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Suppose $A\in M_n(\Bbb C)$ satisfies $A^6-A^3+I=O$. Prove that if a linear transformation $T:M_n(\Bbb C)\rightarrow M_n(\Bbb C)$ is given by $T(B)=AB$, then $T$ is diagonalizable.

How to prove it? Is it related to minimal polynomial? I have no idea..

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The polynomial $f(x)=x^6-x^3+1$ has six distinct complex roots, and annihilates $A$. Hence the minimal polynomial of $A$ has distinct complex roots (we don't know how many, but it it is at most six). Hence $A$ is diagonalizable, because having only linear terms in a minimal polynomial is a characterization of diagonalizability.

Addendum: $f(x)(x^3+1)=x^9+1$, which has nine distinct complex roots (complex ninth roots of $-1$).

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    $\begingroup$ $f(x)(x^3+1)=x^9+1$ and the right hand side has 9 distint roots. Thus, f(x) has six distint roots in $\Bbb C$. And we know that the minimal polynomial divide $f(x)$. Hence the minimal polynomial has at most 6 distinct roots. Thus, A's minimal polynomial is divided in to linear terms over $\Bbb C$. Hence A is diagonaizable. The logic is this, right? But I wonder A's disgonalizability implies T's? Since U(v):=Av for all v in $\Bbb C^n$, A can define another linear transformation. Does $A$'s disgonalizability imply the diagonalizabilities of any other transformations defined by A? $\endgroup$ – Younghoon Jeon Dec 8 '14 at 15:01
  • $\begingroup$ Ah I solved the problem. Thanks for help! $\endgroup$ – Younghoon Jeon Dec 8 '14 at 15:08
  • $\begingroup$ @vadim123: I think the minimal polynomial has the same root as the characteristic polynomial, but each with a multiplicity lesser or equal. $\endgroup$ – enzotib Dec 8 '14 at 15:19

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