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Find $3 \times 3$ real matrix A such that $A^2 \neq I$ and $A^4 = I$, where $I$ is identity matrix.

I first thought that there is no such matrix and tried to show that using determinants, but all I get is that A has to have $detA=1$ or $-1$. Next I actually found such Matrix A= $\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{matrix} \right)$, but I did it through random manipulations and do not have any idea how to coherently write the answer. Any help would be appreciated. Thanks!

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2 Answers 2

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Hint: try to find a $2 \times 2$ matrix $B$ like this. Then, consider the block matrix $$ A = \pmatrix{B & 0\\0&1} $$ Think of rotation matrices.


Your answer is perfect! What you've found is the matrix $$ \pmatrix{1 & 0\\0&B} $$ where $$ B = \pmatrix{0&-1\\1&0} $$ is the rotation in $\Bbb R^2$ by $90^\circ$. Since rotating by $90^\circ$ $4$ times brings you back to where you started, we have $B^4 = I_2$. On the other hand, when we apply the rotation twice, we end up with a $180^\circ$ rotation, which is to say that $B^2 = -I_2$. So, when we take the $4$th power, we find $$ A^4 = \pmatrix{1^4&0\\0&B^4} = \pmatrix{1&0\\0&I_2} = \pmatrix{1&0&0\\0&1&0\\0&0&1} = I_3 $$ On the other hand, $$ A^2 = \pmatrix{1^2&0\\0&B^2} = \pmatrix{1&0\\0&-I_2} = \pmatrix{1&0&0\\0&-1&0\\0&0&-1} \neq I_3 $$ So, we have $A^4 = I_3$ and $A^2 \neq I_3$, which is exactly what we wanted.

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  • $\begingroup$ See my latest edit. $\endgroup$ Commented Jun 4, 2015 at 15:43
  • $\begingroup$ How were you able to yield this at the beginning: $$ \pmatrix{1 & 0\\0&B} $$ where $$ B = \pmatrix{0&-1\\1&0}$$? $\endgroup$
    – Grace
    Commented Aug 8, 2015 at 22:02
  • $\begingroup$ @Grace I don't understand your question $\endgroup$ Commented Aug 8, 2015 at 22:04
  • $\begingroup$ Meaning, what led you to decide to find this: $$ \pmatrix{1 & 0\\0&B} $$ where $$ B = \pmatrix{0&-1\\1&0} $$, when starting the problem? Sorry for the confusion. $\endgroup$
    – Grace
    Commented Aug 8, 2015 at 22:10
  • $\begingroup$ @Grace Are you asking how I knew that this particular matrix would be a solution? I did not decide to use that particular matrix, that was the result that the asker got "through random manipulations". $\endgroup$ Commented Aug 8, 2015 at 22:12
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Hint: the matrix for a rotation by ...

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