Rank of a left-orthogonal decomposition $A^T=UB^T$

Question

I've got a rather simple question, but couldn't find an answer to it: Say $A\in\mathbb R^{m\times n}$ can be decomposed according to $$A^T=UB^T\in\mathbb R^{n\times m}\tag1$$ for some left-orthogonal (i.e. $U^TU=I_k$) $U\in\mathbb R^{n\times k}$ and some $B\in\mathbb R^{m\times k}$. Why do we know that it must hold $k\le m$?

Answer 1

First note that it can hold with $k > m$. Consider the example $$A^T = \begin{bmatrix} 1 & 0 \\ 0&1 \\ 0 & 0\end{bmatrix} = I_3 \cdot \begin{bmatrix} 1 & 0 \\ 0&1 \\ 0 & 0 \end{bmatrix}$$

Also note that it can only hold for $k \leq m$ if $A^T$ is of rank $k$ or less. The reason is because the rank of a product $AB$ of matrices is at most $\min(\text{rank}(A), \text{rank}(B))$. If $\text{rank}(A^T) > k$ and $A^T = U B^T$ with $U \in \mathbb{R}^{n \times k}$, then we arrive at a contradiction, because the rank of the right-hand side can at most be $k$, because the rank of $U$ is $k$.

EDIT: The claim that it must be that $k \leq m$ if $A^T = U B^T$ when $U$ is right-orthogonal is also false. Let $$A^T = [1], ~ U = \begin{bmatrix} 1/2 & 1/2 & 1/2 & 1/2 \end{bmatrix}, ~ B = U^T$$ Note that it is true that $n \leq k$ in this case since otherwise we would have two matrices $U$ and $U^T$ each of at most rank $k$ multiplying to give a rank $n$ matrix.