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- Thread starter LagrangeEuler
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Math_QED

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##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.

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

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

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

##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.

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

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

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Yes. But the other one is not. Is there any example like this where both matrices ##A## and ##B## are hermitian.Yes, look at the edit in my first post. (The identity matrix is hermitian)

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Math_QED

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

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