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Question: Let $A$ be an $n×n$ complex matrix with $n$ distinct eigenvalues. Let $V$ be the set of all $n×n$ complex matrices $B$ that commute with $A$. Prove that V is a vector space and find its dimension (Justify your answer).

My Answer: I know how to show that V is a vector space, but I don't know how to find its dimension. I tried showing that if v is an eigenvector corresponding to some eigenvalue, so is Bv, and got that for all B, Bv = kv for some scalar v. But I'm not sure if this helps.

Arturo Magidin
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Galois Friend
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    (Note that you didn't finish stating the question.) Hint: can you find the dimension of $V$ if you assume that $A$ is diagonal? Further hint: can you reduce the general question to the case where $A$ is diagonal? – Greg Martin Apr 04 '19 at 00:37
  • There should be a post on this site addressing this question. That being said, it helps to start by considering the case where $A$ is a diagonal matrix – Ben Grossmann Apr 04 '19 at 00:37
  • @Omnomnomnom when you say that there should be a post on this site addressing the question, would you be able to link that post? I've also edited the question to make it more clear. – Galois Friend Apr 04 '19 at 00:44
  • I'll link it if I find it, but I haven't found it yet. The question looks familiar though – Ben Grossmann Apr 04 '19 at 00:44
  • Also, we have proven before that if A is a nxn complex matrix with n distinct eigenvalues then A is diagonalizable, so I'm not quite sure how to take your hints. – Galois Friend Apr 04 '19 at 00:46
  • @GaloisFriend We're telling you to focus on the following question: if $A$ is actually a diagonal matrix, which is to say that $$ A = \pmatrix{\lambda_1 \ & \ddots\ && \lambda_n} $$ with the blank entries denoting zeros, then which matrices $B$ commute with $A$? See what the actual matrix-multiplication looks like. – Ben Grossmann Apr 04 '19 at 00:58
  • Alright, so the matrices B must all be diagonal matrices as well in order to commute with A. – Galois Friend Apr 04 '19 at 01:50

1 Answers1

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You have proved that $Bv_i = \mu_i v_i$ for all $i=1,\dots,n$.

Let $p$ be the polynomial such that $p(\lambda_i)=\mu_i$.

Then $p(A)v_i=p(\lambda_i)v_i=\mu_i v_i=Bv_i$.

Therefore, $B=p(A)$.

Thus, $V$ is exactly the subspace of all polynomials in $A$. What is the dimension of that subspace?

lhf
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