How do we prove that if $\{v_1,...,v_m\}$ and $\{w_1,...,w_m\}$ are bases for a real vector space, then there are at most $m$ real numbers $\lambda$ such that $v_1+\lambda w_1,...,v_m+\lambda w_m$ are linearly dependent, using the following fact?
If $A=(a_{ij})$ is an $m\times n$ matrix with no all-zero column and $n>m$, and $k$ is minimal such that the first $k$ columns of A are linearly dependent, then there exist scalars $b_1,...,b_k$ such that $\Sigma_{j=1}^k a_{ij}b_j =0 \:\forall i$ but for any distinct scalars $\lambda_1,...,\lambda_k$, there exists $i$ such that $\Sigma_{j=1}^k a_{ij}b_j\lambda_j \not= 0$.
I think we should assume $n>m$ and let $A=(a_{ij})$ be such that $\Sigma_{i=1}^m a_{ij}(v_i+\lambda w_i) =0 \:\forall j$. Then we've got a matrix $A$ as in the second paragraph, but I can't see how to use the scalars $b_j$. Can they be interpreted as components of $v_i$ or $w_i$? And how do we get around the fact that we're summing over $i$ here and over $j$ in the second paragraph?
(The problem can be solved easily using determinants, but I need to do it using the second paragraph!)
Many thanks for any help with this!