end{pmatrix}
$$
If $\det(M)=0$ is $\det(A)\leq0$? If not, what conditions need to be satisfied?
Also, 
Does A have a non-positive eigenvalue?
 A: Let 
$$
M=
\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\quad
F
=
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
$$
Then 
$$
\det
\begin{pmatrix}
M & F \\
F & M
\end{pmatrix}=1
$$
A: $$\det\begin{pmatrix}
 M  & F \\
 F  & M
\end{pmatrix}=\det(M-F)\det(M+F),$$
So, if, say $M=0$ and $F$ is any matrix of even order, $\det -F=\det F$, and
$$\det\begin{pmatrix}
 0  & F \\
 F  & 0
\end{pmatrix}=(\det F)^2.$$

Edit. Since you added the question about eigenvalues.
Eigenvalues of $A$ are determined by the characteristic polynomial roots:
$$0=\det(\lambda\mathbb{I}-A)=\det\begin{pmatrix}
 \lambda\mathbb{I}-M  & -F \\
 -F  & \lambda\mathbb{I}-M
\end{pmatrix}=\det(\lambda\mathbb{I}-M+F)\cdot\det(\lambda\mathbb{I}-M-F).$$
Therefore, the eigenvalues are the combination of eigenvalues of $M-F$ and $M+F$. Now, we see that if $M=0$ then, indeed, for every eigenvalue of $F$ we have the opposite eigenvalue of $-F$. However, if $M$ is arbitrary, even if $\det M=0$, there is no reason for either $M-F$ or $M+F$ to have non-positive eigenvalues.
For example, let's take
$$M=\begin{pmatrix}
 a  & a \\
 a  & a
\end{pmatrix},
F=\begin{pmatrix}
 0  & -b \\
 b  & 0
\end{pmatrix}$$
$$(M-F)=(M+F)^T=\begin{pmatrix}
 a  & a+b \\
 a-b  & a
\end{pmatrix}\text{ has real eigenvalues }a\pm\sqrt{a^2-b^2}>0\text{ if }0<b<a.$$
And, indeed,
$$\begin{pmatrix}
 a & a & 0 & -b \\
 a & a & b & 0 \\
 0 & -b & a & a \\
 b & 0 & a & a
\end{pmatrix}\text{ has (double) real eigenvalues }a\pm\sqrt{a^2-b^2}>0\text{ if }0<b<a.$$
A: $A$ can possess a full set of positive eigenvalues (and hence a positive determinant). For instance, according to Wolfram Alpha, all four eigenvalues of the following randomly generated matrix are strictly positive:
$$A=\pmatrix{10&0&-1&-9\\ 0&0&1&0\\ -1&-9&10&0\\ 1&0&0&0}.$$
