# what can we say about $\operatorname{im}(A)$ and $\operatorname{im}(B)$?

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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Taking $A$ and $B$ both to be the $n \times m$ zero matrix, we find that $\operatorname{im} A = \operatorname{im} B =$ the trivial subspace, and yet $A^\top B$ is not invertible. Doesn't this refute your first sentence? –  Srivatsan Dec 25 '11 at 4:38
sorry, forgot to include a rank condition. I fixed it. –  harmonic Dec 25 '11 at 4:47

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

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can $im(A)^{\top} \cap im(B) = {0}$ mean $im(B) \subset im(A)$ or anything like that? –  harmonic Dec 25 '11 at 11:27
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. –  Robert Israel Dec 25 '11 at 20:37