Linked Questions

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0answers
63 views

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 ...
32
votes
7answers
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 ...
2
votes
1answer
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 ...
1
vote
1answer
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 ...
2
votes
2answers
142 views

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 ...
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0answers
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 ...
1
vote
3answers
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) ...
3
votes
1answer
139 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 ...
3
votes
0answers
195 views

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 ...
0
votes
3answers
155 views

Matrices that Commute with of a Specific matrix

Let $a$ and $b$ be real numbers. Considere the $2\times 2$ matrix \begin{equation*}A=\left[\begin{array}{cc}a&b\\-b&a\end{array}\right]. \end{equation*} What is the centralizer of the matrix $...
35
votes
3answers
5k views

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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votes
1answer
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 ...
18
votes
3answers
5k views

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 ...
3
votes
1answer
719 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?
0
votes
0answers
41 views

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 ...

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