I would like to solve the following matrix equation:
$ X + a_{0} M + a_{1} Tr(M) M + a_{2} M^{2} = 0 $
Here the X, which is known, and the M, which is unknown, matrices are invertibles, the coefficients $a_{i}$ are non zero, and Tr(M) is the trace of the matrix M. A priori I do not know the trace of M without solving the equation.
So I wonder if anyone can tell me if it is possible to solve this equation or if i have to do it component by component, that i'd prefer to avoid.
Note that $MX=XM$ in $M_n$. In general, $X$ has distinct eigenvalues, that we assume in the sequel; thus $M$ is a polynomial in $X$: $M=\sum_iu_iX^i$ and moreover the $(X^i)_{0\leq i\leq n-1}$ constitute a free system in $M_n$. Finally, if we know the characteristic polynomial of $X$, then we obtain a system of $n$ equations of degree $2$ in the $n$ unknowns $(u_i)_{0\leq i\leq n-1}$.
Example. $n=3,\chi_X(x)=x^3+x+1,a_0=2,a_1=-3,a_2=4$. Then $tr(X)=0,tr(X^2)=-2$.
$M=u_0I+u_1X+u_2X^2$ implies
$M^2=(2u_0u_2+u_1^2-u_2^2)X^2+(2u_0u_1-2u_1u_2-u_2^2)X+(u_0^2-2u_1u_2)I,tr(M)=3u_0-2u_2$.
The system is:
$-u_0u_2+4u_1^2+2u_2^2+2u_2=0,-u_0u_1-2u_1u_2-4u_2^2+2u_1+1=0,$
$-5u_0^2+6u_0u_2-8u_1u_2+2u_0=0$.
There are $8$ solutions in $\mathbb{C}$ (in general, $2^n$ solutions).