Does there exist $X$ such that $A = X^2 + X^t$?

Question

If $A \in M_n( \mathbb{R} )$ , then when does there exist $X \in M_n(\mathbb{R})$ such that $A = X^2 + X^T$?

Answer 1

Of course, we assume that $n\geq 2$.

$\textbf{Proposition 1}$. Let $n$ be an odd integer. Then there is $A\in M_n(\mathbb{R})$ s.t. there are no solutions in $X\in M_n(\mathbb{R})$.

$\textbf{Lemma}$. If $X$ is a solution, then $X^4-AX^2-X^2A+X-A^T+A^2=0$.

$\textbf{Proof}$. Use $X=A^T-{X^T}^2$.

$\textbf{Proof of Proposition 1}$. Since $n$ is odd, if $X$ is a real solution, then $X$ has (at least) a real eigenvalue. We choose (for example) $A=-I_n$. Then the equation is $X^4+2X^2+X+2I=0$. That is contradictory, because the polynomial $x^4+2x^2+x+2$ has no real roots. $\square$

EDIT. $\textbf{Proposition 2}$. Let $n$ be an even integer. If $A$ is normal and has no real eigenvalues, then there is at least one normal solution in $X$.

$\textbf{Proof}$. Up to an orthonormal change of basis, we may assume that $A=diag(U_1,\cdots,U_k)$ where $k=n/2$ and $U_i$ is in the form $\begin{pmatrix}a_i&b_i\\-b_i&a_i\end{pmatrix}$ with $(a_i,b_i)\in\mathbb{R}\times \mathbb{R}^*$. Thus it suffices to solve the equation (in dimension $2$) $X^T+X^2=\begin{pmatrix}a&b\\-b&a\end{pmatrix}$ with $(a,b)\in\mathbb{R}\times \mathbb{R}^*$.

That works with $X=\begin{pmatrix}x&b/(2x-1)\\-b/(2x-1)&x\end{pmatrix}$ where

$x>1/2$ satisfies $f(x)=x^2-b^2/(2x-1)^2+x=a$.

Notice that $x$ exists because $\lim_{x\rightarrow 1/2+0} f(x)=-\infty$ and $\lim_{x\rightarrow +\infty} f(x)=+\infty$. $\square$