Computing a rank

Question

Consider positive, i.e. non-zero, integers $n<m$, an $n$-dimensional vector space with $v_1,...,v_m\in V$ and elements $\alpha_1,...\alpha_m\in V^\ast$.

We define a matrix $A=(a_{ij})$ with $a_{ij}=\alpha_i(v_j)$.

Why do we have $\text{rk}~A\leq n$?

Answer 1

Select a basis $(w_1,\dots,w_n)$ of $V$. Write $$ v_j = \sum_{k=1}^n c_{jk} w_k $$ We note that the $j$th column of $A$ is given by $$ (\alpha_1(v_j),\dots,\alpha_n(v_j)) = \sum_{k=1}^n c_{jk} (\alpha_1(w_k),\dots,\alpha_n(w_k)) $$ Thus, the vectors $(\alpha_1(w_k),\dots,\alpha_n(w_k))$ (for $k = 1,\dots,n$) span the column space of $A$. Thus, the dimension of the column space (and hence the rank of $A$) is at most $n$.