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Suppose that $A$ is an $n\times n$ matrix with distinct eigenvalues. And suppose $B$ commutes with $A$. Show that $B$ is diagonable; i.e., show that $B$ is similar to a diagonal matrix.

I get that $AB=BA$ and that some diagonal matrix $D$ that is similar to $B$ is $D=SBS^{-1}$ and that $S$ is composed of the eigenvecotrs of $B$ I just don't know what to do from their.

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  • $\begingroup$ Hint: "Simultaneously diagonalisable" $\endgroup$
    – Arthur
    Nov 18 '14 at 22:13
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Hint: if $Ax = ax, a$ scalar, then $$a Bx = B(ax) = B(Ax) = BAx= ABx $$ and you know that the subspace $$\{y: Ay = ay\} $$is a line...

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  • $\begingroup$ If I wanted to be difficult, I could point out that the matrices might not be over $\Bbb C$. But the hint is sound. $\endgroup$
    – Arthur
    Nov 18 '14 at 22:18
  • $\begingroup$ @Arthur $\Bbb C$ can be replaced by any field. $\endgroup$
    – mookid
    Nov 18 '14 at 22:19
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The big theorem is that, for a square matrix $A$ for which each eigenvalue occurs in only one Jordan block, all matrices that commute with $A$ can be written as polynomials in $A.$ This includes matrices with distinct eigenvalues, as each Jordan block is one by one. So $$ B = b_0 I + b_1 A + b_2 A^2 + \cdots + b_{n-1} A^{n-1}. $$ You do not need higher degree because of Cayley-Hamilton.

Oh, $A$ itself is diagonalizable because of the distinct eigenvalues. Some $P^{-1}A P = E$ diagonal. What can you say about $P^{-1}B P ?$

Note: I keep collecting equivalent conditions to the Jordan block thing, Given a matrix, is there always another matrix which commutes with it?

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