Inner product space

Positive-definiteness and linearity, respectively, ensure that:

${\displaystyle {\begin{aligned}\langle x,x\rangle &=0\Rightarrow x=\mathbf {0} \\\langle \mathbf {0} ,x\rangle &=\langle 0x,x\rangle =0\langle x,x\rangle =0\end{aligned}}}$

Conjugate symmetry implies that ⟨x, x⟩ is real for all x, because

${\displaystyle \langle x,x\rangle ={\overline {\langle x,x\rangle }}\,.}$

Conjugate symmetry and linearity in the first variable imply

${\displaystyle {\begin{aligned}\langle x,ay\rangle &={\overline {\langle ay,x\rangle }}={\overline {a}}{\overline {\langle y,x\rangle }}={\overline {a}}\langle x,y\rangle \\\langle x,y+z\rangle &={\overline {\langle y+z,x\rangle }}={\overline {\langle y,x\rangle }}+{\overline {\langle z,x\rangle }}=\langle x,y\rangle +\langle x,z\rangle \,\end{aligned}}}$

that is, conjugate linearity in the second argument. So, an inner product is a sesquilinear form.

This important generalization of the familiar square expansion follows:

${\displaystyle \langle x+y,x+y\rangle =\langle x,x\rangle +\langle x,y\rangle +\langle y,x\rangle +\langle y,y\rangle \,.}$

These properties, constituents of the above linearity in the first and second argument:

${\displaystyle {\begin{aligned}\langle x+y,z\rangle &=\langle x,z\rangle +\langle y,z\rangle \,,\\\langle x,y+z\rangle &=\langle x,y\rangle +\langle x,z\rangle \end{aligned}}}$

are otherwise known as additivity.

In the case of ${\displaystyle \mathbb {F} =\mathbb {R} ,}$ conjugate-symmetry reduces to symmetry, and sesquilinearity reduces to bilinearity. Hence an inner product on a real vector space is a positive-definite symmetric bilinear form. That is,

${\displaystyle {\begin{aligned}\langle x,y\rangle &=\langle y,x\rangle \\\Rightarrow \langle -x,x\rangle &=\langle x,-x\rangle \,,\end{aligned}}}$

and the binomial expansion becomes:

${\displaystyle \langle x+y,x+y\rangle =\langle x,x\rangle +2\langle x,y\rangle +\langle y,y\rangle \,.}$

A common special case of the inner product, the scalar product or dot product, is written with a centered dot ${\displaystyle a\cdot b.}$

Some authors, especially in physics and matrix algebra, prefer to define the inner product and the sesquilinear form with linearity in the second argument rather than the first. Then the first argument becomes conjugate linear, rather than the second. In those disciplines, we would write the inner product ${\displaystyle \langle x,y\rangle }$ as ⟨y | x⟩ (the bra–ket notation of quantum mechanics), respectively y^†x (dot product as a case of the convention of forming the matrix product AB, as the dot products of rows of A with columns of B). Here, the kets and columns are identified with the vectors of V, and the bras and rows with the linear functionals (covectors) of the dual space V^∗, with conjugacy associated with duality. This reverse order is now occasionally followed in the more abstract literature,[10] taking ⟨x, y⟩ to be conjugate linear in x rather than y. A few instead find a middle ground by recognizing both ⟨·, ·⟩ and ⟨· | ·⟩ as distinct notations—differing only in which argument is conjugate linear.

There are various technical reasons why it is necessary to restrict the base field to ${\displaystyle \mathbb {R} }$ and ${\displaystyle \mathbb {C} }$ in the definition. Briefly, the base field has to contain an ordered subfield in order for non-negativity to make sense,[11] and therefore has to have characteristic equal to 0 (since any ordered field has to have such characteristic). This immediately excludes finite fields. The basefield has to have additional structure, such as a distinguished automorphism. More generally, any quadratically closed subfield of ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$ will suffice for this purpose (e.g., algebraic numbers, constructible numbers). However, in the cases where it is a proper subfield (i.e., neither ${\displaystyle \mathbb {R} }$ nor ${\displaystyle \mathbb {C} }$), even finite-dimensional inner product spaces will fail to be metrically complete. In contrast, all finite-dimensional inner product spaces over ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} ,}$ such as those used in quantum computation, are automatically metrically complete (and hence Hilbert spaces).

In some cases, one needs to consider non-negative semi-definite sesquilinear forms. This means that ${\displaystyle \langle x,x\rangle }$ is only required to be non-negative. Treatment for these cases are illustrated below.

A simple example is the real numbers with the standard multiplication as the inner product[4]

${\displaystyle \langle x,y\rangle$ :=xy.}

More generally, the real n-space ${\displaystyle \mathbb {R} ^{n}}$ with the dot product is an inner product space,[4] an example of a Euclidean vector space.

${\displaystyle \left\langle {\begin{bmatrix}x_{1}\\\vdots \\x_{n}\end{bmatrix}},{\begin{bmatrix}y_{1}\\\vdots \\y_{n}\end{bmatrix}}\right\rangle$ :=x^{\textsf {T}}y=\sum _{i=1}^{n}x_{i}y_{i}=x_{1}y_{1}+\cdots +x_{n}y_{n},}

where x^T is the transpose of x.

