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How to compute eigenvalues of big $5×5$ matrix (symmetric matrix) .

$\left(\begin{matrix} 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 1 & 0\\ 0 & 1 & 0 & 1 & 0\\ 0 & 1 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 0\\ \end{matrix}\right) $

By characteristics equation will be lengthy process . If you are going to suggest "eigenvalues by inspection", please explain in steps .

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  • $\begingroup$ It took me two minutes to compute the characteristic equation. $\endgroup$ – Michael Hoppe Nov 13 '15 at 13:29
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By inspection you can see that the span $W_2$ of the first and fifth standard basis vectors is an invariant subspace, as is the span $W_3$ of the remaining three standard basis vectors. You therefore have a decomposition $\Bbb R^5=W_2\oplus W_3$ a direct sum of two invariant subspaces (of dimension $2$ and $3$), and you can find the eigenvalues of the restrictions to those subspaces separately, and then combine them.

For $W_2$ you linear operator acts as a (diagonal) reflection interchanging the two coordinates, so you get eigenvalues $1$ and $-1$. For $W_3$ the action is given by the $3\times 3$ sub-matrix at the center, and since you can see that all its rows have the same sum$~2$, it follows that $(1,1,1)$ is an eigenvector with eigenvalue$~2$. Your operator on $W_3$ (still) being given by a symmetric matrix, the orthogonal complement in $W_3$ of the invariant subspace spanned by $(1,1,1)$ will also be an invariant subspace. If you choose any basis of this orthogonal complement (for instance $(1,-1,0),(0,1,-1)$) and compute your operator on its vectors, you find that the operator acts as multiplication by $-1$ on this orthogonal complement. That gives you an eigenvalue $-1$ with multiplicity$~2$ (for $W_3$; remember it already had multiplicity$~1$ for $W_2$, so it will have multiplicity$~3$ overall).

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Hint: Take $v=(x_1,x_2,x_3,x_4,x_5)$ as an eigenvector and compute your matrix (lets call it $A$) times $v$ and compare with $\lambda v$ for some scalar $\lambda$.

It easy to find $v_1=(1,0,0,0,1),v_2=(1,0,0,0,-1)$.

From here you can find the eigenvalues and vectors of the "central" matrix.

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There is a lot of $0$... you can compute it as usual...

$$\left|\begin{matrix}-t&0&0&0&1\\0&-t&1&1&0\\ 0&1&-t&1&0\\0&1&1&-t&0\\1&0&0&0&-t\end{matrix}\right|= -t\left|\begin{matrix} -t&1&1&0\\ 1&-t&1&0\\1&1&-t&0\\0&0&0&-t \end{matrix}\right|+ \left|\begin{matrix} 0&0&0&1\\-t&1&1&0\\ 1&-t&1&0\\1&1&-t&0 \end{matrix}\right|$$

$$=-t^2\left|\begin{matrix} -t&1&1\\1&-t&1\\1&1&-t \end{matrix}\right|- \left|\begin{matrix} -t&1&1\\1&-t&1\\1&1&-t \end{matrix}\right| =-(t^2+1)\left|\begin{matrix} -t&1&1\\1&-t&1\\1&1&-t \end{matrix}\right|=...$$

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    $\begingroup$ Just a doubt... I did not go through your derivation, but one thing I have noticed that two of its eigenvalues are coming $i$ and $-i$ which is quite surprising (because eigenvalues of a symmetric matrix are always real). $\endgroup$ – Rajat Nov 13 '15 at 9:06
  • $\begingroup$ I think , second matrix (after +) is typo ? $\endgroup$ – Mithlesh Upadhyay Nov 13 '15 at 9:12
  • $\begingroup$ I think I tried to fix the extra negatives (sorry if I took a lot of freedom with your answer, clicked edit to copy matrices for math engine, fixed it while cleaning), still looks a mystery! $\endgroup$ – Jesse P Francis Nov 13 '15 at 9:26
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    $\begingroup$ Instead $-t^2$ it should read $-t(-t)=t^2$. $\endgroup$ – Michael Hoppe Nov 13 '15 at 13:32
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Exploit the persymmetry of your matrix, to reduce your problem into several smaller ones. I think this paper shows how to...

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