Eigenvalues of two symmetric $4\times 4$ matrices: why is one negative of the other? Consider the following symmetric matrix:
$$
M_0 = 
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & 4 & 3 \\
2 & 4 & 0 & 1 \\
0 & 3 & 1 & 0
\end{pmatrix}
$$
and a very similar matrix:
$$
M_1 = 
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & -4 & 3 \\
2 & -4 & 0 & 1 \\
0 & 3 & 1 & 0
\end{pmatrix}
$$
To my surprise, the eigenspectrum of $M_0$ and $(-M_1)$ are the same! Why would this be the case?
I also tried playing around with the values a little; for example, if the center block is $\begin{pmatrix}1 & \pm 4 \\ \pm 4 & 1\end{pmatrix}$ instead, then they do not share the same eigenvalues.

Context: I was considering the Hermitian matrix of this form ($M_2$ below) and noted that this has the same property as the matrix $M_0$ from above. Thus, presumably, it has nothing to do with the fact that the middle block is complex.
$$
M_2 = 
\begin{pmatrix}
0 & 1 & 2 & 0 \\
1 & 0 & e^{ix} & 3 \\
2 & e^{-ix} & 0 & 1 \\
0 & 3 & 1 & 0
\end{pmatrix}
$$
ps. I will accept any answer which explains the phenomenon between the real matrices. I think that would give a hint as to why $M_2$ / Hermitian matrices have the same property.
Thanks.
 A: $$-M_1=D^{-1}M_0D$$
where $D=D^{-1}$ is the diagonal matrix with diagonal entries $(-1,1,1,-1)$.
Therefore $M_0$ and $-M_1$ are conjugate, and have the same spectrum. This works
because of the zeroes in the corners of $M_0$. In general,
$$\pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\\
a_{21}&a_{22}&a_{23}&a_{24}\\
a_{31}&a_{32}&a_{33}&a_{34}\\
a_{41}&a_{42}&a_{43}&a_{44}}$$
and
$$-\pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\\
a_{21}&-a_{22}&-a_{23}&a_{24}\\
a_{31}&-a_{32}&-a_{33}&a_{34}\\
-a_{41}&a_{42}&a_{43}&-a_{44}}$$
are conjugate, for precisely the same reason.
A: This is happening because of the somewhat special pattern of zeroes in this matrix. Edit: No it's not. It has everything to do with signature matrices instead, as shown in the other answer.
Let $$M_1 = \begin{bmatrix}0 & a_2 & a_3 & 0\\b_1 & 0 & b_3 & b_4\\c_1 & c_2 & 0 & c_4\\0 & d_2 & d_3 & 0\end{bmatrix}, \quad M_2 = \begin{bmatrix}0 & a_2 & a_3 & 0\\b_1 & 0 & -b_3 & b_4\\c_1 & -c_2 & 0 & c_4\\0 & d_2 & d_3 & 0\end{bmatrix}$$
Let $(\lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = \begin{bmatrix}x_1 & x_2 & x_3 & x_4\end{bmatrix}^T$.
Then we can show that
$\begin{bmatrix}x_1 & -x_2 & -x_3 & x_4\end{bmatrix}^T$
is an eigenvector corresponding to eigenvalue $-\lambda$ for $M_2$.
For,
\begin{align*}
a_2 x_2 + a_3 x_3 = \lambda x_1 & \implies a_2 (-x_2) + a_3(-x_3) = -\lambda x_1\\
b_1 x_1 + b_3 x_3 + b_4 x_4 = \lambda x_2 & \implies b_1 x_1 - b_3(-x_3) + b_4x_4 = (-\lambda)(-x_2).
\end{align*}
And the cases of the third and fourth rows are obviously similar.
A: I'm not sure if what follows is the type of thing you're looking for, but maybe you'll find this useful.
Consider the matrix
$$
M_a = 
\left[\begin{array}{rrrr}
0 & 1 & 2 & 0 \\
1 & 0 & a & 3 \\
2 & a & 0 & 1 \\
0 & 3 & 1 & 0
\end{array}\right]
$$
The characteristic polynomials of $M_a$ and $M_{-a}$ are 
\begin{align*}
\chi_{M_a}(t)
&= t^{4} - \left(a^{2} + 15\right) t^{2} - 10 \, a t + 25 \\
\chi_{M_{-a}}(t)
&= t^{4} - \left(a^{2} + 15\right) t^{2} + 10 \, a t + 25
\end{align*}
Now, note that $\lambda$ is an eigenvalue of $M_a$ if and only if
\begin{align*}
0
&= \chi_{M_a}(t) \\
&= {\lambda}^{4} - {\left(a^{2} + 15\right)} {\lambda}^{2} - 10 \, a {\lambda} + 25\\
&= (-\lambda)^{4} - {\left(a^{2} + 15\right)} (-\lambda)^{2} + 10 \, a (-\lambda) + 25 \\
&= \chi_{M_{-a}}(-\lambda)
\end{align*}
This proves that $M_{a}$ and $M_{-a}$ have eigenvalues related by negation.
Now, suppose that $M$ instead takes the form
$$
M_{a+bi}=\left[\begin{array}{rrrr}
0 & 1 & 2 & 0 \\
1 & 0 & a + i \, b & 3 \\
2 & a - i \, b & 0 & 1 \\
0 & 3 & 1 & 0
\end{array}\right]
$$
In this case, the characteristic polynomials of $M_{a+bi}$ and $M_{-a+bi}$ are
\begin{align*}
\chi_{M_{a+bi}}(t)
&= t^{4} + \left(-a^{2} - b^{2} - 15\right) t^{2} - 10 \, a t + 25 \\
\chi_{M_{-a+bi}}(t)
&= t^{4} + \left(-a^{2} - b^{2} - 15\right) t^{2} + 10 \, a t + 25
\end{align*}
A similiar argument then shows that $M_{a+bi}$ and $M_{-a+bi}$ have eigenvalues related by negation.
