10
$\begingroup$

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?

$\endgroup$
4
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
11
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
5
$\begingroup$

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 $$ D = \left ( \begin{array} {cccc} \lambda_1 & 0 & \cdots & 0\\ 0 & \lambda_2 & \cdots & 0\\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{array}\right) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.