# 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?

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(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?

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1. In this case will be (a-x)^n-b(a-x)^n-2=0</BR> 2. In this case will be a*n</BR> 3. In this case it will be n</BR> 4. No so sure about this one, but I am thinking n.</BR> Also how to join all these 4 to create a proof for this question any hint? Do I make 1 and 2 equal? </BR> Thanks –  newbietolinalg Mar 17 '12 at 21:43
No. I advice you to write down a small $3 \times 3$ matrix $J$. Then subtract $xI$. Then write down your resulting matrix. Then multiply out all the diagonal elements. –  user2468 Mar 17 '12 at 21:44
Please practice using a pen & paper, not using Octave or other softwares. –  user2468 Mar 17 '12 at 21:45
@newbietolinalg if $J = \begin{pmatrix}a & b \\ 0 & a \end{pmatrix}$ then $J - xI = \begin{pmatrix}a-x & b \\ 0 & a-x \end{pmatrix}.$ Hence ${\rm det}(J-xI) = (a-x)^2.$ The roots are $a,a.$ Hence $a$ is the eigenvalue of $J$ with algebraic multiplicity $= 2$. Can you generalize this argument using my $4$ points above, and prove the general case? –  user2468 Mar 17 '12 at 21:52
Thanks J.D. for your help and patience :-) I was able to generalize it to be $(a-x)^n=0$ which is equal to $x$=$a$ if $n$ > 0 –  newbietolinalg Mar 17 '12 at 22:05