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

⟨λv‾1+μv‾2,v‾3⟩=λ⟨v‾1,v‾3⟩+μ⟨v‾2,v‾3⟩  ∀ v‾1,v‾2,v‾3∈V,  ∀ λ,μ∈F\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} ⟨v‾1,λv‾2+μv‾3⟩=λ⟨v‾1,v‾2⟩+μ⟨v‾1,v‾3⟩  ∀ v‾1,v‾2,v‾3∈V,  ∀ λ,μ∈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 V\mathcal{V} satisfies the first bilinear form axiom but is conjugate linear in its second argument:

⟨v‾1,λv‾2+μv‾3⟩=λ‾⟨v‾1,v‾2⟩+μ‾⟨v‾1,v‾3⟩  ∀ v‾1,v‾2,v‾3∈V,  ∀ λ,μ∈F\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∈Cz=a+ib\in\mathbb{C} the complex conjugate is z‾=a−ib∈C.\overline{z}=a-ib\in\mathbb{C}. If F=R\mathbb{F}=\mathbb{R} then conjugation has no effect and sesquilinear and bilinear forms are the same.

A sesquilinear form is conjugate symmetric if ⟨v‾1,v‾2⟩=⟨v‾2,v‾1⟩‾  ∀ v‾1,v‾2∈V.\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 F=R\mathbb{F}=\mathbb{R} conjugate symmetry is simply symmetry: ⟨v‾1,v‾2⟩=⟨v‾2,v‾1⟩.\langle \underline{v}_1, \underline{v}_2 \rangle = \langle \underline{v}_2, \underline{v}_1 \rangle. Conjugate symmetry implies that ⟨v‾,v‾⟩\langle \underline{v}, \underline{v} \rangle is real for every v‾∈V.\underline{v} \in \mathcal{V}.

A conjugate symmetric form is positive definite if ⟨v‾,v‾⟩>0  ∀ v‾∈V∖{0‾},\langle \underline{v}, \underline{v} \rangle > 0 \,\, \forall \, \underline{v} \in \mathcal{V} \setminus \{\underline{0}\}, negative definite if ⟨v‾,v‾⟩<0  ∀ v‾∈V∖{0‾},\langle \underline{v}, \underline{v} \rangle < 0 \,\, \forall \, \underline{v} \in \mathcal{V} \setminus \{\underline{0}\}, positive semi-definite if ⟨v‾,v‾⟩≥0  ∀ v‾∈V∖{0‾}\langle \underline{v}, \underline{v} \rangle \geq 0 \,\, \forall \, \underline{v} \in \mathcal{V} \setminus \{\underline{0}\} and negative semi-definite if ⟨v‾,v‾⟩≤0  ∀ v‾∈V∖{0‾}.\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 V.\mathcal{V}. When F=R\mathbb{F}=\mathbb{R} this is a positive definite, symmetric bilinear form. When F=C\mathbb{F}=\mathbb{C} the form cannot be bilinear: a conjugate symmetric bilinear form would give ⟨iv‾,iv‾⟩=−⟨v‾,v‾⟩,\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 V\mathcal{V} be an inner product space over the field F\mathbb{F} and v‾∈V.\underline{v} \in \mathcal{V}. The norm or length ∥v‾∥\|\underline{v}\| of v‾\underline{v} is the square root of the inner product with itself:

∥v‾∥:=⟨v‾,v‾⟩∈R\|\underline{v}\|:=\sqrt{\langle \underline{v},\underline{v} \rangle} \in \mathbb{R}

The Cauchy-Schwarz inequality states that for an inner product space V\mathcal{V} and v‾1,v‾2∈V:\underline{v}_1,\underline{v}_2\in\mathcal{V}:

∣⟨v‾1,v‾2⟩∣≤∥v‾1∥∥v‾2∥|\langle\underline{v}_1,\underline{v}_2\rangle|\leq\|\underline{v}_1\|\|\underline{v}_2\|

with equality if and only if v‾1\underline{v}_1 and v‾2\underline{v}_2 are linearly dependent.

Let V\mathcal{V} be an inner product space over R.\mathbb{R}. The angle θ\theta between two non-zero vectors u‾,v‾∈V\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:

θ:=cos⁡−1(⟨u‾,v‾⟩∥u‾∥∥v‾∥)∈[0,π]\theta:=\cos^{-1}\left(\frac{\langle \underline{u},\underline{v} \rangle}{\|\underline{u}\|\|\underline{v}\|}\right) \in [0, \pi]

The Cauchy-Schwarz inequality guarantees that the quotient lies in [−1,1].[-1, 1].

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

For a subset of vectors U⊆V,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 nn orthonormal vectors in an nn-dimensional vector space is a basis.

The Gram-Schmidt orthonormalisation procedure is a method to orthonormalise a set of linearly independent vectors u‾1,…u‾n∈U\underline{u}_1,\ldots\underline{u}_n \in \mathcal{U} in an inner product space U.\mathcal{U}. The resultant orthonormalised vectors v‾1,…v‾n\underline{v}_1,\ldots\underline{v}_n span the same subspace as u‾1,…u‾n;\underline{u}_1,\ldots\underline{u}_n; more precisely, v‾1,…v‾k\underline{v}_1,\ldots\underline{v}_k span the same subspace as u‾1,…u‾k\underline{u}_1,\ldots\underline{u}_k for each k.k. If u‾1,…u‾n\underline{u}_1,\ldots\underline{u}_n is a basis of U\mathcal{U} then v‾1,…v‾n\underline{v}_1,\ldots\underline{v}_n is an orthonormal basis of U.\mathcal{U}.

w‾1:=u‾1v‾1:=w‾1∥w‾1∥w‾2:=u‾2−⟨u‾2,v‾1⟩v‾1v‾2:=w‾2∥w‾2∥⋮⋮⋮⋮⋮⋮w‾n:=u‾n−∑i=1n−1⟨u‾n,v‾i⟩v‾iv‾n:=w‾n∥w‾n∥\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}