Projection

If $(V, ⟨ \cdot, \cdot ⟩)$ is an inner product space and $u, v \in V$ , the projection of v onto u is:

{proj}_{u} (v) = \frac{⟨ u, v ⟩}{⟨ u, u ⟩} u

If $U \subseteq V$ is a fdvs with an orthogonal basis $B = {b_{1}, b_{2}, \dots, b_{n}}$ then we define the projection of $V$ onto $U$ as:

{proj}_{u} (v) = \sum_{i = 1}^{n} \frac{⟨ b_{i}, v ⟩}{b_{i}, b_{i}} b_{i} = \sum_{i = 1}^{n} {proj}_{b_{i}} (v)

If $(V, ⟨ \cdot, \cdot ⟩)$ is an inner product space and $U$ is a fdvs with a basis $B = b_{1}, \dots, b_{n}$ then $\forall v \in V$ :

v - {proj}_{u} (v) \in U^{⊥}

\begin{aligned} ⟨ u, v - {proj}_{u} (v) ⟩ & = ⟨ u, v - \sum_{i = 1}^{n} \frac{⟨ v, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} ⟩ \\ = ⟨ u, v ⟩ - \sum_{i = 1}^{n} \frac{⟨ v, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} ⟨ u, b_{i} ⟩ \\ = ⟨ \sum_{n = 1}^{n} \frac{⟨ u, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} b_{i}, v ⟩ - \sum_{i = 1}^{n} \frac{⟨ v, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} ⟨ u, b_{i} ⟩ \\ = \sum_{n = 1}^{n} \frac{⟨ u, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} ⟨ v, b_{i} ⟩ - \sum_{i = 1}^{n} \frac{⟨ v, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} ⟨ u, b_{i} ⟩ \\ = \sum_{i = 1}^{n} \frac{⟨ v, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} ⟨ u, b_{i} ⟩ - \sum_{i = 1}^{n} \frac{⟨ v, b_{i} ⟩}{⟨ b_{i}, b_{i} ⟩} ⟨ u, b_{i} ⟩ \\ = 0 \end{aligned}

see: perp

$(V, ⟨ \cdot, \cdot ⟩)$ is finite with basis $B = {b_{1}, \dots, b_{n}}$ . Define a new basis $F = {f_{1}, \dots, f_{n}}$ inductively as:

$f_{1} = b_{1}$
Suppose $f_{1}, \dots, f_{n}$ has been defined. $U_{k} = span {f_{1}, \dots, f_{k}}$ then $f_{k + 1} = b_{k + 1} - {proj}_{k} (b_{k + 1})$