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Given matrix $A = \begin{bmatrix} 0 & 0 &2 \\ 0 & 2 & 0\\ 2 &0 &0 \end{bmatrix}$

I found its eigenvalues: $\lambda_{1}=2, \lambda_{2}=2, \lambda_{3}=-2$.

With eigenvalue $2$, $A-\lambda I = \begin{bmatrix} -2 & 0 &2 \\ 0 & 0 &0 \\ 2 & 0 & -2 \end{bmatrix}$, and I try to find the eigenvector $x$ such that $\begin{bmatrix} -2 &0 &2 \\ 0& 0 &0 \\ 2& 0 &-2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$

Since two of the eigenvalues are $2$, I should find $2$ corresponding eigenvectors. I came up with $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ (can eigenvectors be the zero vector?), $\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, my question is how do know which two are the correct eigenvectors, or are all of them valid eigenvectors? Did I miss another possible eigenvector?

As for $\lambda_{3}=-2$, $\begin{bmatrix} 2 &0 &2 \\ 0& 4 &0 \\ 2& 0 &2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$, possible eigenvector is just $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$, which is a lot more straightforward.

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Note that eigenvector cannot be zero vector by its definition. Actually number of eigenvectors for a eigenvalue can be infinite but they form a linear space. So in your example, $\lambda=2$ has eigen space of dimension 2, so you can at most find 2 eigenvectors which are linear independent (Any other may produce dependence with previous ones).

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  • $\begingroup$ So would the 2 eigenvectors be (1,0,1) and (0,1,0)? Since those two are linearly independent. And (1,1,1) is a linear combination of (1,0,1) and (0,1,0). $\endgroup$ – Adrian Mar 1 '14 at 8:33
  • $\begingroup$ @Adrian Yes, they are. $\endgroup$ – Shuchang Mar 1 '14 at 8:34
  • $\begingroup$ Technically, pick any two of the three eigenvectors you listed. The third is linear dependent on the first two. $\endgroup$ – nickalh May 6 '15 at 11:26

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