Bdtyk

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 $begin{pmatrix}1 & pm 4 \ pm 4 & 1end{pmatrix}$ 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.

$$-M_1=D^{-1}M_0D$$
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
$$-pmatrix{-a_{11}&a_{12}&a_{13}&-a_{14}\
a_{21}&-a_{22}&-a_{23}&a_{24}\
a_{31}&-a_{32}&-a_{33}&a_{34}\
-a_{41}&-_{42}&a_{43}&-a_{44}}$$
are conjugate, for precisely the same reason.

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 $$M_1 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & b_3 & b_4\c_1 & c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}, quad M_2 = begin{bmatrix}0 & a_2 & a_3 & 0\b_1 & 0 & -b_3 & b_4\c_1 & -c_2 & 0 & c_4\0 & d_2 & d_3 & 0end{bmatrix}$$

Let $(lambda, x)$ be an eigenvalue-eigenvector pair of $M_1$, where
$x = begin{bmatrix}x_1 & x_2 & x_3 & x_4end{bmatrix}^T$.
Then we can show that
$begin{bmatrix}x_1 & -x_2 & -x_3 & x_4end{bmatrix}^T$
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.

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} + 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a}}(t)
&= t^{4} - left(a^{2} + 15right) 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} + 15right)} {lambda}^{2} - 10 , a {lambda} + 25\
&= (-lambda)^{4} - {left(a^{2} + 15right)} (-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} - 15right) t^{2} - 10 , a t + 25 \
chi_{M_{-a+bi}}(t)
&= t^{4} + left(-a^{2} - b^{2} - 15right) 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.