The $3\times3$ matrix is: $$ A=\begin{bmatrix} 2 & -2 & 3\\ -2& -1 & 6\\ 1 & 2 & 0 \end{bmatrix} $$ $\lambda_1=-5, \lambda_2=3$ (repeated) I am concerned about $\lambda=3$

So a $3\times3$ matrix has a repeated eigenvalue and another single eigenvalue. Finding the eigenvectors of the repeated eigenvalue. I just have one equation:

$x+2y-3z=0$ (3 times the same equation)

If I put zero for $x,y,z$ one at each time i got $3$ different independent eigenvectors. but the dimension of eigenspace can't exceed $2$ as the algebraic order of the eigenvalue is only $2$.

What I am doing wrong? Should I just select 2 of the 3 equations I have in random?

Edit: The eigenvectors I get are: (0 3 2), (3 0 1), (-2 1 0)

  • $\begingroup$ Can you actually tell us what the matrix is? It's hard to help you otherwise... $\endgroup$ – 5xum Jun 2 '15 at 6:23
  • $\begingroup$ not sure if you are notified but i added the matrix, thanks for feedback $\endgroup$ – fidias Jun 2 '15 at 6:36
  • $\begingroup$ OK, now one more thing: what are the three eigenvectors you get? What do you mean by ""If I put zero for $x,y,z$"? I think this is where you made your mistake, but I don't know exactly what you did so I can't help you yet. $\endgroup$ – 5xum Jun 2 '15 at 6:40
  • $\begingroup$ Can you prove your claim "got 3 different independent eigenvectors"? $\endgroup$ – wdacda Jun 2 '15 at 6:42
  • $\begingroup$ I think you are right, that's where I am wrong but I am a bit confused if you could help me clarify why I am wrong I would be glad. thank you a lot $\endgroup$ – fidias Jun 2 '15 at 6:44

Your work is correct. Now you have: $x=3z-2y$, and this is the eigenspace (that has dimension $2$). You can choose: $y=0,z=1$ and you have the eigenvector $(3,0,1)^T$ or $y=1,z=0$ and you have the eigenvector $(-2,1,0)^T$.

  • $\begingroup$ Okay, thank you, so the dimension is 2 and I am not allowed to put x=0 $\endgroup$ – fidias Jun 2 '15 at 6:52

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