= Solution
Assume for contradiction that $n>m$. Apply part b to the <matrix> of the given relation <coefficients>, whose columns are nonzero. For its first dependent prefix choose $b_1,\ldots,b_k$ with
$$
\sum_{j=1}^ka_{ij}b_j=0\quad\text{for all }i,
\qquad z_i:=\sum_{j=1}^ka_{ij}\lambda_jb_j\text{ not all zero}.
$$
Multiply the $j$th given vector relation by $b_j$ and sum. Interchanging the finite sums gives
$$
0=\sum_{i=1}^m\left(\sum_{j=1}^ka_{ij}b_j\right)v_i
+\sum_{i=1}^m\left(\sum_{j=1}^ka_{ij}\lambda_jb_j\right)w_i
=\sum_{i=1}^mz_iw_i.
$$
The $w_i$ are a <basis>, so all $z_i$ must be zero, contradicting part b. Therefore \b[$n\le m$].
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