Find eigenvalues and eigenvectors of weird matrix

  • #1
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Homework Statement


find eigenvalues and eigenvectors for the following matrix

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

Homework Equations




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.
 

Answers and Replies

  • #2
vela
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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
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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
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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?

The

0 1 | | 1 (a - lamnda)|

should be shifted accordingly to match the top row
 
  • #5
vela
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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
Ray Vickson
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Homework Statement


find eigenvalues and eigenvectors for the following matrix

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

Homework Equations




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.

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}##.
 

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