I encounter the following question when I read the paper "Quadratic Optimization with Orthogonality Constraints" (https://arxiv.org/pdf/1510.01025.pdf).
Simply speaking, let $X\in \mathbb{R}^{m\times n}$ with $m \geq n$ be a semi-orthogonal matrix, i.e. $X^\top X = I_n$. $A\in \mathbb{R}^{m\times m}$ and $B\in \mathbb{R}^{n\times n}$ are real symmetric matrix.
I wonder why the following holds: $$X^\top AXB = B X^\top AX.$$ In other word, we can commute the multiplication between $X^\top AX$ and $B$.
I have stuck here for a while. Any help or hint is highly appreciated.
Thank you!