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Given: $$A \in M_{n\times n} (\mathbb C) \; , \; A \; \text{is diagonalizable}$$

We need to prove that:

$$ \exists B \in M_{n\times n} (\mathbb C) \; : B^{2012} = A$$

What I said so far:

If $A$ is diagonalizable, then $\exists P$ and $D$ such that $A = P D P^{-1}$ where $D = \begin{pmatrix} \lambda_1 & 0 &\cdots& 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}$ , $P = [[v_1], [v_2], \dotsc ,[v_n]]$, $\det(P) \neq 0$ and also $\lambda_1 , \lambda_2 , \dotsc , \lambda_n$ are all distinct.

also,

$A^{2012} = P D^{2012} P^{-1}$

But where do I exactly go from there?

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    $\begingroup$ It need not be true that the $\lambda_i$ are all distinct; for example $A$ could be the identity. The important thing is that the $\lambda_i$ all have $2012$-th roots, as they are elements of $\mathbb{C}$. $\endgroup$ – mdp May 2 '13 at 9:31
  • $\begingroup$ Thanks for the heads up. Can you please explain why all $\lambda_i$ all have 2012th roots? They are a "number" afterall.. $\endgroup$ – TheNotMe May 2 '13 at 9:33
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    $\begingroup$ Consider the polynomial $x^{2012}-\lambda_i$, and the fundamental thereom of algebra. The key property here is that $\mathbb{C}$ is algebraically closed. $\endgroup$ – mdp May 2 '13 at 9:34
  • $\begingroup$ Any non-zero complex number has $2012$ distinct $2012$th roots. You only need one of them for each $\lambda_i$. $\endgroup$ – Arthur May 2 '13 at 9:35
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The crucial step is that if you have a matrix of the form $ABA^{-1}$ and you take some power of it, this is equal to $(ABA^{-1})^n=AB^nA^{-1}$.

In the case you mention, you have $A=PDP^{-1}$. If you define $\tilde{D}$ to be the one obtained from $D$ by replacing every $\lambda_i$ by $\sqrt[2012]{\lambda_i}$ (where the choice of root doesn't really matter), this results in $(P\tilde{D}P^{-1})^{2012} = P\tilde{D}^{2012}P^{-1} = PDP^{-1}=A$ and so $B=P\tilde{D}P^{-1}$ would solve your question.

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  • $\begingroup$ In that case I will be able to prove that $B^n=A$ for any n? $\endgroup$ – user67773 May 2 '13 at 9:36
  • $\begingroup$ Wow, thank you so much, very simple! And @Umakant I believe so, but only if the matrixes are both $\in \mathbb C$ $\endgroup$ – TheNotMe May 2 '13 at 9:38
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    $\begingroup$ Not quite, since $B$ is constructed for $n=2012$ explicitly. Just taking an arbitrary power won't give you $A$, in general. However, this says you can construct such a matrix $B$ with $B^n=A$ for any $n$, since the matrices are over $\mathbb{C}$ indeed. $\endgroup$ – HSN May 2 '13 at 9:38
  • $\begingroup$ Got it. Thanks. $\endgroup$ – user67773 May 2 '13 at 9:40
  • $\begingroup$ Good question Uma! $\endgroup$ – TheNotMe May 2 '13 at 9:45
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Consider $B = P D_1 P^{-1}$ such that $$ D_1 = \left ( \begin{array} {cccc} \lambda_1^{\frac 1{2012}} & 0 & \cdots & 0\\ 0 & \lambda_2^{\frac 1{2012}} & \cdots & 0\\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & 0 & \lambda_n^{\frac 1{2012}} \end{array}\right) $$ then $$ B^{2012} = PD_1^{2012}P^{-1} = PDP^{-1} $$ where $$ D = \left ( \begin{array} {cccc} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{array}\right) $$

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