[Math] Eigenvalue or Eigenvector for a bidiagonal $n\times n$ matrix

Question

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!!!

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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?

Answer 1

(homework) so some hints:

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

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

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

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