5
$\begingroup$

An $n\times n$ complex matrix $A$ satisfies $A^k=I_n$, where $I_n$ is $n\times n$ identity matrix and $k$ is positive integer $\gt1$. Suppose that 1 is not an eigen value of $A$. Then which of the fallowing is necessarily true?

1) $A$ is diagonalizble

2) $A+A^2+...+A^{k-1}=0$

3) $tr(A)+tr(A^2)+...+tr(A^{k-1})=-n$

4) $A^{-1}+A^{-2}+...+A^{-(k-1)}=-I_n$

I think 1) and 2) are ture.

Improve this question
$\endgroup$
  • 1
    $\begingroup$ For $2)$, consider $A = \left(\begin{array}{rr} -1 & 0 \\ 0 & -1\end{array}\right)$. $\endgroup$ – Cameron Williams Jun 23 '15 at 1:16
  • $\begingroup$ @cameron this is already a diagonal maatrix such that $A^2=I_2$ and eigen values are -1,-1, its characteristic poly is $(x+1)^2$ $tr(A)=-2$ $\endgroup$ – aryan Jun 23 '15 at 1:25
  • $\begingroup$ @CameronWilliams you just buldozen my answer :( $\endgroup$ – aryan Jun 23 '15 at 1:28
  • $\begingroup$ @CameronWilliams according the your hint ans are 1),3),4). can we reach to general result by a simple example? $\endgroup$ – aryan Jun 23 '15 at 1:34
  • 1
    $\begingroup$ @aryan : No, you can not generalize the result but you can discard the options, like by the given matrix option 2) is not right. $\endgroup$ – Chiranjeev_Kumar Jun 23 '15 at 2:09
3
$\begingroup$

If $A^k = I$, then we have $A^k - I = 0$. Factoring, we have $$ (A - I)(I + A + A^2 + \cdots + A^{k-1}) = 0 $$ Since $1$ is not an eigenvalue, $A-I$ is invertible so that $$ I + A + A^2 + \cdots + A^{k-1} = 0 $$ In fact, $A$ satisfies the hypothesis of your question if and only if the above equality holds. Let $p(x)$ denote the polynomial $$ p(x) = 1 + x + x^2 + \cdots + x^{k-1} $$

1) necessarily holds. Note that $p(x)$ factors as $$ p(x) = \prod_{j=1}^{k-1} (x - e^{2 \pi ij/k}) $$ since the minimal polynomial of $A$ divides $p$, we may conclude that $A$ is diagonalizable.

2) Never holds. In particular, since $p(A) = 0$, we have $$ A + A^2 + \cdots + A^{k-1} = -I $$

3) holds by the above equality

4) is true. Note that $A$ is necessarily invertible, and that $$ A + A^2 + \cdots + A^{k-1} = -I \implies\\ A^{-k}(A + A^2 + \cdots + A^{k-1}) = -A^{-k} \implies\\ A^{-1} + A^{-2} + \cdots + A^{-k+1} = -A^{-k} = -[A^{k}]^{-1} = -I^{-1} = -I $$\

Improve this answer
$\endgroup$

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.