Prove that if $A$ is diagonalizable then there is a matrix $B$ such that $B^{2012} = A$ 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?
 A: 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
$$
D = \left ( \begin{array} {cccc}
\lambda_1 & 0 & \cdots  & 0\\
0 & \lambda_2 & \cdots  & 0\\
\vdots & \ddots & \vdots & \vdots \\
0 & \cdots & 0 & \lambda_n
\end{array}\right)
$$
A: 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.
