Rank of matrix product invariant under change with matrix with same span

Question

Let $A,B \in \mathbb{R}^{n,k}$ with $\text{Im}(A)=\text{Im}(B)$, where $\text{Im}(A)$ denotes the image or column space.

Then does the following hold for $U \in \mathbb{R}^{n,k}$?

\begin{equation} \text{rank}(U^TA) = \text{rank}(U^TB) \end{equation}

Answer 1

Yes, it holds.

Indeed, we know that for any matrix $X\in \Bbb R^{n\times k}$, its image is defined by $\operatorname{Im}(X)=\{X\cdot v| v\in \Bbb R^k\}$. Thus \begin{align*}\operatorname{Im}(U^TX) & =\{(U^T X)\cdot v| v\in \Bbb R^k\} =\{U^T \cdot (X\cdot v)| v\in \Bbb R^k\} \\ & =U^T\cdot(\{ X\cdot v| v\in \Bbb R^k\})=U^T\cdot \operatorname{Im}(X).\end{align*} In your case this implies that $$ \operatorname{Im}(U^TA)=U^T \cdot (\operatorname{Im}(A)) = U^T \cdot (\operatorname{Im} (B)) = \operatorname{Im} (U^TB), $$and thus their dimensions must be equal.