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Given this matrix:

$\\\ A=\left( \begin{array}{cccccccc} 3 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\\ 1 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\\ 0 & 3 & 0 & 0 & 0 & 1 & 0 & 0 \\\ 0 & 1 & 0 & 0 & 0 & -1 & 0 & 0 \\\ 0 & 0 & 3 & 0 & 0 & 0 & 1 & 0 \\\ 0 & 0 & 1 & 0 & 0 & 0 & -1 & 0 \\\ 0 & 0 & 0 & 3 & 0 & 0 & 0 & 1 \\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -1 \\\ \end{array} \right) $

Is there any way to find the eigenvalues and/or the charactaristic polynomial without computing $\det (A-\lambda I)$? It's clear that $\det (A) = 2^8$ and $Tr(A)=2$. This indicates that the characteristic polynomial is:

$x^8 -2x^7 + a_{6}x^6 + \ldots + a_1x -2^8$

and that $\sum \lambda _i =2$ and $\prod \lambda_i=2^8$, but that's not enough to actually determine the $\lambda _i$'s is it?

Anyone have any thoughts? Thanks!

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    $\begingroup$ You could calculate $A^2,A^3,\dots,A^8$ and then find a linear dependence relation among $I,A,\dots,A^8$. $\endgroup$
    – Gerry Myerson
    Commented Feb 15, 2013 at 22:04
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    $\begingroup$ Or you could compute the traces of $A$, $A^2$, ..., $A^8$ to obtain the power sums of the eigenvalues, and then use the Newton formula to express the elementary symmetric functions of the eigenvalues (which up to signs are the coefficients of the characteristic polynomial) in terms of the power sums. $\endgroup$
    – Peter Mueller
    Commented Feb 15, 2013 at 22:11
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    $\begingroup$ You could spot that $A^8 - 2A^7 - 4A^6 + 8A^5 - 16A^4 - 32A^3 - 64A^2 + 128A + 256=0$ and then check that $1,A,A^2,\ldots,A^7$ are linearly independent, and hence that this must be the char poly. $\endgroup$
    – user30035
    Commented Feb 15, 2013 at 23:00
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    $\begingroup$ In view of Qiaochu's observation: where did this matrix arise? $\endgroup$
    – Yemon Choi
    Commented Feb 16, 2013 at 0:53
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    $\begingroup$ Bemao -- I just typed the matrix into my computer and asked it what the char poly was. Whatever did you think I did?? $\endgroup$
    – user30035
    Commented Feb 20, 2013 at 21:23

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