# Find 3x3 real matrix A such that $A^2 \neq I$ and $A^4 = I$, where I is identity matrix.

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!

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.

• See my latest edit. Commented Jun 4, 2015 at 15:43
• How were you able to yield this at the beginning: $$\pmatrix{1 & 0\\0&B}$$ where $$B = \pmatrix{0&-1\\1&0}$$? Commented Aug 8, 2015 at 22:02
• @Grace I don't understand your question Commented Aug 8, 2015 at 22:04
• 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. Commented Aug 8, 2015 at 22:10
• @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". Commented Aug 8, 2015 at 22:12

Hint: the matrix for a rotation by ...