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If $A$ be a $n\times n$ matrix over complex field that $n>1$ and $A^{2n}=2A$ then:
A)$A$ is not triangularisable
B)$A$ is not diagonalizable
C)$A$ is triangularisable
D)$A$ is triangularisable but isn't diagonalizable

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closed as too localized by Did, Davide Giraudo, Paul, Stefan Hansen, Brandon Carter Jan 25 '13 at 14:18

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

What is the question? –  Did Jan 25 '13 at 7:45
this is a multiple choice question!!! –  rese Jan 25 '13 at 7:47
Thanks, I can see that, but what are YOU asking about it? Surely not for its flat full solution, without showing what you know, what you tried, or what you think about it? Please refer to the comments made on your previous questions, at the moment your modus operandi is not suited to this site. –  Did Jan 25 '13 at 7:52

2 Answers 2

up vote 1 down vote accepted

Hint: Can you find a triangular respectively diagonal matrix that fits the bill? Replacing a matrix by a similar one will not alter the condition $A^{2n}=2A$, so you can ignore the "isable/izable" parts.

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Hint: Compute the roots of $X^{2n}-2X$ and recall the definition of the minimal polynomial.

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