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I know this is kind of stupid, but does anyone know that is there any theorem actually proved the uniqueness of eigenvector?

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What do you mean by that? –  Qiaochu Yuan Jan 8 '12 at 9:28
    
If $\mathbf x$ is an eigenvector of $\mathbf A$, then $c\mathbf x$, when $c\neq 0$, is an eigenvector as well. –  J. M. Jan 8 '12 at 9:30
    
@QiaochuYuan Maybe my question is not so clear. For uniqueness here I mean no other matrix can produce the same eigenvector. –  Rein Jan 8 '12 at 9:35
    
@Rein: That's false. For example, both $\left(\begin{array}{cc}2&0\\0&3\end{array}\right)$ and $\left(\begin{array}{cc}5&0\\0&9\end{array}\right)$ have the exact same eigenspaces. And of course, any conjugate of a matrix will have the same eigenspaces with the same eigenvalues as the original matrix. So , no. Even if you require that the matrices have the same eigenspaces with the same eigenvalues and that they not be conjugate, there are examples: both of the following matrices have the same eigenspaces with the same eigenvalues (cont) –  Arturo Magidin Jan 8 '12 at 9:41
    
(cont) $\left(\begin{array}{cccc}1 & 1 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1\end{array}\right)$ and $\left(\begin{array}{cccc} 1 & 1 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{array}\right)$. But they are not conjugate. –  Arturo Magidin Jan 8 '12 at 9:43
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As J. M. noted, any eigenvector may be multiplied by a scalar, hence it's not unique strictly speaking. Plus in case of degeneracy, situation where several eigenvectors correspond to the same eigenvalue, any linear combination of them is the eigenvector as well.

So that eigenvectors for a specific eigenvalue actually span a corresponding sub-space. This sub-space however is strictly defined.

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