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### Can anyone explain why commuting matrices share common eigenvector? [duplicate]

If two matrices $A$ and $B$ commute with each other, why would they share some eigenvector? Does that mean that an eigenvector for $A$ is also an eigenvector for $B$ and vice-versa?
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### If $A,B$ are commuting diagonalizable complex matrices , then $A,B$ have a common eigen-basis ? [duplicate]

If $A,B$ are diagonalizable complex matrices of size $n$ such that $AB=BA$ , then is there a common eigen-basis of $A$ and $B$ ?
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### Self-Adjoint Operators Basis of Eigenfunctions [duplicate]

I encountered the following statement in the context of the Hecke algebra on the space of cusp forms, left without proof: Let there be a family of commutative self-adjoint operators on a finite ...
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### Do commuting matrices share the same eigenvectors?

In one of my exams I'm asked to prove the following Suppose $A,B\in \mathbb R^{n\times n}$, and $AB=BA$, then $A,B$ share the same eigenvectors. My attempt is let $\xi$ be an eigenvector ...
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### Simultaneous Jordanization?

IN a comment to Qiaochu's answer here it is mentioned that two commuting matrices can be simultaneously Jordanized (sorry that this sounds less appealing then "diagonalized" :P ), i.e. can be brought ...
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### If matrices $A$ and $B$ commute, $A$ with distinct eigenvalues, then $B$ is a polynomial in $A$

If $A\in M_{n}$ has $n$ distinct eigenvalues and if $A$ commutes with a given matrix $B\in M_{n}$, how can I show that $B$ is a polynomial in $A$ of degree at most $n-1$? I think first I need to show ...
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### If $AB=BA$, show that $B$ is diagonalizable.

We were asked to prove the following: Let $A$ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable. It was suggested I show that an ...
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### How do I prove that in every commuting family there is a common eigenvector?

The proof given by my textbook is highly non-satisfying. The author adopted some magic-like "reductio ad absurdum" and the proof (although is correct) didn't reveal the nature of this problem. I made ...
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### If $A$ and $B$ commute, show that they have a common eigenvector? [duplicate]

That is, if $AB = BA$. I can see that it is true if all the eigenvalues of $A$ or $B$ are distinct. Let $(\lambda, \textbf{x})$ be an eigenpair of $A$ (or whichever one has distinct eigenvalues). So ...
### If $A$ and $B$ are commuting Hermitian matrices, then they have the same eigenvectors?
If $AB = BA$ and both $A$ and $B$ are Hermitian matrices, then I can show that if $Av = \lambda v$, then $ABv = BAv = B\lambda v = \lambda Bv$. So $Bv$ is an eigenvector of $A$ as well. Where I am ...