Solving matrix equations using characteristic or minimal polynomials

Question

Which of the following matrix equations have a real matrix solution?

$$X^3 = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 2 & 3 & 0 \end{bmatrix}$$

$$2X^5+X = \begin{bmatrix} 3 & 5 & 0 \\ 5 & 1 & 9 \\ 0 & 9 & 0 \end{bmatrix} $$

$$X^6+2X^4+10X = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$

$$X^4 = \begin{bmatrix} 3 & 4 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -3 \end{bmatrix} $$

Answer 1

For the first, $$A = \pmatrix{0 & 0 & 0\cr 1 & 0 & 0\cr 2 & 3 & 0\cr}$$ satisfies $A^3 = 0$ but $A^2 \ne 0$, so if $X^3 = A$ you'd have $X^9 = 0$ but $X^6 \ne 0$. Thus the minimal polynomial of $X$ would have degree at least $7$, which is impossible for a $3 \times 3$ matrix.

Answer 2

A general principle which you may find useful, is when solving an equation $$a_nX^n+\cdots+a_1X+a_0I=B$$ where $a_j$ are scalars, is that any solution $X$ commutes with $B$, that is $XB=BX$. This narrows down the solution space considerably. If say $B$ has a basis of eigenvectors with distinct eigenvalues, then $X$ has the same eigenvectors (equivalently a matrix diagonalising $B$ will also diagonalise $X$).