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Let $A\in M_{n}(\Bbb C)$ and $A^{n}=2A$ with $n>1$. Is $A$ diagonalizable?

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    $A$ is diagonalizable if and only if its minimal polynomial can be written as $ m(x)=(x-x_1)(x-x_2) \cdots (x-x_s) $ over the field $\mathbb{F}$,here is $\mathbb{C}$. where $x_i \not= x_j$ when $i \not= j$2012-12-08

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Your matrix satisfies the polynomial $p(x) = x(x^{n-1} - 2)$. The roots of the polynomial are $0$ and the $(n-1)$th roots of $2$. What can you say about the diagonalizability of the matrix if it satisfies a polynomial with non-repeating roots?

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To answer the title, there are several ways to characterise diagonalisable matrices, one of which is useful here: a matrix $A$ is diagonalisable if and only if there exists some polynomial $P\in K[X]$ that is split and without multiple roots, and such that $A$ is annihilated by substitution for $X$ into $P$ (often written $P(A)=0$). In case of diagonalisable matrices you can take for $P$ their minimal polynomial, but you don't have to; in the example given, trying whether $P=X^n-2X\in\Bbb C[X]$ works would be a natural option. Can you find its roots?

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    yes, it is clear that $x^n-2x=0$ has $n$ roots in $C$.2012-12-08