Prove that this matrix equation has no roots if a matrix meets certain conditions

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

Set $$ -C=(-c_{ij})_{3\times 3}=\begin{bmatrix} -1&5&3\\-2&1&2\\0&-4&-3\end{bmatrix} $$

Solve the no linear sistem

\begin{cases} 2\cdot \Big( \sum_{k=1}^3x_{ik}\cdot x_{kj}\Big) +x_{ij}=-c_{ij} \quad i=1,2,3 \mbox{ and } j=1,2,3. \end{cases} or set $$ A=\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix} \quad B=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \quad C=\begin{bmatrix} -1&5&3\\-2&1&2\\0&-4&-3\end{bmatrix} $$ We have that $M_{3\times 3}(\mathbb{R})$ is a ring. Then \begin{align} 0=AX^2+BX+C= &4A\big( AX^2+BX+C\big)\\ =& (2AX)^2+2B(2AX)+4AC \\ =& \big[ 2AX+B \big]^2 -B^2+4AC\\ \end{align} implies $$ \left[2AX+B\right]^2=B^2-4AC $$ if $\Delta=B^2-4AC$ is no neqative and simetric exists $U\in M_{3\times 3}(\mathbb{R})$ such that $U^2=B^2-4AC$. Then $$ X=(2A)^{-1} \Big[ -B \pm U\Big] $$ If $A,B,C$ and $X$ commute it's more easy.