Let \(V\) be a finite-dimensional vector space and let \(W_1\) be any subspace of \(V\). Prove that there is a subspace \(W_2\) of \(V\) such that \(V = W_1 \oplus W_2\).
Solution
1. Extend a basis of \(W_1\) to a basis of \(V\).
Choose a basis \[ w_1,\ldots,w_r \] of \(W_1\). Since \(V\) is finite dimensional, the basis-extension theorem gives vectors \[ v_{r+1},\ldots,v_n \] such that \[ w_1,\ldots,w_r,v_{r+1},\ldots,v_n \] is a basis of \(V\) [@HK71]. Define \[ W_2=\operatorname{span}\{v_{r+1},\ldots,v_n\}. \]
2. The sum \(W_1+W_2\) is all of \(V\) and is direct.
The basis from step 1 spans \(V\), so every vector in \(V\) is a sum of a linear combination of the \(w_i\) and a linear combination of the \(v_j\). Hence \[ V=W_1+W_2. \]
If \(x\in W_1\cap W_2\), then for some scalars \(a_i,b_j\), \[ x=\sum_{i=1}^r a_iw_i =\sum_{j=r+1}^n b_jv_j. \] Thus \[ \sum_{i=1}^r a_iw_i-\sum_{j=r+1}^n b_jv_j=0. \] Linear independence of the extended basis forces all \(a_i\) and \(b_j\) to be zero. Therefore \(x=0\) and \[ W_1\cap W_2=\{0\}. \] Consequently \[ \boxed{V=W_1\oplus W_2}. \]