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Let $A$ and $B$ be two matrices of dimension $n \times m$ where $m < n$ and the ranks of $A$ and $B$ are $m$. One can show that if $\operatorname{im}(A) = \operatorname{im}(B)$ then $A^{\top} B$ is invertible ($\mathrm{im}(A)$ is the image of $A$). However, the reverse is not true -- $A^{\top} B$ does not mean that $\operatorname{im}(A) = \operatorname{im}(B)$.

Still, if we know $A^{\top}B$ is invertible, what can we say about the relationship between $A$ and $B$, or some relationship between images of $A$ and $B$?

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    sorry, forgot to include a rank condition. I fixed it.2011-12-25

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$A^T B$ is invertible iff $B$ is injective and $\text{im}(A)^\perp \cap \text{im}(B) = \{0\}$.

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    In general, if $\text{im}(B) \subset \text{im}(A)$, then $\text{im}(A)^\perp \cap \text{im}(B) \subset \text{im}(B)^\perp \cap \text{im}(B) = \{0\}$. It doesn't go the other way, except that if $\text{im}(A)$ is a proper subset of $\text{im}(B)$, $\text{im}(A)^\perp \cap \text{im}(B) \ne \{0\}$. Think of two one-dimensional subspaces $L_1, L_2$ of ${\mathbb R}^2$ (lines through the origin in the plane): $L_1^\perp \cap L_2 = \{0\}$ unless the lines are at right angles.2011-12-25