Is a complex matrix A such that A^k = I; the n×n identity matrix and where k>1 diagonalizable provided that 1 is not an eigenvalue of A.
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1What are your thoughts? – Ted Shifrin Jun 24 '15 at 20:42
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If $A=I$, it is diagonalised, but its only eigenvalue is $1$. – Bernard Jun 24 '15 at 20:43
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Same question asked here – Ben Grossmann Jun 24 '15 at 21:11
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The same question was asked again here. – Ben Grossmann Jun 24 '15 at 21:14
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If you find the answers provided insufficient, then you should post a new question that specifies exactly what you've tried and where you're running into trouble. – Ben Grossmann Jun 24 '15 at 21:16
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I'm curious as to whether you three who posted the same question happen happen to be in the same class, or whether all three accounts are just the same person reposting the same question for some odd reason – Ben Grossmann Jun 24 '15 at 21:18
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The question came in a nation-wide exam. I hope this does answer your curiosity. And thanks for the answers, it was a simple one. Just did not think over it.Sorry for posting the same question. – Jagdeep Singh Jun 25 '15 at 05:59
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The polynomial $p(x) = x^k - 1$ has simple roots. Then any matrix $A$ such that $p(A) = 0$ is diagonalizable. Indeed, a non difficult theorem of linear algebra says that $A$ is diagonalizable iff its minimal polynomial $m_A(x)$ has simple roots. Then if $p(x)$ has simple roots and $p(A) = 0$ we get that also $m_A(x)$ has simple roots since $m_A(x)|p(x)$.
Holonomia
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But in case if k=2 won't the fact that A does not have eigenvalue 1 play a role ? – Jagdeep Singh Jun 24 '15 at 20:53
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1The fact that $A$ has 1 as eigenvalue or not is irrelevant if you want to know if a matrix which satisfies $A^k = I$ is diagonalizable: https://en.wikipedia.org/wiki/Minimal_polynomial_%28linear_algebra%29 – Holonomia Jun 24 '15 at 20:59