Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $B$?

share|improve this question
2  
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

1 Answer 1

up vote 1 down vote accepted

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.