Complex cyclic matrix is diagonalizable?

Question

A matrix $A$ is cyclic if there exists an integer $m$, s.t. $A^m=I$, where $I$ is the identity matrix. Then it's said that every complex cyclic matrix is diagonalizable.

I know if a matrix has distinct eigen-vectors, then it's diagonalizable. And the minimal polynomial of $A$ should be $X^m-1$, which means it may have different eigen-values, so it may have distinct eigen-vectors. But I don't know whether it's right. Hope someone could help. Thanks!

You are right, $x^m-1$ cannot have a multiple root since the derivation is $m\cdot x^{m-1}$ which can only have root $0$. — Peter, Nov 10 '19 at 14:38
A matrix is diagonalizable iff its minimal polynomial has not repeated factors. — Ben Grossmann, Nov 10 '19 at 14:39

score 1 · Accepted Answer · answered Nov 10 '19 at 14:40

1

Indeed, every matrix satisfying $X^m = I$ will be diagonalizable. It suffices to note that a matrix is diagonalizable iff its minimal polynomial has no repeated factors (see this post for instance).

answered Nov 10 '19 at 14:40

Ben Grossmann

225,327

Complex cyclic matrix is diagonalizable?

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