Let \(\mathcal{V}\) be a vector space over the field \(\mathbb{F}.\) A bilinear form on \(\mathcal{V}\) is a function from pairs of vectors in \(\mathcal{V}\) to \(\mathbb{F}\) written \(\langle \cdot, \cdot \rangle : \mathcal{V}\times\mathcal{V} \rightarrow \mathbb{F}\) and satisfies the bilinear form axioms:

\[\langle \lambda\underline{v}_1+\mu\underline{v}_2, \underline{v}_3 \rangle = \lambda\langle \underline{v}_1, \underline{v}_3 \rangle + \mu\langle \underline{v}_2, \underline{v}_3 \rangle \,\, \forall \, \underline{v}_1, \underline{v}_2, \underline{v}_3 \in \mathcal{V},\,\,\forall \, \lambda, \mu \in \mathbb{F}\] \[\langle \underline{v}_1, \lambda\underline{v}_2+\mu\underline{v}_3 \rangle = \lambda\langle \underline{v}_1, \underline{v}_2 \rangle + \mu\langle \underline{v}_1, \underline{v}_3 \rangle \,\, \forall \, \underline{v}_1, \underline{v}_2, \underline{v}_3 \in \mathcal{V},\,\,\forall \, \lambda, \mu \in \mathbb{F}.\]

A sesquilinear form on \(\mathcal{V}\) satisfies the first bilinear form axiom but is conjugate linear in its second argument:

\[\langle \underline{v}_1, \lambda\underline{v}_2+\mu\underline{v}_3 \rangle = \overline{\lambda}\langle \underline{v}_1, \underline{v}_2 \rangle + \overline{\mu}\langle \underline{v}_1, \underline{v}_3 \rangle \,\, \forall \, \underline{v}_1, \underline{v}_2, \underline{v}_3 \in \mathcal{V},\,\,\forall \, \lambda, \mu \in \mathbb{F}\]

where for a complex number \(z=a+ib\in\mathbb{C}\) the complex conjugate is \(\overline{z}=a-ib\in\mathbb{C}.\) If \(\mathbb{F}=\mathbb{R}\) then conjugation has no effect and sesquilinear and bilinear forms are the same.

A sesquilinear form is conjugate symmetric if \(\langle \underline{v}_1, \underline{v}_2 \rangle = \overline{\langle \underline{v}_2, \underline{v}_1 \rangle} \,\, \forall \, \underline{v}_1, \underline{v}_2 \in \mathcal{V}.\) If \(\mathbb{F}=\mathbb{R}\) conjugate symmetry is simply symmetry: \(\langle \underline{v}_1, \underline{v}_2 \rangle = \langle \underline{v}_2, \underline{v}_1 \rangle.\) Conjugate symmetry implies that \(\langle \underline{v}, \underline{v} \rangle\) is real for every \(\underline{v} \in \mathcal{V}.\)

A conjugate symmetric form is positive definite if \(\langle \underline{v}, \underline{v} \rangle > 0 \,\, \forall \, \underline{v} \in \mathcal{V} \setminus \{\underline{0}\},\) negative definite if \(\langle \underline{v}, \underline{v} \rangle < 0 \,\, \forall \, \underline{v} \in \mathcal{V} \setminus \{\underline{0}\},\) positive semi-definite if \(\langle \underline{v}, \underline{v} \rangle \geq 0 \,\, \forall \, \underline{v} \in \mathcal{V} \setminus \{\underline{0}\}\) and negative semi-definite if \(\langle \underline{v}, \underline{v} \rangle \leq 0 \,\, \forall \, \underline{v} \in \mathcal{V} \setminus \{\underline{0}\}.\)

An inner product is a positive definite, conjugate symmetric sesquilinear form on \(\mathcal{V}.\) When \(\mathbb{F}=\mathbb{R}\) this is a positive definite, symmetric bilinear form. When \(\mathbb{F}=\mathbb{C}\) the form cannot be bilinear: a conjugate symmetric bilinear form would give \(\langle i\underline{v}, i\underline{v} \rangle = -\langle \underline{v}, \underline{v} \rangle,\) contradicting positive definiteness. A vector space is an inner product space if it is equipped with the inner product.

