3
$\begingroup$

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?

1 Answers 1

3

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

  • 0
    Now I understand! Thanks!2012-11-11