The general form of an inner product on ${\displaystyle \mathbb {C} ^{n}}$ is known as the Hermitian form and is given by

${\displaystyle \langle x,y\rangle$ :=y^{\dagger }\mathbf {M} x={\overline {x^{\dagger }\mathbf {M} y}},}

where M is any Hermitian positive-definite matrix and y^† is the conjugate transpose of y. For the real case, this corresponds to the dot product of the results of directionally-different scaling of the two vectors, with positive scale factors and orthogonal directions of scaling. It is a weighted-sum version of the dot product with positive weights—up to an orthogonal transformation.

The article on Hilbert spaces has several examples of inner product spaces, wherein the metric induced by the inner product yields a complete metric space. An example of an inner product space which induces an incomplete metric is the space ${\displaystyle C([a,b])}$ of continuous complex valued functions ${\displaystyle f}$ and ${\displaystyle g}$ on the interval ${\displaystyle [a,b].}$ The inner product is

${\displaystyle \langle f,g\rangle$ :=\int _{a}^{b}f(t){\overline {g(t)}}\,\mathrm {d} t.}

This space is not complete; consider for example, for the interval [−1, 1] the sequence of continuous "step" functions, { f_k}_k, defined by:

${\displaystyle f_{k}(t)={\begin{cases}0&t\in [-1,0]\\1&t\in \left[{\tfrac {1}{k}},1\right]\\kt&t\in \left(0,{\tfrac {1}{k}}\right)\end{cases}}}$

This sequence is a Cauchy sequence for the norm induced by the preceding inner product, which does not converge to a continuous function.

For real random variables X and Y, the expected value of their product

${\displaystyle \langle X,Y\rangle$ :=\operatorname {E} (XY)}

is an inner product.[12][13][14] In this case, ⟨X, X⟩ = 0 if and only if Pr(X = 0) = 1 (i.e., X = 0 almost surely). This definition of expectation as inner product can be extended to random vectors as well.

For real square matrices of the same size, ⟨A, B⟩ ≝ tr(AB^T) with transpose as conjugation

${\displaystyle \left(\langle A,B\rangle =\left\langle B^{\textsf {T}},A^{\textsf {T}}\right\rangle \right)}$

is an inner product.

On an inner product space, or more generally a vector space with a nondegenerate form (hence an isomorphism V → V^∗), vectors can be sent to covectors (in coordinates, via transpose), so that one can take the inner product and outer product of two vectors—not simply of a vector and a covector.

Suppose that ${\displaystyle \langle \cdot ,\cdot \rangle }$ is an inner product on V (so it is antilinear in its second argument). The polarization identity shows that the real part of the inner product is

${\displaystyle \operatorname {Re} \langle x,y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)}$

If ${\displaystyle V}$ is a real vector space then ${\displaystyle \langle x,y\rangle =\operatorname {Re} \langle x,y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right)}$ and the imaginary part (also called the complex part) of ${\displaystyle \langle \cdot ,\cdot \rangle }$ is always 0.

Assume for the rest of this section that V is a complex vector space. The polarization identity for complex vector spaces shows that

${\displaystyle {\begin{alignedat}{4}\langle x,\ y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}+i\|x+iy\|^{2}-i\|x-iy\|^{2}\right)\\&=\operatorname {Re} \langle x,y\rangle +i\operatorname {Re} \langle x,iy\rangle .\\\end{alignedat}}}$

The map defined by ${\displaystyle \langle x\mid y\rangle$ :=\langle y,x\rangle } for all ${\displaystyle x,y\in V}$ satisfies the axioms of the inner product except that it is antilinear in its first, rather than its second, argument. The real part of both ${\displaystyle \langle x\mid y\rangle }$ and ${\displaystyle \langle x,y\rangle }$ are equal to ${\displaystyle \operatorname {Re} \langle x,y\rangle }$ but the inner products differ in their complex part:

${\displaystyle {\begin{alignedat}{4}\langle x\mid y\rangle &={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}-i\|x+iy\|^{2}+i\|x-iy\|^{2}\right)\\&=\operatorname {Re} \langle x,y\rangle -i\operatorname {Re} \langle x,iy\rangle .\\\end{alignedat}}}$

The last equality is similar to the formula expressing a linear functional in terms of its real part.

Real vs. complex inner products

Let ${\displaystyle V_{\mathbb {R} }}$ denote ${\displaystyle V}$ considered as a vector space over the real numbers rather than complex numbers. The real part of the complex inner product ${\displaystyle \langle x,y\rangle }$ is the map ${\displaystyle \langle x,y\rangle _{\mathbb {R} }:=\operatorname {Re} \langle x,y\rangle ~:~V_{\mathbb {R} }\times V_{\mathbb {R} }\to \mathbb {R} ,}$ which necessarily forms a real inner product on the real vector space ${\displaystyle V_{\mathbb {R} }.}$ Every inner product on a real vector space is symmetric and bilinear.

