> [!NOTE] Definition (Orthogonal complement) > Let $V=\mathbb{R}^{n}$ equipped with [[Dot Product in Real n-Space|dot product]], and let $U\subset V$ be a [[Vector subspace|subspace]]. We define the orthogonal complement $U^{\perp}$ of $U$ in $V$ to be the set of elements of $V$ that are orthogonal to every element of $U:$ $U^{\perp} = \{ v\in V \mid u \cdot v = 0 \quad \text{for all } u\in U \}$ # Properties > [!NOTE] Theorem (Orthogonal complements) > Then $U^{T}$ is a subspace of $V,$ and moreover it is a [[Existence of complement of subspace of finite-dimensional vector space|complement]] to $U$ > >*Proof*. Choose any basis $B$ of $U$ and apply [[Gram-Schmidt orthogonalisation in real n-space|Gram-Schmidt]]. The result is an orthonormal basis $v_{1},\dots,v_{s}$ of $U.$ We can extend this set to an orthonormal basis $v_{1},\dots,v_{n}$ of $V.$ Claim: $U^{\perp} = \langle v_{s+1},\dots,v_{n}\rangle$which completes the proof. Clearly $v_{j}$ for $j\geq s+1$ lies in $U^{\perp},$ while conversely if $v=\lambda_{1}v_{1}+\dots+\lambda_{n}v_{n}\in U^{\perp},$ then for all $j=1,\dots,s$ $0=v_{j}\cdot v = v_{j} \cdot (\lambda_{1}v_{1}+\dots+\lambda_{n}v_{n})=\lambda_{j}$so in fact $v\in\langle v_{s+1},\dots,v_{n}\rangle$ as claimed.