Is it possible to find this unitary matrix?

Question

If we are given a matrix $A^2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$

Is it possible to find a matrix $A (2 \times 2)$ which isn't unitary? (A is a matrix in complex numbers)

Answer 1

All matrices with that property are unitary.

In general, it is not true that if ${\bf A}^{\! 2}$ is unitary then ${\bf A}$ is unitary.

Let $\displaystyle{{\bf A} = \left(\begin{array}{cc} a & b \\ c & d \end{array}\right)}$ and consider $${\bf A}^{\! 2} = \left(\begin{array}{cc} a^2+bc & b(a+d) \\ c(a+d) & bc+d^2 \end{array}\right)$$ If we solve the equations $a^2+bc=0$, $b(a+d)=1$, $c(a+d)=-1$ and $bc+d^2=0$ then $$a=\pm\frac{1}{\sqrt{2}}, \ b=\pm\frac{1}{\sqrt{2}}, \ c=\mp\frac{1}{\sqrt{2}}, \ d=\pm\frac{1}{\sqrt{2}}$$ $$a=\pm\frac{\mathrm{i}}{\sqrt{2}}, \ b=\mp\frac{\mathrm{i}}{\sqrt{2}}, \ c=\pm\frac{\mathrm{i}}{\sqrt{2}}, \ d=\pm\frac{\mathrm{i}}{\sqrt{2}}$$ If you follow through with the calculation then you'll see that, in all cases, $$\overline{{\bf A}}{\bf A}^{\!\top}={\bf I}$$