# Find eigenvalues and eigenvectors of weird matrix

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1. Oct 3, 2015

### dukemiami

1. The problem statement, all variables and given/known data
find eigenvalues and eigenvectors for the following matrix

|a 1 0|
|1 a 1|
|0 1 a|

2. Relevant equations

3. The attempt at a solution
I'm trying to find eigenvalues, in doing so I've come to a dead end at 1 + (a^3 - lambda a^2 -2a^2 lambda + 2a lambda^2 + lambda^2 a - lambda^3 - a + lambda)

This is because i have to put in a - lamnda across the board, then it gets tricky when trying to find eigenvalues with all these variables, can someone please help in this bizarre question.

2. Oct 3, 2015

### vela

Staff Emeritus
The polynomial you came up with isn't correct. I suggest you resist the temptation to multiply everything out immediately. You should find that $(a-\lambda)$ is a factor.

3. Oct 3, 2015

### dukemiami

Ok so before i multiplied everything out, it came to:

det|1 (a - lamnda)| + (a - lamnda)*det|(a - lamnda) 1 | = 0
|0 1 | | 1 (a - lamnda)|

which becomes 1 + (a - lamnda)((a - lamnda)(a - lamnda) - 1)

I think i see now, you're right with the a - lamnda

Thus a represents an eigenvalue, and is the only one then?

4. Oct 3, 2015

### dukemiami

The

0 1 | | 1 (a - lamnda)|

should be shifted accordingly to match the top row

5. Oct 3, 2015

### vela

Staff Emeritus
I don't think your expansion is correct. To get matrices to appear correctly, use LaTeX. It's pretty easy to learn. If you reply to this post, you can see an example.
$$\begin{vmatrix} a-\lambda & 1 & 0 \\ 1 & a-\lambda & 1 \\ 0 & 1 & a-\lambda \end{vmatrix}$$

6. Oct 3, 2015

### Ray Vickson

If you set $b = a - \lambda$ the determinant of $A - \lambda I$ becomes $b^3 -2b$, so equating this to zero gives roots $b = 0$ and $b = \pm \sqrt{2}$.