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I need to calculate the characteristic polynomial and eigenvalues of the following matrix. It's been a long time since my linear algebra courses, so I have pretty much lost the ability to compute such things by hand. Can someone perhaps help me?

$$\left( \begin{array}{cc} r & p & 0 & & & & & &q\\ q & r & p & 0 \\ 0 & q & r & p & 0 \\ 0 & 0 & q & r & p & 0 \\ ... & & & ... & ... & ... \\ p & 0 & & & & & 0 & q & r \end{array} \right) $$

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@mvw I know the definition. The clever calculation is the part where I need help. – Balerion_the_black May 26 '14 at 22:14
Permutating the last two columuns to be the first two columns and then permutating the last row to be the first row gives a nice tridiagonal-Toeplitz-matrix which has the same determinant (except maybe sign) like the original one. – mvw May 26 '14 at 22:44
This article lists eigenvalues $\lambda_k = r + 2\sqrt{pq}\cos\frac{\pi\,k}{n+1}$. Seems not to look like the answer by @GeorgeShakan. Hm, where do I err? – mvw May 26 '14 at 22:55
That article is computing the eigenvalues for a different matrix :) notice that there is a p in the bottom left and a q in the top right. – George Shakan May 26 '14 at 23:25
up vote 2 down vote accepted

Call your matrix $A$ which is $pA_1 + qA_2 + rI$ where $A_1$ and $A_2$ are defined accordingly. You are lucky that $A_1 , A_2$ and $I$ have the same eigenvectors. The reason for this is $A_1$ and $A_2$ are adjacency matrices for isomorphic directed graphs. Indeed $A_1^{n-1} =A_2$.

Now let's see the eigenvalues of $A_1$. It's characteristic polynomial is $t^n -1$ (where $A$ is a $n \times n$ matrix), so the eigenvalues are the $n^{th}$ roots of unity! Now if $x$ is an eigenvector of $A_1$ with eigenvalue $\lambda$, then $$(pA_1 +qA_2 + rI)x = (pA_1 +qA_1^{n-1} + rI)x = p\lambda x + q \lambda^{n-1} x + rx = (p \lambda + q \lambda^{n-1} + r)x.$$

Let $\xi$ be any primitive $n^{th}$ root of unity. Then the eigenvalues are $$(p \lambda + q \lambda^{n-1} + r),$$ where $\lambda = \xi^i$ for $0 \leq i \leq n-1$.

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I think you missed an $r$ coefficient at some point in the calculation, but otherwise this seems correct. Thanks :) – Balerion_the_black May 26 '14 at 22:52
Yes I had missed several! – George Shakan May 26 '14 at 23:09
The argument makes sense, and I also checked empirically on a few examples just to make sure and the formula works correctly. Thank you a lot! – Balerion_the_black May 27 '14 at 21:50
Glad it helped! – George Shakan May 27 '14 at 23:46

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