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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?2012-01-08
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    If $\mathbf x$ is an eigenvector of $\mathbf A$, then $c\mathbf x$, when $c\neq 0$, is an eigenvector as well.2012-01-08
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    @QiaochuYuan Maybe my question is not so clear. For uniqueness here I mean no other matrix can produce the same eigenvector.2012-01-08
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    @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)2012-01-08
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    (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.2012-01-08
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    @ArturoMagidin Ok I get it. But how about if I consider the principal eigenvector only not the entire eigenspace?2012-01-08
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    @Rein: In the example I gave you above, the two matrices have *the exact same* eigenvectors, with the *exact same* eigenvalues, but the two matrices are not conjugate. So it doesn't matter if you consider the principal eigenvector or not, that's still a counterexample.2012-01-08
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    @ArturoMagidin Thanks for the example, I think I am not familiar enough with the eigen-space. I am going to grab some readings.2012-01-08
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    @ArturoMagidin Your statement that conjugate matrices have the same eigenspaces is incorrect. If $v$ is an eigenvector for $A$ and $M$ is invertable, then it is $Mv$ which is an eigenvector for $MAM^{-1}$, not $v$. However, conjugate matrices do have the same list of eigenvalues each occurring with the same multiplicity.2012-01-08
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    The best result in this general direction is related to the Jordan normal form. http://en.wikipedia.org/wiki/Jordan_normal_form2012-01-08
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    @Adam: What I meant is that the same linear transformation may be represented by several different matrices (depending on the basis), but of course you are absolutely correct.2012-01-08

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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.