Linear-2 8

Linear independence of orthogonal sets #lemma

\begin{matrix} S \subseteq V is orthogonal and 0 \notin S ⟹ S is a linear independence \end{matrix}

\begin{matrix} Let V be an inner product space over F \\ Let B be an orthogonal basis of V \\ Let v \in V \\ \exists {α_{i}}_{i \in [1, n]} : v = \sum_{i = 1}^{n} α_{i} v_{i} \\ ⟹ \forall i \in [1, n] : α_{i} = \frac{⟨ v, v_{i} ⟩}{‖ v ‖^{2}} \\ Or in other words: [v]_{B} = \frac{1}{‖ v ‖^{2}} (\begin{matrix} ⟨ v, v_{1} ⟩ \\ ⋮ \\ ⟨ v, v_{n} ⟩ \end{matrix}) \\ Proof: \\ Let {α_{i}}_{i \in [1, n]} : v = \sum_{i = 1}^{n} α_{i} v_{i} \\ \forall i \in [1, n] : ⟨ v, v_{i} ⟩ = ⟨ \sum_{k = 1}^{n} α_{k} v_{k}, v_{i} ⟩ = \sum_{k = 1}^{n} α_{k} ⟨ v_{k}, v_{i} ⟩ = α_{i} ⟨ v_{i}, v_{i} ⟩ = α_{i} ‖ v_{i} ‖^{2} \\ \forall i \in [1, n] : v_{i} \neq 0 ⟹ ‖ v_{i} ‖ > 0 \\ ⟹ \forall i \in [1, n] : α_{i} = \frac{⟨ v, v_{i} ⟩}{‖ v_{i} ‖^{2}} \end{matrix}

\begin{matrix} Let B be an orthogonal basis of V \\ Let v \in V \\ v = \sum_{i = 1}^{n} α_{i} v_{i} \\ ⟹ {‖ \sum_{i = 1}^{n} α_{i} v_{i} ‖}^{2} = \sum_{i = 1}^{n} ‖ α_{i} v_{i} ‖^{2} = \sum_{i = 1}^{n} {| α_{i} |}^{2} ‖ v_{i} ‖^{2} \\ Proof: \\ {‖ \sum_{i = 1}^{n} α_{i} v_{i} ‖}^{2} = ⟨ \sum_{i = 1}^{n} α_{i} v_{i}, \sum_{i = 1}^{n} α_{i} v_{i} ⟩ = \sum_{i = 1}^{n} \sum_{j = 1}^{n} ⟨ α_{i} v_{i}, α_{j} v_{j} ⟩ = \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{n} α_{i} \overset{―}{α_{j}} ⟨ v_{i}, v_{j} ⟩ = \sum_{i = 1}^{n} α_{i} \overset{―}{α_{i}} ⟨ v_{i}, v_{i} ⟩ = \sum_{i = 1}^{n} {| α_{i} |}^{2} ‖ v_{i} ‖^{2} \end{matrix}

\begin{matrix} Let V be an inner product space over F \\ Let S \subseteq V \\ Set of vectors that are orthogonal to all vectors in S is then called an \\ orthogonal complement and denoted \\ S^{⊥} = {v \in V | \forall s \in S : ⟨ v, s ⟩ = 0} \end{matrix}

\begin{matrix} Let V be an inner product space over F \\ S \subseteq V ⟹ S^{⊥} is a subpspace of V \\ Proof: \\ \forall s \in S : ⟨ 0, s ⟩ = 0 ⟹ 0 \in S^{⊥} \\ Let v, u \in S^{⊥}, α \in F \\ \forall s \in S : ⟨ v + α u, s ⟩ = ⟨ v, s ⟩ + α ⟨ u, s ⟩ = 0 + α \cdot 0 = 0 \\ ⟹ v + α u \in S^{⊥} ⟹ S^{⊥} is a subspce of V \end{matrix}

\begin{matrix} S \subseteq (S^{⊥})^{⊥} \\ Proof: \\ (S^{⊥})^{⊥} = {v \in V | \forall s^{'} \in S^{⊥} : ⟨ v, s^{'} ⟩ = 0} \\ Let s \in S \\ \forall s^{'} \in S^{⊥} : ⟨ s^{'}, s ⟩ = 0 ⟹ ⟨ s, s^{'} ⟩ = 0 ⟹ s \in (S^{⊥})^{⊥} ⟹ S \subseteq (S^{⊥})^{⊥} \end{matrix}