For example, if ${\displaystyle V=\mathbb {C} }$ with inner product ${\displaystyle \langle x,y\rangle =x{\overline {y}},}$ where ${\displaystyle V}$ is a vector space over the field ${\displaystyle \mathbb {C} ,}$ then ${\displaystyle V_{\mathbb {R} }=\mathbb {R} ^{2}}$ is a vector space over ${\displaystyle \mathbb {R} }$ and ${\displaystyle \langle x,y\rangle _{\mathbb {R} }}$ is the dot product ${\displaystyle x\cdot y,}$ where ${\displaystyle x=a+ib\in V=\mathbb {C} }$ is identified with the point ${\displaystyle (a,b)\in V_{\mathbb {R} }=\mathbb {R} ^{2}}$ (and similarly for ${\displaystyle y}$). Also, had ${\displaystyle \langle x,y\rangle }$ been instead defined to be the symmetric map ${\displaystyle \langle x,y\rangle =xy}$ (rather than the usual antisymmetric map ${\displaystyle \langle x,y\rangle =x{\overline {y}}}$) then its real part ${\displaystyle \langle x,y\rangle _{\mathbb {R} }}$ would not be the dot product.

The next examples show that although real and complex inner products have many properties and results in common, they are not entirely interchangeable. For instance, if ${\displaystyle \langle x,y\rangle =0}$ then ${\displaystyle \langle x,y\rangle _{\mathbb {R} }=0,}$ but the next example shows that the converse is in general not true. Given any ${\displaystyle x\in V,}$ the vector ${\displaystyle ix}$ (which is the vector ${\displaystyle x}$ rotated by 90°) belongs to ${\displaystyle V}$ and so also belongs to ${\displaystyle V_{\mathbb {R} }}$ (although scalar multiplication of ${\displaystyle x}$ by i is not defined in ${\displaystyle V_{\mathbb {R} },}$ it is still true that the vector in ${\displaystyle V}$ denoted by ${\displaystyle ix}$ is an element of ${\displaystyle V_{\mathbb {R} }}$). For the complex inner product, ${\displaystyle \langle x,ix\rangle =-i\|x\|^{2},}$ whereas for the real inner product the value is always ${\displaystyle \langle x,ix\rangle _{\mathbb {R} }=0.}$

If ${\displaystyle V=\mathbb {C} }$ has the inner product mentioned above, then the map ${\displaystyle A:V\to V}$ defined by ${\displaystyle Ax=ix}$ is a non-zero linear map (linear for both ${\displaystyle V}$ and ${\displaystyle V_{\mathbb {R} }}$) that denotes rotation by 90° in the plane. This map satisfies ${\displaystyle \langle x,Ax\rangle _{\mathbb {R} }=0}$ for all vectors ${\displaystyle x\in V_{\mathbb {R} },}$ where had this inner product been complex instead of real, then this would have been enough to conclude that this linear map ${\displaystyle A}$ is identically ${\displaystyle 0}$ (i.e. that ${\displaystyle A=0}$), which rotation is certainly not. In contrast, for all non-zero ${\displaystyle x\in V,}$ the map ${\displaystyle A}$ satisfies ${\displaystyle \langle x,Ax\rangle =-i\|x\|^{2}\neq 0.}$

If V is a vector space and ⟨·, ·⟩ a semi-definite sesquilinear form, then the function:

${\displaystyle \|x\|={\sqrt {\langle x,x\rangle }}}$

makes sense and satisfies all the properties of norm except that ||x|| = 0 does not imply x = 0 (such a functional is then called a semi-norm). We can produce an inner product space by considering the quotient W = V/{x : ||x|| = 0}. The sesquilinear form ⟨·, ·⟩ factors through W.

This construction is used in numerous contexts. The Gelfand–Naimark–Segal construction is a particularly important example of the use of this technique. Another example is the representation of semi-definite kernels on arbitrary sets.

Alternatively, one may require that the pairing be a nondegenerate form, meaning that for all non-zero x there exists some y such that ⟨x, y⟩ ≠ 0, though y need not equal x; in other words, the induced map to the dual space V → V^∗ is injective. This generalization is important in differential geometry: a manifold whose tangent spaces have an inner product is a Riemannian manifold, while if this is related to nondegenerate conjugate symmetric form the manifold is a pseudo-Riemannian manifold. By Sylvester's law of inertia, just as every inner product is similar to the dot product with positive weights on a set of vectors, every nondegenerate conjugate symmetric form is similar to the dot product with nonzero weights on a set of vectors, and the number of positive and negative weights are called respectively the positive index and negative index. Product of vectors in Minkowski space is an example of indefinite inner product, although, technically speaking, it is not an inner product according to the standard definition above. Minkowski space has four dimensions and indices 3 and 1 (assignment of "+" and "−" to them differs depending on conventions).

Purely algebraic statements (ones that do not use positivity) usually only rely on the nondegeneracy (the injective homomorphism V → V^∗) and thus hold more generally.