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Let $ T $ be the matrix $$ \begin{pmatrix} 4 & -1 & 1 \\ 4 & 0 & 1 \\ 0 & 0 & 3 \\ \end{pmatrix} $$ I need to find the eigenvectors.

What I have done so far:

The Characteristic polynomial is $ (\lambda - 3)(\lambda -2)^2 $. So the eigenvalues are $ 2 $ and $ 3 $. I have found the eigenvector for $ \lambda = 3 $ but I'm having trouble finding the eigenvector for $ \lambda = 2 $.

$ \begin{pmatrix} -2 & 1 & -1 \\-4 & \ \ 2 & -1 \\ \ \ \ 0 & \ \ 0 & -1 \end{pmatrix} \left( \begin{array}{c} x \\ y \\ z \end{array} \right)$ $ = \left( \begin{array}{c} 2x \\ 2y \\ 2z \end{array} \right)$ which gives us the equations, $ 3z =0,-4x-z=0, -4x+y-z=0 $ so $ x=y=z=0 $. I'm trying to figure out where I went wrong.

Thanks in advance for any replies.

Edited once.

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  • $\begingroup$ $(1,2)$-entry of $2I - T$. $\endgroup$ – Catalin Zara Apr 4 '17 at 2:38
  • $\begingroup$ That was stupid of me. Thank you!! $\endgroup$ – R Squared Apr 4 '17 at 2:49
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To find eigenvectors, you should solve either the homogeneous equation $(\lambda I-T)\mathbf v=0$ or the equation $T\mathbf v=\lambda\mathbf v$. You appear to have combined the two and are trying to solve $(\lambda I-T)\mathbf v=\lambda\mathbf v$.

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were does $\begin{pmatrix} -2 & -1 & -1 \\-4 & \ \ 2 & -1 \\ \ \ \ 0 & \ \ 0 & -1 \end{pmatrix} \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 2x \\ 2y \\ 2z \end{array} \right)$ come from?

You can either say: $\begin{pmatrix} -4 & -1 & 1 \\-4 & 0 & 1 \\ 0 & 0 & -3 \end{pmatrix} \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 2x \\ 2y \\ 2z \end{array} \right)$

or you can say

$\begin{pmatrix} -2 & -1 & 1 \\-4 & -2 & 1 \\ 0 & 0 & -1 \end{pmatrix} \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \mathbf 0$

I think the second one is easier to solve.

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