Prove: if $\bf{AB^T}$ is skew-symmetric and $\bf A$ full-rank, then $\bf{AX}=\bf B$ has unique solution $\bf X$

Question

I've run into this statement while trying to prove that the energy of a rotating body in $N$ dimensions is conserved, it's the last puzzle piece I'm missing.

Let $\bf A$ and $\bf B$ be two $M \times N$ matrices, with $ M \geq N $. Furthermore, $\bf A$ is full rank ($N$) and $\bf{AB^T}$ is skew-symmetric.

Then (I believe), there exists a (unique?) $N \times N$ matrix $\bf X$ such that $\bf{AX}=\bf B$.

How would I go about proving this? And am I correct that the solution is unique?

P.S. In my problem I also know that, for each matrix separately, the sum of the rows is zero, but I haven't included it above because I believe it's irrelevant for this proof. That is to say, $\bf{1^T A} = \bf 0 , \bf{1^T B} = \bf 0$, where $\bf 1$ and $\bf 0$ are the vectors of all ones and all zeros respectively.

Answer 1

Since $A$ is full column rank, then $A^TA$ is invertible. Now let $X=-B^TA(A^TA)^{-1}$, then given the assumptions: $$ AX=-AB^TA(A^TA)^{-1} = BA^TA(A^TA)^{-1}=B. $$

The proof of uniqueness is trivial: if $AX_1=AX_2 = B$ then $$A(X_1-X_2) = 0 \implies R(X_1 - X_2)\subseteq N(A)=\{0\}\implies X_1-X_2=0\implies X_1=X_2,$$ since $A$ is full column rank and $m\geq n$.

Answer 2

I'll write how we can derive the formula for the unique solution of the equation $AX = B$.

Let's assume that there exists such a solution $X$. Then we have the following:

\begin{align} AX=B &\implies X^TA^T = B^T \\ &\implies AX^TA^T=AB^T \\ &\implies (AX^TA^T)^T = (AB^T)^T \\ &\implies AXA^T = -AB^T \\ &\implies A^TAXA^TA = -A^TAB^TA \\ &\implies (A^TA)^{-1}(A^TA)X(A^TA)(A^TA)^{-1} = -(A^TA)^{-1}(A^TA)B^TA(A^TA)^{-1} \\ &\implies X = -B^TA(A^TA)^{-1} \end{align}

where we used the fact that $A^TA$ is invertible since $A$ has full column rank and that $AB^T$ is skew-symmetric.

So, if a solution exists, it is unique and given by the above formula. By plugging it in we can verify that $X = -B^TA(A^TA)^{-1}$ indeed is a solution:

$$AX = -AB^TA(A^TA)^{-1} = (AB^T)^TA(A^TA)^{-1} = B(A^TA)(A^TA)^{-1} = B.$$