Define:
Let \(\boldsymbol{v}, \boldsymbol{w} \in \mathbb{R}^n\). The inner product of \(\boldsymbol{v}\) and \(\boldsymbol{w}\) is defined as: \(<\boldsymbol{v}, \boldsymbol{w}> = \boldsymbol{v}^T\boldsymbol{w} = \boldsymbol{w}^T \boldsymbol{v} = <\boldsymbol{w}, \boldsymbol{v}>\)
Define:
The norm (length) of a vector \(\boldsymbol{v} \in R^n\) is \(\|\boldsymbol{v}\| = \sqrt{\boldsymbol{v}^T\boldsymbol{v}}\), \(\|\boldsymbol{v}\| = 0\) if only if \(\boldsymbol{v} = \boldsymbol{0}\).
Define:
\(\boldsymbol{v}\) and \(\boldsymbol{w}\) are orthiginal if only if \(\|\boldsymbol{v} + \boldsymbol{w}\|^2 = \|\boldsymbol{v}\|^2 + \|\boldsymbol{w}\|^2\), i.e. \(<\boldsymbol{v}, \boldsymbol{w}> = 0\)
Define:
Two subspaces \(V\) and \(W\) are within a vector space and orthogoanl if every \(\boldsymbol{v} \in V\) is orthogonal to every \(\boldsymbol{w} \in W\).
Define:
The orthogonal complement of a subspace \(V\) contains every vector that is orthogonal every vector in \(V\), which is denoted by \(V^\perp\)
(i)
\(V^\perp\) is a subspace.
(ii)
\(V^\perp\) is unique to \(V\).
Hint for Proof:
(\(V^\perp\) contains every vector that is orthogonal to vectors in \(V\).)
Orthogonality is impossible when \(dim(V_1) + dim(V_2) > dim(V)\), where \(V_1\) and \(V_2\) are subspaces of vector space \(V\).
Proof:
If \(dim(V_1) + dim(V_2) > dim(V)\), there exists \(\boldsymbol{v} \in V_1, \boldsymbol{v} \neq \boldsymbol{0}\) such that it is a linear combination of vectors of in \(V_2\), i.e. \(\boldsymbol{v} \in V_2\)
\(\because\) When a vector is in two orthogonal subspaces, it must be \(\boldsymbol{0}\)
\(\therefore V_1 \not\perp V_2\)