> [!NOTE] Lemma (Existence of subspace of fdvs complement) > Let $V$ be a [[Vector spaces|FDVS]] and $U\subset V$ a subspace. Then there exists a complement $U'\subset V$ to $U$. *Proof*. Let $u_{1},\dots,u_{s}$ be a basis of $U$. We can extend it, by [[Basis of Vector Space#^df1788|sifting lemma]], to a basis $u_{1},\dots, u_{s},v_{1},\dots,v_{r}$ of $V$. Let $U'=\langle v_{1},\dots,v_{r}\rangle$. Clearly $U+U'=V$. If $w\in U\cap U'$ then $w=\lambda_{1}u_{1}+\dots+\lambda_{s}u_{s} \quad \text{and}\quad w = \mu_{1} u_{1}+\dots+\mu_{r} v_{r} $for unique scalars $\lambda_{i},\mu_{j}$. Subtracting these two expressions gives $\lambda_{1}u_{1}+\dots+\lambda_{s}u_{s} - \mu_{1}v_{1}-\dots-\mu_{r}v_{r}=0_{V}$so by linear independence, all the coefficients are zero and so $w=0_{V}$. Thus $U \cap U'=\{ 0_{V} \}$ and $U'$ is a complement to $U$.