# intersection of row spaces

Assume you are given two full rank matrices with the same number of columns $$A$$ and $$B$$. That is, $$A$$ is $$n\times m$$, $$B$$ is $$k\times m$$, and rank$$(A)=n$$, rank$$(B)=k$$ (where we have assumed $$n,k\leq m$$). Is it possible to determine the dimension of the intersection of their rowspaces?

For me it is tempting to reason as follows: Let $$I$$ and $$J$$ be the pivot (see below what I mean) positions of $$A$$ and $$B$$ respectively. Does the dimension in question equal to $$|I\cap J|$$?

By $$\textbf{pivot positions}$$ I mean the columns in which the identity matrix shows up after necessary row operations.

I am primarily interested on binary matrices, but the question can be asked in any field.

• " two full rank matrices with the same number of columns " ? To me "full rank" means the number of columns and rows are the same? In which case the result is truly elementary. – rrogers Aug 13 at 14:07
• @rrogers No, it doesn't mean that. To elaborate, you are given two matrices $n\times m$ and $k\times m$ respectively, each of which is full rank. But I made an edit anyway. – user 1987 Aug 13 at 14:22
• I would use exterior products on the rows and compare the subsequent volumes/subspaces. Alternately I think that using Psuedo-inverses can be carried out. – rrogers Aug 13 at 14:46
• The advantage of exterior products is that it is a steamroller without any particular decisions about contents. – rrogers Aug 13 at 15:29
• By $|I\cap J|$ do you mean rank $(I \cap J)$? – Marc Bogaerts Aug 13 at 17:23

Label as $$a_k$$ the row vectors of $$A$$ that don't belong to $$I \cap J$$, $$i_l$$ the vectors in $$I \cap J$$ and $$b_m$$ the vectors of $$B$$ not in $$I \cap J$$. Then it is not hard to see that the vectors $$a_k, i_l, b_m$$ form an lineary independent set. If $$v$$ is a linear combination of the $$i_l$$ then it is clear that $$v \in \operatorname{span}(A)$$ and also $$v \in \operatorname{span}(B)$$. If on the other hand $$v \in \operatorname{span}(A) \cap \operatorname{span}(A)$$ then $$v = \sum_k \alpha_k a_k + \sum_l \beta_l i_l$$ and also $$v = \sum_m \gamma_m b_m + \sum_l \delta_l i_l$$ for some $$\alpha_k, \beta_l, \gamma_m, \delta_l$$ from which $$\sum_k \alpha_k a_k-\sum_m \gamma_m b_m + \sum_l \beta_l i_l -\sum_l \delta_l i_l = 0$$ and, because of linear independency $$\alpha_k = \gamma_m = 0$$ and $$\beta_l = \delta_l$$. This shows that the $$i_l$$ forms a basis of the intersection of the rowspaces and that your assertion about the dimension is correct.