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Let $A$ and $B$ be $n\times n$ matrices such that $AB=BA$. Show that $A$ and $B$ have a common eigenvector.

I am not able to prove this. Can anyone help?

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    BTW A similar statement for Hilbertspaces is the reason for the Heisenberg uncertainty principle in QM, if I remember right. – 2017-02-10
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    What do you mean by "characteristic vector"? – 2017-02-10
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    Are $A,B$ normal matrices, i.e. does their EVD exist? – 2017-02-10
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    @Test123 Eigen vectors – 2017-02-10
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    @user218931 I don't understand the question. Do you mean if A and B have same eigen spaces? – 2017-02-10
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    I mean that if $\lambda$ is an eigenvector for $B$ and $V_\lambda = \{v\in \mathbb C^n\,|\, Bv = \lambda v\}$ is its eigenspace, then show that $A(V_\lambda) \subseteq V_\lambda$. – 2017-02-10
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    @user218931 No I haven't been able to show this – 2017-02-10
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    @user218931 What does the second line mean? Does it mean that eigen space of A is a subset of eigen space of B? – 2017-02-10
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    It means that $A$ applied to the eigenspace of $B$ lies in the eigenspace of $B$ (viewing $A$ as a linear map $\mathbb C^n\rightarrow \mathbb C^n$). – 2017-02-10
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    @user218931 What does "applied" mean? – 2017-02-10
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    Let $$A=B=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$ – 2017-02-10
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    Possible duplicate of [$AB=BA$ with same eigenvector matrix](http://math.stackexchange.com/questions/1022082/ab-ba-with-same-eigenvector-matrix) – 2017-02-13

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Let $\lambda$ eigenvalue of $B$ and $v\in V_\lambda$ an eigenvector i.e. $Bv=\lambda v$. Then, since $BA=AB$ we have: $$ BAv=ABv=\lambda Av $$ This implies that $Av$ is also an eigenvector of $B$ for the eigenvalue $\lambda$, namely $Av\in V_\lambda$. We deduce that $A V_\lambda \subset V_\lambda$ which is what you are asking as mentioned in the comments (even though the post doesn't clearly mention this).

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    But how does this prove that Av is also an eigen vector of A? – 2017-02-13
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    @AnimeshTiwari $Av$ is an eigenvector of $B$, not $A$. – 2017-02-13
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    yes but one has to prove that A and B have a common eigen vector. Av is an eigen vector of B, but how is it an eigen vector of A as well? – 2017-02-13
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    @AnimeshTiwari Not true in general. Check the accepted answer here: http://math.stackexchange.com/questions/1022082/ab-ba-with-same-eigenvector-matrix It's assumed though that $A,B$ are diagonalizable. – 2017-02-13
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    will go throught that. Thank you! – 2017-02-13