find value of $x$ in a way that the matrix has square root with real entries

Question

I want to find all possible real values of $x$ in a way that $X=\begin{pmatrix}x & -x & -1 & 0 \\x & -x & 0 & -1 \\ 1 & 0 & x & -x\\ 0 & 1 & x & -x \end{pmatrix}$ is the perfect square of a matrix with real entries.

Here is my try: I wanted to find square root of $X$. When $X$ is diagonalizable the question is much easier. But the given matrix is not diagonalizable. So I was looking for another way to find the square root of $X$. I guess it's related to Jordan form. But finding a basis for that seems so complicated for me. Also I guess it should work for all values of $x$ as the characteristic polynomial of $X$ does not depend on $x$'s value.

Answer 1

Let $$ A=\begin{pmatrix} 1 & -1 & 1 & -1 \\ 1 & -1 & 1 & -1 \\ -1 & 1 & 1 & -1 \\ -1 & 1 & 1 & -1 \end{pmatrix} \;\;\text{and}\;\; B=\begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 0 & -1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{pmatrix} $$ Then $$ A^2=0,\;\; AB+BA = \begin{pmatrix} 4 & -4 & 0 & 0 \\ 4 & -4 & 0 & 0 \\ 0 & 0 & 4 & -4 \\ 0 & 0 & 4 & -4 \end{pmatrix} ,\;\; B^2 = \begin{pmatrix} 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & -2 \\ 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \end{pmatrix} $$ Therefore $$ (aA+bB)^2 = a^2A^2+ab(AB+BA)+b^2B^2 = \begin{pmatrix} 4ab & -4ab & -2b^2 & 0 \\ 4ab & -4ab & 0 & -2b^2 \\ 2b^2 & 0 & 4ab & -4ab \\ 0 & 2b^2 & 4ab & -4ab \end{pmatrix} $$ So we just have to set $$ a = \frac{x}{2\sqrt{2}} \;\; \text{and}\;\; b=\frac{1}{\sqrt{2}} $$ and then $(aA+bB)^2=X.$

Answer 2

We have $$ (t^3-t^2+5t+3)^2 \\ =t^6-2t^5+11t^4-4t^3+19t^2+30t+9 \\ =(t^2+1)^2(t^2-2t+9)+32t $$ As $(X^2+1)^2=0$, we have $(X^3-X^2+5X+3I)^2 = 32X$ or $$ \left( \frac{1}{4\sqrt{2}} (X^3-X^2+5X+3I) \right)^2 = X $$