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Find all diagonal 3 x 3 matrices A with complex entries such that $A^2$=$$\begin{bmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} $$

Can anyone please help me how to start this? I'm not sure how I should approach this question. Any help would be greatly appreciated!

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  • $\begingroup$ A $3\times 3$ diagonal matrix has the form $$A=\begin{bmatrix} a & 0 & 0\\ 0 & b & 0\\ 0 & 0 &c\end{bmatrix}$$ for some $a,b,c\in\mathbb{C}$. Now what must $A^2$ look like? $\endgroup$ – Zack Cramer Mar 13 '17 at 4:09
  • $\begingroup$ does that mean that$$A^2=\begin{bmatrix} a^2 & 0 & 0 \\ 0 & b^2 & 0 \\ 0 & 0 & c^2 \\ \end{bmatrix}$$? so does complex entries mean that entries should be in the a+bi form??? $\endgroup$ – user425030 Mar 13 '17 at 4:14
  • $\begingroup$ Yes, each entry has that form. Now compare this to the matrix $$\begin{bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & -1\end{bmatrix}.$$ $\endgroup$ – Zack Cramer Mar 13 '17 at 4:24
  • $\begingroup$ So I can do, something like (a+bi)^2 = 2 and same with -1? $\endgroup$ – user425030 Mar 13 '17 at 4:28
  • $\begingroup$ Yup! That should do it. $\endgroup$ – Zack Cramer Mar 13 '17 at 4:29
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@Zack Cramer has done the heavy lifting.

In the complex plane, the $n$th root of unity has produces $n$ points on the unit circle. For the square root, $n=2$: $\theta_1 = 0$, $\theta_2 = \pi$. When $z = 2$, $$ \sqrt{z} = \pm \sqrt{2} $$

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