Prove that this matrix equation has no roots if a matrix meets certain conditions Could you explain to me how to solve matrix equations?
Here is an example:
Prove that:
$$2X^2 + X = \begin{bmatrix} -1&5&3\\-2&1&2\\0&-4&-3\end{bmatrix}$$
has no solutions in $M(3,3;\mathbb{R})$, where $M(3,3;\mathbb{R})$ is the space of all $3\times3$ matrices with real entries.
 A: First note that $X$ is a $3\times 3$ real matrix and hence it must have at least one real eigenvalue. 
Consider the characteristic polynomial of $$A=\begin{pmatrix} -1&5&3\\-2&1&2\\0&-4&-3\end{pmatrix}$$
which is $$p(\lambda) = -\lambda^3 -3\lambda^2 -17\lambda -11$$
Taking $A=2X^2 + X$ in $p$ gives 
$$p(A) = -8X^6 - 12X^5 -18X^4 -13X^3 - 37X^2 - 17X - 11 = 0$$
Note that in particular,
$$q(x) = -8x^6 - 12x^5 -18x^4 -13x^3 - 37x^2 - 17x - 11 = 0$$
is an annihilating polynomial of $X$. Therefore the eigenvalues of $X$ must be amongst the roots of $q$. But all of $q$'s roots are complex, a contradiction.
A: Since $A = \pmatrix{-1 & 5 & 3\cr -2 & 1 & 2\cr 0 & -4 & 3\cr}$ is equal to a polynomial is $X$, it commutes with $X$.  Now $0 = 2 X^2+ X - A = 2 (X + I/4)^2 - I/8 - A$.
Now $A$ has a simple  eigenvalue $\lambda \approx -.715923208$.  If $v$ is an eigenvector of $A$ for this eigenvalue, it is also an eigenvector of $X$ for eigenvalue $\mu$, where $2 (\mu + 1/4)^2 = 1/8 + \lambda$.  But since $1/8 + \lambda < 0$, $\mu$ can't be real.  Now $v$ can be taken to be real (since $A$ and $\lambda$ are real), 
so $X v = \mu v$ is impossible if $X$ is real. 
