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I find this exercise in my textbook.

Find all Hermitian matrices $A\in M_n(\mathbb{C})$ satisfying $$A^5+A^3+A-3I=0$$

I have two questions.

1) How do I solve a matrix polynomial? If I simply factorize it, I can only get those answers with the form $\lambda I$.

2) How a matrix being Hermitian (basically it means a matrix is "complexly" symmetric) makes it special in this problem?

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There is no general way by abel-ruffini (for the 'easy' case $n=1$). In this case try Jordan normal form, maybe it will help –  Belgi Sep 19 '12 at 19:34
    
Hermitian matrices are diagonalizable, so this reduces to a question about scalars satisfying the given polynomial constraint. –  copper.hat Sep 19 '12 at 19:34
    
@Belgi: That is not true, $x=1$ satisfies the equation, reducing to $(x-1)(x^4+x^3+2x^2+2x+3)$ wihch can be solved :-). –  copper.hat Sep 19 '12 at 19:38
    
@copper.hat - I said "in general" because q1 of the OP is general –  Belgi Sep 19 '12 at 19:41

1 Answer 1

up vote 6 down vote accepted

The only real root of $x^5+x^3+x-3$ is $1$. Hence the only eigenvalue of $A$ is $1$, so $A$ is the identity matrix.

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but the matrix can have non-real eigenvalues to, no ? –  Belgi Sep 19 '12 at 19:42
    
@Belgi: Being Hermitian ensures all of its eigenvalues are real. –  Voldemort Sep 19 '12 at 19:43
    
+1 Nice answer. –  copper.hat Sep 19 '12 at 20:01

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