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How do I find the eigenvalues, eigenvectors and determinant of the matrix

$$ \begin{matrix} a & b & b & ... & b \\ c & a & 0 & ... & 0 \\ c & 0 & a & ... & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ c & 0 & 0 & ... & a \\ \end{matrix} $$

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I think it has something to do with adding a matrix of form $cI$, which alters the eigenvalues and makes it easier to find the eigenvalues. But I'm not sure how to do this. Would really appreciate any help!

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Let $A$ be your given matrix. Then $A=B+aI$ where $$B=\pmatrix{0 & b & b & \dots & b \\ c & 0 & 0 & \dots & 0 \\ c & 0 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots &\vdots \\ c & 0 & 0 & \dots & 0}$$ If you can find the eigenvalues and eigenvectors of this simpler matrix $B$, then you can easily find the eigenvalues and eigenvectors of $A$. Namely, the eigenvectors of $A$ will be identical to those of $B$, while the eigenvalues of $B$ will be $\lambda_1+a,\dots,\lambda_n+a$ where $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $B$.

To see why this works, suppose that $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, so $Av=\lambda v$. Then $$Bv = (A+aI)v = Av + aIv = \lambda v + av = (\lambda +a)v$$ so $v$ is an eigenvector of $B$ with eigenvalue $\lambda+a$.

From here, to get the eigenvalues and eigenvectors of $B$, you could probably just do it directly (i.e., compute the characteristic polynomial, etc.). Another approach though, which might be easier, is to first apply a similarity transformation: Define $C=SBS^{-1}$, where $$S=\pmatrix{1 & 0 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & 0& \dots & 0 \\ 0 & -1 & 1 & 0 & \dots & 0 \\0 & -1 & 0 & 1&\dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots&\vdots \\ 0 & -1 & 0 & 0 & \dots & 1}$$ The idea is that $S$ performs the necessary row operations to $B$ in order to cancel out all but one of instances of $c$. You should then be able to find the characteristic polynomial of $C$ directly and find its eigenvalues and eigenvectors of $C$. The eigenvalues of $B$ will be the same, and the eigenvectors of $B$ will have the form $S^{-1}v$, as $v$ ranges over the eigenvectors of $C$.

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    $\begingroup$ yes I got this, but I was wondering how do you find the eigenvalues and eigenvectors of matrix B? $\endgroup$ – inggumnator Feb 27 '15 at 21:08
  • $\begingroup$ I've added a bit more to my answer now to explain how this can be done. $\endgroup$ – Brent Kerby Feb 27 '15 at 21:45
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I can offer some help for the determinant (sorry not high enough to just comment). Use Block matrices!

Let: $A = a$ (a $1x1$ matrix)
$B = [b \ b \ b \ ...]$ (your top row less the first $a$)

$C = \begin{bmatrix} c\\ c\\ c\\ .\\ .\\ \end{bmatrix}$

and $D = aI$

Then from a formula in the above link:

$Det\begin{pmatrix} A & B\\ C & D \end{pmatrix} = Det(A)Det(D-CA^{-1}B)$

Hope this helps (with the determinant at least).

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