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I have a code which is spitting out matrices of the form

$$\left(\matrix{0&a&-a\\a&+\gamma&0\\-a&0&-\gamma}\right)$$

It has trace $0$ and thus its eigenvalues are of the form $0,\pm\lambda$ so that they sum to $0$.

This pattern seems to hold for larger matrices, where we get eigenvalues of the form $\{\pm\lambda_1,\pm\lambda_2,\dots\}$ with an additional $0$ if it has odd dimensions

For example the $8\times8$ matrix

Or for a size of $8\times8$ we have

$$\left(\matrix{0&0&0&0&0&0&a&0\\0&0&-a&0&0&0&0&0\\0&-a&0&a&0&0&0&0\\0&0&a&0&0&-a&0&0\\0&0&0&0&0&0&0&0\\0&0&0&-a&0&0&0&0\\a&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0}\right)$$

gives eigenvalues $\lambda={\pm a,\pm\frac 1 2 a \left(1-\sqrt 5\right),\pm\frac 1 2 a \left(1+\sqrt 5\right),0,0}$

Or for an alternative size of $8\times8$ we have

$$\left(\matrix{0&a&-a&a&b&-a&0&-b\\a&c&0&b&d&0&0&0\\-a&0&-c&0&e&-b&-e&-d\\a&b&0&c&d&0&e&-e\\b&d&e&d&g&0&0&0\\-a&0&b&0&0&-c&0&-d\\0&0&-e&e&0&0&0&0\\-b&0&-d&-e&0&-d&0&-g}\right)$$

Edit: For simplicity the $8\times 8$ case could be reduced to the form

$$\left(\matrix{c&0&b&d&0&0\\0&-c&0&e&-b&-d\\b&0&c&d&0&-e\\d&e&d&g&0&0\\0&b&0&0&-c&-d\\0&-d&-e&0&-d&-g}\right)$$

and we still get eigenvalues of the form mentioned. For example seeing the variables to iterating integers.

What is the general form of these matrices, and why do we get this repeated $\pm$ pattern?

Is it just that any symmetric matrix with trace $0$ will give eigenvalues of this form? (No, proven by generating a random one and it doesn't hold!)

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    $\begingroup$ Sorry, you are wrong. $0$ is not an eigenvalue unless $\gamma = 0$ or $a^2 = b^2$. $\endgroup$ – Robert Israel Feb 22 at 17:36
  • $\begingroup$ @RobertIsrael Ah sorry you are correct, a=-b in my specific case. Is there some name for the matrices following this pattern. I'll update the question with a 5x5 as well $\endgroup$ – James Feb 22 at 18:55
  • $\begingroup$ @RobertIsrael I was unable to provide a 5x5 example but I have an 8x8 $\endgroup$ – James Feb 22 at 19:13
  • $\begingroup$ @James for your $8 \times 8$, every eigenvalue should be a multiple of $a$, so I think you went wrong somewhere $\endgroup$ – Omnomnomnom Feb 22 at 20:02
  • $\begingroup$ If we delete the $0$ rows and columns of your $8 \times 8$ matrix, we get $$ \pmatrix{0&0&0&0&0&a\\ 0&0&-a&0&0&0\\ 0&-a&0&a&0&0\\ 0&0&a&0&-a&0\\ 0&0&0&-a&0&0\\ a&0&0&0&0&0\\} $$ which will have the same non-zero eigenvalues $\endgroup$ – Omnomnomnom Feb 22 at 20:06
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The condition for a matrix $A$ to have $-\lambda$ as an eigenvalue whenever $\lambda$ is an eienvalue (and with the same multiplicity) is that its characteristic polynomial should be an even or odd function, i.e. either all the nonzero coefficients are odd or all the nonzero coefficients are even.

A more "geometric" condition is that $M$ is similar to $-M$. Thus in your first example, if you switch the second and third rows and the second and third columns, i.e. conjugate with $\pmatrix{1 & 0 & 0\cr 0 & 0 & 1\cr 0 & 1 & 0\cr}$, you transform $M$ to $-M$. Your other examples don't seem to work with a permutation matrix, though.

EDIT: Omnomnomnom's $6 \times 6$ example works with conjugation by the antidiagonal matrix $$ \left[ \begin {array}{cccccc} 0&0&0&0&0&1\\ 0&0&0&0 &1&0\\ 0&0&0&-1&0&0\\ 0&0&1&0&0&0 \\ 0&-1&0&0&0&0\\ -1&0&0&0&0&0 \end {array} \right] $$

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  • $\begingroup$ Why does the first argument imply that M is similar to -M? $\endgroup$ – James Feb 22 at 22:09
  • $\begingroup$ Two normal matrices are similar if they have the same characteristic polynomial. $\endgroup$ – Robert Israel Feb 22 at 22:14
  • $\begingroup$ Sorry, maybe I should have asked: does M being similar to -M imply that it will have $\pm$ eigenvalues? And why? $\endgroup$ – James Feb 22 at 22:17
  • $\begingroup$ Otherwise is there some property that causes the all odd or all even characteristic polynomial? $\endgroup$ – James Feb 22 at 22:32
  • $\begingroup$ oh.. because the eigenvalues of M are the negative of the eigenvalues of -M. So if M is similar to -M they must have a set of eigenvalues equal to the negative of the same set? $\endgroup$ – James Feb 23 at 11:29

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