$$\mathrm{X^2} = \left[\begin{matrix} 1 & 0 & 1\\ 0 &-1 & 0\\ 0 & 0 & -1\\ \end{matrix}\right]$$ 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.
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 $u^2+vw=-1$.
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 $a^2 + bc = -1$.
