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Let $A$ be a $3\times 3$ complex matrix such that $A^3=-I$

How to show that $A$ has distinct eigenvalues?

What if i consider $A=-I?$ Isn't then -1 becoming the only eigen value?

ddabir
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    You just have found a counterexample. But maybe you meant to say: complex matrix such that there is at least one non-real entry. Or something to that effect. – Raskolnikov Jul 12 '13 at 17:49
  • Let $m_A(x)$ be the minimal polynomial of $A$, then $m_A(x)|x^3 + 1$, may be this can help. – jouge Jul 12 '13 at 17:56
  • @Raskolnikov the question just tell A be a complex $3\times 3$ matrix such that A^3=-I and i need to verify whether the statement "A has three distinct eigenvalues" is correct. It is incorrect right? – ddabir Jul 12 '13 at 18:11
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    @Raskolnikov: Your suggested reformulation doesn't change much since $A = e^{i\pi/3} I$ has $A^3 = - I$ but only one eigenvalue. In fact, any polynomial is satisfied by a matrix $\lambda I$ where $\lambda$ is one of the roots of the polynomial... – Mike F Jul 12 '13 at 18:20
  • @Mike: Nice catch. – Raskolnikov Jul 12 '13 at 18:48
  • @Mike Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Sep 15 '13 at 11:52

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This question was answered (more like confirmed, actually) in the comments.

For any $\lambda \in \mathbb{C}$ the matrix equation $A^3 = \lambda I$ has $A = \mu I$ as a solution where $\mu$ is one of the cube roots of $\lambda$. More generally, for any nonzero polynomial $p(z) \in \mathbb{C}[z] $, the matrix equation $p(A) = 0$ has a solution $A = \mu I$ where $\mu$ is one of the roots of $p(z)$. Matrices of the form $\mu I$, far from having distinct eigenvlaues, have only one eigenvalue, that is to say $\mu$.

Mike F
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