Find the matrix X such that $A\cdot X = X\cdot A^T$

Question

I am working at some linear transformations and I need to know if it is possible given a square real matrix $A$, to find a real square invertible matrix $X$ such that $$ X^{-1} \cdot A \cdot X = A^T$$ where $A^T$ means the transposed of $A$

Answer 1

A matrix and its transpose have the same characteristic polynomial, minimal polynomial and elementary divisors etc. so they have the same Jordan normal form, hence are similar.

Answer 2

We assume that $A$ is real. Of course, we do not look for the Jordan forms of $A,A^T$ (because, practically, we cannot find them when $n\geq 5$).

The solution is in the OP's title; it suffices to solve the LINEAR equation $AX=XA^T$;

EDIT. (Answer to C Marius). The set $V$ of the solution is a vector space that has same dimension as the vector space of the matrices that commute with $A$ (in particular, its dimension $k$ is $\geq n$). Any software working on the Sylvester equation gives to you (with complexity $O(n^3)$) a basis of $V$: $(X_i)_i$. Since $A,A^T$ are similar over $\mathbb{R}$, we know that among the $\sum_i a_i X_i\in V$, there exists at least an invertible matrix.

We assume that each $a_i$ follows an absolutely continuous distribution law of probability (for example the normal law on $\mathbb{R}$ or the uniform law on $[a,b]\subset\mathbb{R}$) and that these previous laws are independent..

$\textbf{Proposition}$. If we randomly choose the $(a_i)$, then we obtain an invertible solution with probability $1$.

$\textbf{Proof}$. Consider the function $f:a=(a_i)_i\in\mathbb{R}^k\rightarrow \det(\sum_i a_iX_i)$. There is $a$ s.t. $f(a)\not= 0$; then $f^{-1}(0)$ is a Zariski closed proper subset of $\mathbb{R}^k$. Thus $\{a;f(a)\not= 0\}$ is a Zariski open dense Zariski subset of $\mathbb{R}^k$ and we are done.

Answer 3

As loup blanc mentioned that it's a Sylvester equation, here is an addendum utilizing vec operator:

\begin{align} AX-XA^T = 0 \\ AXI - I X A^T = 0\\ \textrm{vec} \left( AXI \right) - \textrm{vec} \left(I X A^T \right) = 0\\ \left(I \otimes A\right)\textrm{vec}\left(X\right) - \left(A \otimes I\right)\textrm{vec}\left(X\right) = 0\\ \left(I \otimes A - A \otimes I\right)\textrm{vec}\left(X\right) = 0 \end{align}

Then one can find a solution which lies in the nullspace of $\left(I \otimes A - A \otimes I\right)$.