2
$\begingroup$

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..

|cite|improve this question
$\endgroup$
1
$\begingroup$

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 $-1$).

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ $f(x)(x^3+1)=x^9+1$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.