Eigenvalue or Eigenvector for a bidiagonal $n\times n$ matrix Let $$J =
\begin{bmatrix} a & b & 0 & 0 & \cdots & \cdots\\\\ 0 & a & b & 0 & \cdots & \cdots\\\\ \vdots & \vdots & \ddots & \cdots & \cdots & \cdots \\\\ \vdots & \vdots & \vdots & \ddots & \cdots & \cdots \\\\ \vdots & \vdots & \vdots   &\ddots & a & b \\\\ \vdots & \vdots & \vdots & \vdots & 0 & a \\ \end{bmatrix}$$
I have to find eigenvalues and eigenvectors for $J$.
My thoughts on this...
a = 
    2  3
    0  2
octave-3.2.4.exe:2> b=[2,3,0;0,2,3;0,0,2]
b =
   2   3   0
   0   2   3
   0   0   2

octave-3.2.4.exe:3> eig(a)
ans =

   2
   2

octave-3.2.4.exe:4> eig(b)
ans =

   2
   2
   2

octave-3.2.4.exe:5>

I can see that the eigenvalue is $a$ for $n \times n$ matrix. 
Any idea how I can prove it that is the diagonal for any $N \times N$ matrix.
Thanks!!!

I figured out how to find the eigenvalues. But my eigenvector corresponding to the eigenvalue a comes out to be a zero vector... if I try using matlab, the eigenvector matrix has column vctors with 1 in the first row and zeros in rest of the col vector...
what am I missing? can someone help me figure out that eigenvector matrix?

(homework) so some hints:


*

*The eigenvalues are the roots of ${\rm det}(A-xI) = 0.$

*The determinant of a triangular matrix is the product of all diagonal entries.

*How many diagonal entries does an $n\times n$ matrix have?

*How many roots does $(a - x)^n = 0$ have?
