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$$\mathrm{X^2} = \left[\begin{matrix} 1 & 0 & 1\\ 0 &-1 & 0\\ 0 & 0 & -1\\ \end{matrix}\right]$$ Here $\mathrm{X}$ is a $3\times3$ matrix with real entries

I wish to know whether the above equation has a solution. Wolfram Alpha says no solutions exist. Can you please explain why there are no solutions?

Thank you for any help.

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Try $$X=\pmatrix{1&-1/2&1/2\\0&0&-1\\0&1&0}.$$

With Wolfram Alpha, you get what you pay for.

If my calculations are right, the general solution is $$X=\pm\pmatrix{1&-w/2&(1+u)/2\\0&u&v\\0&w&-u}$$ where $u$, $v$, $w$ satisfy $u^2+vw=-1$.

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Let $D = diag(1,-I_2)$, and let $X \in M_3(\Bbb{R})$ be such that $X^2 = D$. I claim that $X$ is a solution if and only if

$$X = \begin{bmatrix} \pm 1 & 0 & 0 \\ 0 & a & b \\ 0 & c & -a \\ \end{bmatrix},$$

where $a^2 + bc=-1$.

The equation $X^2 = D$ implies $\det(X) = 1$ and therefore $X^{-1}$ exists. From the equation, and the invertibility of $X$, we can deduce $X = X^{-1}D$ and $X = DX^{-1}$ and therefore $X^{-1} D = DX^{-1}$, from which we get $DX=XD$. This itself implies that $X$ is of the form

$$X = \begin{bmatrix} x & 0 & 0 \\ 0 & a & b \\ 0 & c & d \\ \end{bmatrix}$$

Call the lower block $Z$, which will be a matrix in $M_2(\Bbb{R})$. Then the equation $X^2 = D$ now implies that $x = \pm 1$ and $Z^2 =-I_2$. The latter equation implies that $Z^2$ has eigenvalues $-1$, implying that $Z$ has eigenvalues $\pm i$. This implies that $Z$ is trace-zero, which gives us $d = -a$, and $a(-a) - bc = det(Z) = -i \cdot i = 1$ or $a^2 + bc = -1$.

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  • $\begingroup$ There's a $1$ in the top right corner of dark32's matrix. $\endgroup$ – Lord Shark the Unknown Jun 4 '17 at 17:52
  • $\begingroup$ @LordSharktheUnknown Whoops...I completely missed that...Sometimes I read a little too quickly. $\endgroup$ – Eli Bashwinger Jun 4 '17 at 17:55
  • $\begingroup$ Your matrix is similar to the original matrix, so your effort isn't wasted. $\endgroup$ – Lord Shark the Unknown Jun 4 '17 at 17:59

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