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So for linear algebra, I either need to prove that, for all square matrices, if $A^4 = I$, then $A^2 = \pm I$, or find a counterexample of this statement. Can anyone help please?

Thanks!

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  • $\begingroup$ Welcome to this site! Unlike some sites, we look for more than just a statement of a problem. This question is missing context or other details: Please improve the question by providing additional context, which ideally includes its source and motivation, your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ Commented Feb 25, 2018 at 18:34
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    $\begingroup$ You may also want to clarify whether you want to work with real vector spaces or complex ones; either way there is a counterexample, such as a 90 degree rotation. $\endgroup$ Commented Feb 25, 2018 at 18:35
  • $\begingroup$ @CarlMummert: Your comment deserves to be the best answer! Almost all answers directly assumed complex values but you game an easy to check and intuitive answer for the real case! $\endgroup$
    – MrYouMath
    Commented Feb 25, 2018 at 18:37
  • $\begingroup$ @MrYouMath: well, it still takes some work to start with a 90 degree rotation and end up with something that is not a 180 degree rotation when you square it (so that $A^2 \not = -I$). $\endgroup$ Commented Feb 25, 2018 at 18:42

6 Answers 6

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Check $$A=\begin{bmatrix} i & 0 \\0 & 1 \end{bmatrix}$$

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Or $$A=\begin{bmatrix} 0 & -1 & 0 \\1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

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Let consider $$A^2=\begin{bmatrix} 0 & 1 \\1 & 0 \end{bmatrix}$$

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  • $\begingroup$ Which has complex but no real solutions. $\endgroup$ Commented Feb 25, 2018 at 18:39
  • $\begingroup$ @LordSharktheUnknown I was thinking the same, but how would you prove it ? Resolving the system with $4$ unknowns ? $\endgroup$
    – krirkrirk
    Commented Feb 25, 2018 at 18:41
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    $\begingroup$ @krirkrirk determinants.... $\endgroup$ Commented Feb 25, 2018 at 18:50
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This is not true. Take, for instance, the matrix

$\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 &0 \\ 0& 0 & i & 0 \\ 0& 0& 0& -i \end{bmatrix}$

which is maybe a little too much, but I liked it.

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Completing the previous answers, it's enough to look for matrices $n \times n$ that satisfy $$(A^{2}) = (A^{2})^{-1}$$ where $A^{2} \neq I$. Check this in the previous answers.

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This answer expands and expounds upon Botund's answer above; if $S= \{ \pm 1, \pm i \}$, then $x^4 = 1$ for every $x \in S$.

Consequently, if $D$ is any $n$-by-$n$ diagonal matrix with diagonal entries from $S$, then $D^4 = I_n$; however, the diagonal entries can be chosen such that $D^2 \ne \pm I_n$. For example, select any diagonal matrix containing at least one element of $\{\pm i\}$ and at least one element of $\{ \pm 1 \}$ in the diagonal.

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