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Let $A$ be the Banach algebra of $n \times n$ matrices over $\mathbb C$. Then the subset consisting of all diagonal matrices is an abelian subalgebra. (correct me if I'm wrong).

Now I want to show that it is in fact maximal. (I am quite sure it is but maybe it isn't?).

How can I show that the diagonal matrices are a maximal abelian subalgebra? Of course the way to prove it is to assume that there exists an abelian subalgebra containing the diagonal matrices. But what then?

Student
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  • Let $A$ be a matrix that isn't diagonal. Construct a diagonal matrix that doesn't commute with it. Note: A matrix $A$ is not diagonal if there exists $i\ne j$ with $A_{ij}\ne 0$. – PVAL-inactive Jun 24 '14 at 07:00

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Prove that if $A$ is a diagonal matrix with pairwise distinct diagonal entries and $AB=BA$, then $B$ is diagonal.

Vladimir
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