Let \(\mathcal{V}\) be an inner product space over the field \(\mathbb{F}\) and \(\underline{v} \in \mathcal{V}.\) The norm or length \(\|\underline{v}\|\) of \(\underline{v}\) is the square root of the inner product with itself:

\[\|\underline{v}\|:=\sqrt{\langle \underline{v},\underline{v} \rangle} \in \mathbb{R}\]

Let \(\mathcal{V}\) be an inner product space over \(\mathbb{R}.\) The angle \(\theta\) between two non-zero vectors \(\underline{u},\underline{v} \in \mathcal{V}\) is defined as the inverse cosine of the quotient of the inner product of the two vectors and the product of their norms:

\[\theta:=\cos^{-1}\left(\frac{\langle \underline{u},\underline{v} \rangle}{\|\underline{u}\|\|\underline{v}\|}\right) \in [0, \pi]\]

The Cauchy-Schwarz inequality below guarantees that the quotient lies in \([-1, 1].\)

Two vectors \(\underline{u}, \underline{v} \in \mathcal{V}\) are orthogonal to each other \(\underline{u} \perp \underline{v}\) if \(\langle \underline{u},\underline{v} \rangle=0.\) If \(\underline{u}\) and \(\underline{v}\) are orthogonal unit vectors \(\|\underline{u}\|=\|\underline{v}\|=1\) then they are orthonormal.

For a subset of vectors \(U \subseteq \mathcal{V},\) they are said to be an orthogonal set or orthonormal set if all vectors contained within the sets are pairwise orthogonal or orthonormal respectively. Orthonormal sets are linearly independent and a set of \(n\) orthonormal vectors in an \(n\)-dimensional vector space is a basis.

The Gram-Schmidt orthonormalisation procedure is a method to orthonormalise a set of linearly independent vectors \(\underline{u}_1,\ldots\underline{u}_n \in \mathcal{U}\) in an inner product space \(\mathcal{U}.\) The resultant orthonormalised vectors \(\underline{v}_1,\ldots\underline{v}_n\) span the same subspace as \(\underline{u}_1,\ldots\underline{u}_n;\) more precisely, \(\underline{v}_1,\ldots\underline{v}_k\) span the same subspace as \(\underline{u}_1,\ldots\underline{u}_k\) for each \(k.\) If \(\underline{u}_1,\ldots\underline{u}_n\) is a basis of \(\mathcal{U}\) then \(\underline{v}_1,\ldots\underline{v}_n\) is an orthonormal basis of \(\mathcal{U}.\)

\[\begin{matrix} \underline{w}_1 & := & \underline{u}_1 & \underline{v}_1 & := & \frac{\underline{w}_1}{\|\underline{w}_1\|} \\ \underline{w}_2 & := & \underline{u}_2-\langle \underline{u}_2, \underline{v}_1 \rangle\underline{v}_1 & \underline{v}_2 & := & \frac{\underline{w}_2}{\|\underline{w}_2\|} \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ \underline{w}_n & := & \underline{u}_n-\sum_{i=1}^{n-1}\langle \underline{u}_n, \underline{v}_i \rangle\underline{v}_i & \underline{v}_n & := & \frac{\underline{w}_n}{\|\underline{w}_n\|} \\ \end{matrix}\]

The Cauchy-Schwarz inequality states that for an inner product space \(\mathcal{V}\) and \(\underline{v}_1,\underline{v}_2\in\mathcal{V}:\)

\[|\langle\underline{v}_1,\underline{v}_2\rangle|\leq\|\underline{v}_1\|\|\underline{v}_2\|\]

with equality if and only if \(\underline{v}_1\) and \(\underline{v}_2\) are linearly dependent.