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### Understanding “matrices commute iff they share a common basis of eigenvectors”

I am trying to understand the answer to this question. It was pointed out that this doesn't necessarily hold for non-diagonalizable matrices, so the top answer starts by assuming $A$ and $B$ are ...
41 views

### Dimension of the space of matrices which is commutative to a given matrix.

Suppose I have a matrix $A$ in the space $V$ of $n$ by $n$ matrices. Then it is quite clear that $S=\{B : AB=BA\}$ form a subspace. I want to find out its dimension. I think it depends on the ...
49 views

### 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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### 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 ...
146 views

### Show that every eigenvector of A is an eigenvector of B

Let $A, B\in\mathcal M_n(\mathbb{R})$ such that $AB=BA$ with $n$ distinct eigenvalues. 1) Show that if $v\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}$, $Av=\lambda v\implies ABv=\lambda Bv$ 2) ...
27 views

### To show orthogonal basis exist consisting common eigenvectors

Let $h$ be a positive definite hermitian form on $E$ and $A,B: E\rightarrow E$ be two hermitian endomorphisms which commute, $AB = BA$. Prove that there exists a orthogonal basis for $E$ consisting of ...
140 views

### Regarding a proof of: “if $A,B \in M_n(\mathbb{k})$ are diagonalizable and commute, they are simultaneously diagonalizable”.

As the title states, I'm looking for a proof of the following, Proposition. Let $A, B \in M_n(\mathbb{k})$ be commuting diagonalizable matrices, so that $AB = BA$. Therefore, $A$ and $B$ can be ...
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### It is said that two matrices commute “if and only if” they share eigenvectors

It is for homework. But I am not asking the answer, since it seems to be already given in many posts like this one... I am pretty sure that I will get an A just copying and pasting it. My question is ...
1k views

### 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 ...
721 views

### 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 $\bigoplus\limits_{i=1}^{k} Eig(A,\lambda_i)=\bigoplus\limits_{i=1}^{k} Eig(B,\mu_i)$ then $A$ and $B$ commute

Let $A,B$ be two $n\times n$-matrices over $\mathbb K$ and assume $\bigoplus\limits_{i=1}^{k} Eig(A,\lambda_i)=\bigoplus\limits_{i=1}^{k} Eig(B,\mu_i)=\mathbb K^n$, where $\lambda_i, \mu_i$ are ...
28k views

### 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 ...
117 views

### 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 $B$ ?
If two operators $A$ and $B$ commute then any eigenvector of $A$ is an eigenvector of $B$? I know that if that happens there is a basis in which the eigenvectors of $A$ and $B$ are equal, but I don't ...
### Set of commuting Matrices $\Rightarrow$ Common Eigenvector [duplicate]
I am trying to prove that if we have an arbitrary set of commuting matrices in $M_n(\mathbb C)$ then they have a common eigenvector. Well, if we have only 2 matrices, the answer is easy and it has ...