\begin{matrix} A \subseteq B ⟹ A^{⊥} \supseteq B^{⊥} \\ Proof: \\ Let v \in B^{⊥} \\ \forall b \in B : ⟨ v, b ⟩ = 0 \underset{A \subseteq B}{⟹} \forall a \in A : ⟨ v, a ⟩ = 0 ⟹ v \in A^{⊥} \\ ⟹ B^{⊥} \subseteq A^{⊥} \end{matrix}

\begin{matrix} S^{⊥} = (s p (S))^{⊥} \\ Proof: \\ S \subseteq s p (S) ⟹ (s p (S))^{⊥} \subseteq S^{⊥} \\ Let v \in S^{⊥} \\ Let u \in s p (S) \\ u = \sum_{i = 1}^{k} α_{i} s_{i} \\ ⟨ v, u ⟩ = ⟨ v, \sum_{i = 1}^{k} α_{i} s_{i} ⟩ = \sum_{i = 1}^{k} \overset{―}{α_{i}} ⟨ v, s_{i} ⟩ = 0 \\ ⟹ v \in (s p (S))^{⊥} ⟹ S^{⊥} \subseteq (s p (S))^{⊥} \\ ⟹ S^{⊥} = (s p (S))^{⊥} \end{matrix}

\begin{matrix} Let V be an inner product space \\ Let W be a subspace of V \\ Let B be an orthogonal basis of W \\ Let v \in V \\ Then projection P_{W} (v) = \sum_{i = 1}^{k} \frac{⟨ v, w_{i} ⟩}{‖ w_{i} ‖^{2}} w_{i} \\ is a vector such that \forall w \in W : ‖ v - w ‖ \geq ‖ v - P_{W} (v) ‖ \end{matrix}

\begin{matrix} \forall v \in V : P_{W} (v) \in W \end{matrix}

\begin{matrix} v \in W ⟺ P_{W} (v) = v \\ Proof: \\ ⟹ Let v \in W \\ B is an orthogonal basis ⟹ v = \sum_{i = 1}^{k} \frac{⟨ v, w_{i} ⟩}{‖ w_{i} ‖^{2}} w_{i} \\ ⟹ v = P_{W} (v) \\ ⟸ Let P_{W} (v) = v \\ P_{W} (v) \in W ⟹ v \in W \end{matrix}

\begin{matrix} P_{W} (v) = 0 ⟺ v \in W^{⊥} \\ Proof: \\ ⟹ Let P_{W} (v) = 0 \\ B is a linear independence ⟹ \forall w_{i} \in W : \frac{⟨ v, w_{i} ⟩}{‖ w_{i} ‖^{2}} = 0 \\ ⟹ \forall w_{i} \in B : ⟨ v, w_{i} ⟩ = 0 ⟹ v \in B^{⊥} = (s p (B))^{⊥} = W^{⊥} \\ ⟸ Let v \in W^{⊥} \\ ⟹ \forall w_{i} \in B : ⟨ v, w_{i} ⟩ = 0 ⟹ P_{W} (v) = 0 \end{matrix}

\begin{matrix} w \in W ⟺ \forall v \in V : ⟨ v - P_{W} (v), w ⟩ = 0 \\ Proof: \\ w \in W ⟹ w = \sum_{i = 1}^{k} α_{i} w_{i} \\ ⟨ v - P_{W} (v), w ⟩ = 0 ⟺ ⟨ v, w ⟩ = ⟨ P_{W} (v), w ⟩ \\ ⟨ v, w ⟩ = ⟨ v, \sum_{i = 1}^{k} α_{i} w_{i} ⟩ = \sum_{i = 1}^{k} \overset{―}{α_{i}} ⟨ v, w_{i} ⟩ \\ ⟨ P_{W} (v), w ⟩ = ⟨ P_{W} (v), \sum_{i = 1}^{k} α_{i} w_{i} ⟩ = \sum_{i = 1}^{k} β_{i} ⟨ w_{i}, w ⟩ = \\ = \sum_{i = 1}^{k} \sum_{j = 1}^{k} β_{i} \overset{―}{α_{j}} ⟨ w_{i}, w_{j} ⟩ = \sum_{i = 1}^{k} \frac{⟨ v, w_{i} ⟩}{‖ w_{i} ‖^{2}} \overset{―}{α_{i}} ⟨ w_{i}, w_{i} ⟩ = \sum_{i = 1}^{k} \overset{―}{α_{i}} ⟨ v, w_{i} ⟩ \\ ⟹ ⟨ v, w ⟩ = ⟨ P_{W} (v), w ⟩ \end{matrix}