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Let $A=\begin{bmatrix}2&1&2\\4&2&4\\2&1&2\end{bmatrix}$

Its eigenvalues are $0,0,6$. I want to find its eigenvectors.

My solution:

when $\lambda_1=0$

$$(A-\lambda_1I)x_1=\begin{bmatrix}2&1&2\\4&2&4\\2&1&2\end{bmatrix}x_1=0$$

$x_1=(-\frac{1}{2}s-t s t)=-\frac{1}{2}s(1, -2, 0)+t(-1, 0, 1)$ since 2nd and 3rd vairables are free variables. Thus $(1, -2, 0)$ and $(-1, 0, 1)$ are the eigenvectors.

BUT the answer is $(1, -2, 0)$ and $(0, -2, 1)$.

I can't understand why these are the true answers. What is wrong with my solution?

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1 Answer 1

up vote 3 down vote accepted

Nothing, you just chose a different basis of the eigenspace corresponding to $0$. Note that $(0,-2,1)=(1,-2,0)+(-1,0,1)$.

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Oh, then both are right? Then it means that eigenvectors are not unique? –  existence Nov 11 '12 at 12:19
1  
Yes. Suppose $v_1$ and $v_2$ are eigenvectors to the eigenvalue $\lambda$, then $A(v_1+v_2)=Av_1+Av_2=\lambda v_1+\lambda v_2=\lambda(v_1+v_2)$ –  Julian Kuelshammer Nov 11 '12 at 12:22
    
Now I understand! Thanks! –  existence Nov 11 '12 at 12:31

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