I have a equation $X^2 = \begin{bmatrix}9 & 1\\ 0 & 9 \end{bmatrix}$. The textbook says that the Jordan matrix of $X$ can be $\begin{bmatrix} 3 & 1 \\ 0 & 3\end{bmatrix}$ or $\begin{bmatrix} -3 & 1 \\ 0 & -3\end{bmatrix}$. Therefore, this equation has 2 non-similar solutions.
My question is that why $\begin{bmatrix} -3 & 1 \\ 0 & 3\end{bmatrix}$ can not be the Jordan matrix of $X$?
If $X$ had distinct eigenvalues (which must be $\pm 3$ in this case), then there would be a matrix of eigenvectors such that $V^{-1} X V = \operatorname{diag}(3,-3)$. Then $X^2 = V \operatorname{diag}(3,-3) V^{-1} V \operatorname{diag}(3,-3) V^{-1} = 9I$. Hence $X$ cannot have distinct eigenvectors.