Inner product space

In mathematics, an inner product space (or, rarely, a

Hausdorff

pre-Hilbert space) is a real vector space or a complex vector space with a binary operation called an inner product. The inner product of two vectors in the space is a scalar, often denoted with angle brackets, as in $\langle a, b \rangle$. Inner products allow formal definitions of intuitive geometric notions, such as lengths,

angle

s, and

orthogonality

(zero inner product) of vectors. Inner product spaces generalize Euclidean vector spaces, in which the inner product is the dot product or ''scalar product'' of

Cartesian coordinates

. Inner product spaces of infinite dimension are widely used in functional analysis. Inner product spaces over the field of

complex number

s are sometimes referred to as unitary spaces. The first usage of the concept of a vector space with an inner product is due to

Giuseppe Peano

, in 1898.

An inner product naturally induces an associated norm, (denoted $, x,$ and $, y,$ in the picture); so, every inner product space is a normed vector space. If this normed space is also complete (that is, a Banach space) then the inner product space is a Hilbert space. If an inner product space is not a Hilbert space, it can be ''extended'' by completion to a Hilbert space $\overline.$ This means that $H$ is a linear subspace of $\overline,$ the inner product of $H$ is the restriction of that of $\overline,$ and $H$ is dense in $\overline$ for the topology defined by the norm.

Definition

In this article, the field of scalars denoted is either the field of real numbers $\R$ or the field of

complex number

s $\Complex$.

Formally, an inner product space is a vector space over the field together with an ''inner product'', i.e., with a map
:$\langle \cdot, \cdot \rangle : V \times V \to F$

that satisfies the following three properties for all vectors and all scalars :

* Conjugate symmetry:A bar over an expression denotes complex conjugation; e.g., $\overline$ is the complex conjugation of $x$. For real values, $x = \overline$ and conjugate symmetry is just symmetry.
*: $\langle x, y \rangle = \overline$
* Linearity in the first argument:By combining the ''linear in the first argument'' property with the ''conjugate symmetry'' property you get ''conjugate-linear in the second argument'': $\langle x,by \rangle = \langle x,y \rangle \overline$. This is how the inner product was originally defined and is still used in some old-school math communities. However, all of engineering and computer science, and most of physics and modern mathematics now define the inner product to be ''linear in the second argument'' and ''conjugate-linear in the first argument'' because this is more compatible with several other conventions in mathematics. Notably, for any inner product, there is some hermitian, positive-definite matrix $M$ such that $\langle x,y \rangle = x^* M y$. (Here, $x^*$ is the conjugate transpose of $x$.)
*: $\begin \langle ax, y \rangle &= a \langle x, y \rangle \\ \langle x + y, z \rangle &= \langle x, z \rangle + \langle y, z \rangle \end$
* Positive-definite:
*: $\langle x, x \rangle > 0,\quad x \in V \setminus \.$

If the positive-definite condition is replaced by merely requiring that $\langle x, x \rangle \geq 0$ for all ''x'', then one obtains the definition of ''positive semi-definite Hermitian form''. A positive semi-definite Hermitian form $\langle \cdot, \cdot \rangle$ is an inner product if and only if for all ''x'', if $\langle x, x \rangle = 0$ then ''x = 0''.

Elementary properties

Positive-definiteness and linearity, respectively, ensure that:
:$\begin \langle x, x \rangle &= 0 \Rightarrow x = \mathbf \\ \langle \mathbf, \mathbf \rangle &= \langle 0x, 0x \rangle = 0 \langle x, 0x \rangle = 0 \end$

Notice that conjugate symmetry implies that is real for all , since we have:
: $\langle x, x \rangle = \overline \,.$

Conjugate symmetry and linearity in the first variable imply

:$\begin \langle x, a y \rangle &= \overline = \overline \overline = \overline \langle x, y \rangle \\ \langle x, y + z \rangle &= \overline = \overline + \overline = \langle x, y \rangle + \langle x, z \rangle \, \end$

that is, conjugate linearity in the second argument. So, an inner product is a sesquilinear form. Conjugate symmetry is also called Hermitian symmetry, and a conjugate-symmetric sesquilinear form is called a ''Hermitian form''. While the above axioms are more mathematically economical, a compact verbal definition of an inner product is a ''positive-definite Hermitian form''.

This important generalization of the familiar square expansion follows:
: $\langle x + y, x + y \rangle = \langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle \,.$

These properties, constituents of the above linearity in the first and second argument:
:$\begin \langle x + y, z \rangle &= \langle x, z \rangle + \langle y, z \rangle \,, \\ \langle x, y + z \rangle &= \langle x, y\rangle + \langle x, z \rangle \end$

are otherwise known as ''additivity''.

In the case of = $\R$, 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,

:$\begin \langle x, y \rangle &= \langle y, x \rangle \\ \Rightarrow \langle -x, x \rangle &= \langle x, -x \rangle \,, \end$

and the

binomial expansion

becomes:
: $\langle x + y, x + y \rangle = \langle x, x \rangle + 2\langle x, y \rangle + \langle y, y \rangle \,.$

Convention variant

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.

