Find $X \in \mathbb{M}_n $ such that $ AX + X^TA = 0 $.

Question

We want $A$ (of the same dimensions as $X$) to be arbitrary and non-trivial. Working on it so far, I suspect that this can only mean $X = 0$ ($X$ is the zero matrix). But how can I fully prove or disprove this?

I'm thinking I need to rule out that the two terms could somehow cancel (add up to the zero matrix) -- i.e. $AX = -X^TA$ for some $X$ or $A$...because then only $X= 0$ works for any $A$. But how to show this?

Answer 1

Let $A\in M_n$ be a fixed matrix over a field with characteristic $0$. We consider the equation

$(*)$ $AX+X^TA=0$ in the unknown $X\in M_n$.

The set of solutions of $(*)$ is a vector-space.

i) $A$ is symmetric invertible.

Then $AX=K$ where $K$ is skew symmetric, that is, $X=A^{-1}K$; thus the vector-space of the solutions has dimension $n(n-1)/2$.

ii) $A$ is generic (for example, choose a random matrix $A$).

It's much more difficult; several numerical tests "show" that, for a generic matrix $A$, the set of solutions has dimension $int(n/2)$ (the integer part of $n/2$), which is much smaller than the result in case i).

EDIT 1. $\textbf{Proposition 1}$. i) For complex matrices, the set of solutions of $(*)$ has minimal dimension $int(n/2)$.

ii) If $A$ is a generic matrix , then the set of solutions of $(*)$ has dimension $int(n/2)$.

$\textbf{Proof}$. cf. Theorems 3 and 4 in this paper by De Teran and Dopico

https://reader.elsevier.com/reader/sd/pii/S0024379510004131?token=6F5EF56CBD9E9ACCCF72D75351D3D5D255ABDCE584234CAAFBB7A62126151A57326592D205A4A9960D87B4338A22933A

EDIT 2. Let $A\in M_n(\mathbb{C})$. We consider the equation

$(**)$ $X^TAX=A$ where the unknown is $X\in M_n(\mathbb{C})$. Let $Z_A$ be the algebraic set of the solutions of $(**)$. According to

What are the solutions to $X$ for $X^{T} A X = A$? Knowing that, what are the solutions to Y for $Y= (I−X)(I+X)^{-1}$ (with $\det(I+X) \neq 0$)?

there is an algebric quasi-isomorphism between the solutions of $(*)$ and the solutions of $(**)$ (the "quasi" does not prevent the conservation of the dimensions). Thus

$\textbf{Proposition 2}$. i) When $A$ is symmetric invertible, $Z_A$ has dimension $n(n-1)/2$.

ii) When $A$ is generic, $Z_A$ has dimension $int(n/2)$.

Answer 2

If $A$ is allowed to be arbitrary, then allowing $A=I$, we see that $X+X^T=0$, so $X^T=-X$. Thus, what we have is that $AX-XA=0$ for all $A$, or $AX=XA$ for all $A$. But the center of the matrix ring consists of central scalar multiples of the identity, so in particular $X$ is diagonal. Thus $X^T=X$. And therefore $X=-X$, or $2X=0$.

Thus if we are working over a base (unital, not necessarily commutative) ring $A$, then matrices with this property are precisely those of the form $aI$, where $a$ is a scalar in the center of $A$ with $2a=0$.

If you are working over any field of characteristic not two, such as the real or complex numbers, then there are no nonzero scalars with that property, so $X=0$ is the only matrix with this property.