Matrices commute & Eigenvectors question

[LagrangeEuler]

Is it possible to find matrices that commute but eigenvectors of one matrix are not the eigenvectors of the other one. Could you give me example for it?

 

[member 587159]

Is it possible to find matrices that commute but eigenvectors of one matrix are not the eigenvectors of the other one. Could you give me example for it?

$A = \pmatrix{0&1\\0&0}, B = \pmatrix{1&0\\0&1}$ commute. $B$ has an eigenvector $(0,1)$, which $A$ doesn't have. They do have a common eigenvector.

EDIT: Take $A = I_n$ and $B \in M_{n,n}(\mathbb{F})$ where $\mathbb{F}$ is a field. Then you can see that this claim is obviously false.

 

[LagrangeEuler]

Matrix A practically do not have eigenvectors. Right? Because it is not diagonalizable.

What about two hermitian matrix. Is there any posiibility like this. Is it some easy way to construct this?

 

[member 587159]

Matrix A practically do not have eigenvectors. Right? Because it is not diagonalizable.

What about two hermitian matrix. Is there any posiibility like this. Is it some easy way to construct this?

Matrix A does have eigenvectors.. (but yes, A is not diagonalizable because we do not have enough linearly independent eigenvectors)

$det(A - \lambda I_n) = 0 \iff \lambda = 0$

Thus $0$ is an eigenvalue. We deduce that the eigenvectors are $V_0 = span(\{(1,0)\})$

As for your question with the hermitian matrices, look at the edit in my reply to your first post.

 

[LagrangeEuler]

Thanks a lot. Is it possible to construct similar example for to Hermitian matrices?

 

[member 587159]

Yes, look at the edit in my first post. (The identity matrix is hermitian)

 

[LagrangeEuler]

Yes, look at the edit in my first post. (The identity matrix is hermitian)

Yes. But the other one is not. Is there any example like this where both matrices $A$ and $B$ are hermitian.

 

[member 587159]

Yes. But the other one is not. Is there any example like this where both matrices $A$ and $B$ are hermitian.

Take $A = I_n$ and $B$ such that $B$ is hermitian. You should do some effort yourself (almost any hermitian matrix for $B$ will give you what you search for).