Matrix equation $X+AX-XA^2=0$ has only zero solution

Question

Let $A$ be a $n\times n$ complex matrix, with even numbers as eigenvalues. Show that the matrix equation $X+AX-XA^2=0$ has only zero solution. My attempt is just consider an eigenvector $v$ of $A$, then $Xv+AXv-4m^2Xv=0.$ Then what to do? Or just $v'(X+AX-XA^2)v=0$. But $A$ is not symmetric, how to proceed?

Answer 1

Let eigenvalues of $A$ be denoted by $2m$ with $m \in \mathbb Z$ and let $u,v$ be the left/right eigenvector respectively and note that $A^2 v = 4m^2 v$. Then, $$ u^H X v + 2m u^H X v - 4m^2 u^H X v = 0 $$ This implies that $1 + 2m - 4m^2 = 0$ or $u^H X v = 0$. Note that the $m$ satisfying the first equality will not be an integer and so $u^H X v = 0$. Can you finish it from here?

Answer 2

Hint: If $\mu, \lambda$ are eigenvalues of $A$, then we can find a vector $v$ for which $Av = \lambda v$ and a vector $w$ for which $A'w = \bar \lambda w \implies w'A = \lambda w'$. Let $\Phi$ denote the linear map defined by $$ \Phi(X) = X + AX + XA^2. $$ The matrix $X = vw'$ is an eigenvector of $\Phi$ assocaited with the eigenvalue $1 + \lambda + \mu^2$.

Answer 3

$(I+A)X+X(-A^2)= IX+AX-XA^2=\mathbf 0$
s a Sylvester Equation and is isomorphic to

$\big((I+A)\oplus (-A^2)^T\big)\text{vec}\big(X\big) = \text{vec}\big(\mathbf 0\big)= \mathbf 0_{n^2}$
where $\oplus$ denotes Kronecker Sum.

The eigenvalues of $(I+A)$ are necessarily odd, and the eigenvalues of $(-A^2)^T$ are necessarily even since $A$ has real even eigenvalues. Thus each eigenvalue of $\big((I+A)\oplus (-A^2)^T\big)$ is an odd number plus an even number $\neq 0$. Therefore $\det\big((I+A)\oplus (-A^2)^T\big)\neq \mathbf 0$ and $\text{vec}\big(X\big) =\mathbf 0_{n^2}$ is the only solution $\implies X=\mathbf 0$ is the only solution for the original problem.