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I Need to compute the Eigenvalues of the following General Matrix. Let $b\geqslant a$. Consider the $(a+b+1)\times (a+b+1)$-Matrix $C$ with the following entries. $$ c_{1,1}=c_{a+b+1,0}=1 $$ and $$ c_{i,j}=1~\text{ for }1\leqslant i\leqslant a+b, j=i+1 $$ and all other entries are $0$.

Example: a=2, b=3

$$ \begin{pmatrix}1 & 1 & 0 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0 & 0\\0 & 0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 0 & 1\\1 & 0 & 0 & 0 & 0 & 0\end{pmatrix} $$

Sorry that I asked the question so confusing before. Note, that I Need the Eigenvalues, not the determinant as I asked before, sorry.

mathfemi
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  • Your notation is not clear. Do you mean $a_{0,0}=a_{a+b,0}=1$? Also, standard notation begins at $a_{1,1}$, not $a_{0,0}$. – Rory Daulton May 17 '15 at 19:41
  • Yes, i mean $a_{0,0}=a_{a+b,0}=1$. Sorry, I started with $a_{0,0}$ not with $a_{1,1}$. – mathfemi May 17 '15 at 19:43
  • No, there is a 1 in the left low Corner... I'll edit my post to make it more clear- – mathfemi May 17 '15 at 19:46
  • Perhaps you should make your question more clear, and include an explicit example: say for $a=2,\ b=3$. – Rory Daulton May 17 '15 at 19:47
  • @mathfemi Do you want to get $\left\lvert \begin{array}{ccccc} 1&1&0&\cdots&0\ 1&1&1&\ddots&\vdots\ 1&1&1&\ddots&0\ \vdots&\vdots&\vdots&\ddots&1\ 1&1&1&\cdots&1

    \end{array} \right\rvert$ ?

     – Joelafrite May 17 '15 at 19:47

2 Answers2

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If I've well understood, the matrix is as follows: $$A-\lambda I=\begin{bmatrix} 1-\lambda&1&0&0&\dots &0\\ 0&-\lambda&1&0&\dots&0\\0&0&-\lambda&1&\dots&0\\\vdots&\vdots&\vdots&\vdots&&\vdots\\ 0&0&0&0&\dots&1\\ 1&0&0&0&\dots&-\lambda \end{bmatrix}$$ You just have to develop the determinant along the first column: \begin{align*} \det(A-\lambda I)&=(1-\lambda)\begin{bmatrix} -\lambda&1&0&\dots&0\\0&-\lambda&1&\dots&0\\\vdots&\vdots&\vdots&&\vdots\\ 0&0&0&\dots&1\\ 0&0&0&\dots&-\lambda \end{bmatrix}+(-1)^{a+b}\begin{bmatrix} 1&0&0&\dots &0\\ -\lambda&1&0&\dots&0\\0&-\lambda&1&\dots&0\\\vdots\\ 0&0&0&\dots&1 \end{bmatrix}\\[1.5ex]&=(-1)^{a+b}\bigl((1-\lambda)\lambda^n+1\bigr)=(-1)^{a+b+1}\bigl(\lambda^{n+1}-\lambda^n-1\bigr).\end{align*}

Bernard
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For eigenvalues you need to calculate $|A-\lambda I|$. So you get $$ |A-\lambda I|=\begin{vmatrix} 1-\lambda&1&0&\dots&0\\0&-\lambda&1&\dots&0\\\vdots&\vdots&\vdots&&\vdots\\ 0&0&0&\dots&1\\ 1&0&0&\dots&-\lambda \end{vmatrix} = $$

$ = (1-\lambda)*(-\lambda)^{a+b} +(-1)^{a+b} = (-1)^{a+b}*((1-\lambda)\lambda^{a+b}+1) = 0$

so the eigenvalues are the solutions of $(1-\lambda)\lambda^{a+b}+1 = 0$

Avi
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