Matrix multiplication with transpose Let $A,B\in\mathbb{F}^{n\times n}$ be two $n\times n$ matrices over
the underlying field $\mathbb{F}$. In addition, $A$ is guaranteed
to be a symmetric matrix, i.e, $A=A^{T}$. We assume complexity measure to be the number of field operations over $\mathbb{F}$. Now consider at the following problem:-


  Given $A,B$ and $AB$ as input, can we compute $AB^{T}$ in $o(n^{\omega})$ time?


Here $\omega$ is the exponent of matrix multiplication. Note that $AB^{T}=A^{T}B^{T}=(BA)^{T}$.
Thus computing $BA$ in $o(n^{\omega})$ time is also enough.
No.
Consider block matrices $A = \left(\begin{matrix} 0 &amp; 0 \\ 0 &amp; X \end{matrix}\right)$ (with symmetric $X$) and $B = \left(\begin{matrix} 0 &amp; Y \\ 0 &amp; 0 \end{matrix} \right)$. Computing $AB^T$ from $A$, $B$ and $AB$ means computing $XY^T$ from $X$ and $Y$, since $AB = 0$