<s>[INST] 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.
 [/INST] 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.
</s>