Basis Calculation (11.1.1)

Dummit and Foote Abstract Algebra, section 11.1, exercise 1:

Let $V=\mathbb{R}^n$ and let $(a_1,...,a_n)∈V$ be fixed. Let $W \subseteq V$ be the set of vectors $(x_1,...,x_n)$ such that $x_1a_1+...+x_na_n=0$. Prove $W$ is a subspace and find a basis for $W$.

If all $a_i=0$ then clearly $W=V$ and $e_1,...,e_n$ suffice as a basis. Otherwise, let $a_m≠0$. For $(x_1,...,x_n),(y_1,...,y_n)∈W$ we see$$(x_1,...,x_n)-r(y_1,...,y_n)=(x_1-ry_1,...,x_n-ry_n)$$as well as$$(x_1-ry_1)a_1+...+(x_n-ry_n)a_n=$$$$(x_1a_1+...+x_na_n)-r(y_1a_1+...+y_na_n)=0$$so that $W$ is a subspace. We claim $e_i-(a_i/a_m)e_m$ for all $i∈\{1,...,m-1,m+1,...,n\}$ is a basis for $W$. Foremost, all of these are seen to be vectors of $W$. Linear independence:$$b_1(e_1-(a_1/a_m)e_m)+...+b_{m-1}(e_{m-1}-(a_{m-1}/a_m)e_m)+$$$$b_{m+1}(e_{m+1}-(a_{m+1}/a_m)e_m)+...+b_n(e_n-(a_n/a_m)e_m)=0⇒$$$$b_1e_1+...+b_{m-1}e_{m-1}+b_{m+1}e_{m+1}+...+b_ne_n-(\sum_{i≠m} b_i a_i/a_m)e_m=0⇒$$$$b_1,...b_{m-1},b_{m+1},...,b_n=0$$This implies $W$ is either of dimension $n-1$ or $n$. If it were the latter, then we would have $W=V$ despite $e_m∉W$ so that necessarily $W$ is of dimension $n-1$ and now this set must be a basis.$~\square$