What are the Eigenvalues of this matrix?

Question

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.

Answer 1

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

Answer 2

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*}