# Thread: Eigenvalues & eigenvectors of an n*n matrix

1. ## Eigenvalues & eigenvectors of an n*n matrix

Hello,

I need some help with this problem:
$\begin{pmatrix} 1 & 0 & 1 & 0 & \cdots \\ 0 & 1 & 0 & 1 & \cdots \\ 1 & 0 & 1 & 0 & \cdots \\ 0 & 1 & 0 & 1 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}$

The instructions are to find the eigenvalues and eigenvectors of this n by n matrix. I've tried to use $det(A -\lambda I_n) = 0$ to get the characteristic polynomial, but I can't seem to finish it.

I'd be grateful for any hints.

2. ## Re: Eigenvalues & eigenvectors of an n*n matrix

Here are a few observations that may be useful.

First, The $\text{Rank}(A)=\text{Rank}(A-0\cdot I)=2$ for all n.

This gives that $\text{Null}(A)=n-2$ so zero is an eigenvalue of multiplicity n-2.

When you compute $A-0\cdot I=A$ and reduce to row echelon form you can see a pattern for the eigenvectors. Just write out a few small cases and look for the pattern.

For the case when $n$ is even we get

$A^2=nA \iff A^2-\frac{n}{2}A=0$ This implies that the polynomial $p(x)=x^2-\frac{n}{2}x=x(x-\frac{n}{2})$ is in the annihilating Ideal, so the minimum polynomial must divide $p(x)$

so the only possible eigenvalues are 0 and $\frac{n}{2}$. The two cases I checked admitted 2 eigenvectors for the eigenvalue $\frac{n}{2}$

When I computed the the matrix $A-\frac{n}{2}I$ and reduced to row echelon form for n=4 and n=6 there seems to be a pattern for the eigenvectors . I hope this helps.

I haven't discovered anything useful for the case when n is odd.

3. ## Re: Eigenvalues & eigenvectors of an n*n matrix

Thanks for help, I managed to finish it