# 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.

• It's because of all the conveniently placed zeroes. – M. Vinay Apr 7 at 2:48
• @M.Vinay Yes, seems that way. Is there a name for such matrices or any property sticking out to you right now which would explain why this is true for symmetric matrices of this kind? – Troy Apr 7 at 2:50
• In my answer as currently written, I've shown that this holds for a slightly more general case (the matrix doesn't have to be symmetric/Hermitian, and may be real or complex). But I'd like to generalise still further, to higher orders. And also try to find a more big-picture explanation, as you say. – M. Vinay Apr 7 at 3:04
• In case this helps: this would be "hollow" (zeroes at the diagonal) "pentadiagonal" or "band" symmetric matrix. – leonbloy Apr 7 at 3:04
• @leonbloy that certainly narrows down the search for me, thanks for the input! – Troy Apr 7 at 3:06

$$-M_1=D^{-1}M_0D$$ 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.

• Of course, signature matrix. This is the answer. – M. Vinay Apr 7 at 3:13
• okay, this is amazing.. (there's a small typo on the last line of the matrix, I can't edit since it's <6 characters long) – Troy Apr 7 at 3:18

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.

• oh this is promising. let me mull on this a little before I accept. thanks! – Troy Apr 7 at 3:04
• The would imply that the property has no obvious generalization for larger sizes, no? – leonbloy Apr 7 at 3:07
• @leonbloy I think it can be done with careful placement of zeroes, but I don't know if those generalisations would be naturally interesting or too contrived. Probably the latter. – M. Vinay Apr 7 at 3:09

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.

• thanks for the attempt; yes this is a tad too "high-level" for my use-case -- I need a slightly more general/abstracted explanation. +1 nonetheless. – Troy Apr 7 at 3:00
• This does not explain if the property depends on having those non-zero elements. – leonbloy Apr 7 at 3:03
• @leonbloy I mean, if someone wants to edit the question so that it is more rigorously posed, then we can take a stab at it. As it stands, it's unclear what's actually being asked here. – Brian Fitzpatrick Apr 7 at 3:07