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Is there a real matrix $3 \times 3$ Matrix $B$ such that $B^2=A$ where $$A=\begin{bmatrix} 1 & 0& 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{bmatrix}$$

I have written $A$ as

$$A=C-I$$ where

$$C= \begin{bmatrix} 2 & 0& 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$$

So $$B^2+I=C$$ and obviously

$$Det(B^2+I)=0$$

Any help here?

Ekaveera Gouribhatla
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  • The matrix $J_{2 \times 2} = \left(\begin{matrix} 0 & -1 \ 1 & 0 \end{matrix}\right)$ works as the imaginary number. Try yourself to square it. The matrix $A$ is a block matrix $\left(\begin{matrix} 1_{1 \times 1} & 0_{1 \times 2} \ 0_{2 \times 1} & -1_{2 \times 2}\end{matrix}\right)$ and can be written as $\left(\begin{matrix} 1_{1 \times 1} & 0_{1 \times 2} \ 0_{2 \times 1} & J_{2 \times 2}\end{matrix}\right)^2$. – md2perpe Jul 01 '17 at 10:28

1 Answers1

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HINT Use Cayley Hamilton theorem. You have $B^2=A$ and $B^3=AB$. Also $\det(B)=\pm 1$