Some examples

Real and complex numbers

Among the simplest examples of inner product spaces are $\R$ and $\Complex.$
The real numbers $\R$ are a vector space over $\R$ that becomes a real inner product space when endowed with standard multiplication as its real inner product:
$\langle x, y \rangle := x y \quad \text x, y \in \R.$

The

complex number

s $\Complex$ are a vector space over $\Complex$ that becomes a complex inner product space when endowed with the complex inner product
$\langle x, y \rangle := x \overline \quad \text x, y \in \Complex.$
Unlike with the real numbers, the assignment $(x, y) \mapsto x y$ does define a complex inner product on $\Complex.$

Euclidean vector space

More generally, the real $n$-space $\R^n$ with the dot product is an inner product space, an example of a Euclidean vector space.
$\left\langle \begin x_1 \\ \vdots \\ x_n \end, \begin y_1 \\ \vdots \\ y_n \end \right\rangle = x^\textsf y = \sum_^n x_i y_i = x_1 y_1 + \cdots + x_n y_n,$
where $x^$ is the

transpose

of $x.$

A function $\langle \,\cdot, \cdot\, \rangle : \R^n \times \R^n \to \R$ is an inner product on $\R^n$ if and only if there exists a symmetric positive-definite matrix $\mathbf$ such that $\langle x, y \rangle = x^ \mathbf y$ for all $x, y \in \R^n.$ If $\mathbf$ is the

identity matrix

then $\langle x, y \rangle = x^ \mathbf y$ is the dot product. For another example, if $n = 2$ and $\mathbf = \begin a & b \\ b & d \end$ is positive-definite (which happens if and only if $\det \mathbf = a d - b^2 > 0$ and one/both diagonal elements are positive) then for any x := \left _1, x_2\right, y := \left _1, y_2\right \in \R^2,
\langle x, y \rangle
:= x^ \mathbf y
= \left _1, x_2\right\begin a & b \\ b & d \end \begin y_1 \\ y_2 \end
= a x_1 y_1 + b x_1 y_2 + b x_2 y_1 + d x_2 y_2.
As mentioned earlier, every inner product on $\R^2$ is of this form (where $b \in \R, a > 0$ and $d > 0$ satisfy $a d > b^2$).

Complex coordinate space

The general form of an inner product on $\Complex^n$ is known as the Hermitian form and is given by
$\langle x, y \rangle = y^\dagger \mathbf x = \overline,$
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.

Hilbert space

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 $C($, b of continuous complex valued functions $f$ and $g$ on the interval $$, b The inner product is
$\langle f, g \rangle = \int_a^b f(t) \overline \, \mathrmt.$
This space is not complete; consider for example, for the interval the sequence of continuous "step" functions, $\_k,$ defined by:
f_k(t) = \begin 0 & t \in 1, 0\\ 1 & t \in \left tfrac, 1\right\\ kt & t \in \left(0, \tfrac\right) \end

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

Random variables

For real random variables $X$ and $Y,$ the expected value of their product
$\langle X, Y \rangle = \operatorname(XY)$
is an inner product. In this case, $\langle X, X \rangle = 0$ if and only if $\Pr (X = 0) = 1$ (that is, $X = 0$ almost surely), where $\Pr$ denotes the

probability

of the event. This definition of expectation as inner product can be extended to random vectors as well.

Complex matrices

The inner product for complex square matrices of the same size is the Frobenius inner product $\langle A, B \rangle := \operatorname\left(AB^\right)$. Since trace and transposition are linear and the conjugation is on the second matrix, it is a sesquilinear operator. We further get Hermitian symmetry by,
$\langle A, B \rangle = \operatorname\left(AB^\right) = \overline = \overline$
Finally, since for $A$ nonzero, $\langle A, A\rangle = \sum_ \left, A_\^2 > 0$, we get that the Frobenius inner product is positive definite too, and so is an inner product.

Vector spaces with forms

On an inner product space, or more generally a vector space with a nondegenerate form (hence an isomorphism $V \to V^*$), 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.

Basic results, terminology, and definitions

Norm

Every inner product space induces a norm, called its , that is defined by
$\, x\, = \sqrt.$
With this norm, every inner product space becomes a normed vector space.

As for every normed vector space, an inner product space is a metric space, for the distance defined by
$d(x, y) := \, y - x\, = \, x - y\, .$
The axioms of the inner product guarantee that the map above forms a norm, which will have the following properties.

Orthogonality

Real and complex parts of inner products

Suppose that $\langle \cdot, \cdot \rangle$ 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
$\operatorname \langle x, y \rangle = \frac \left(\, x + y\, ^2 - \, x - y\, ^2\right).$

If $V$ is a real vector space then
$\langle x, y \rangle = \operatorname \langle x, y \rangle = \frac \left(\, x + y\, ^2 - \, x - y\, ^2\right)$
and the imaginary part (also called the ) of $\langle \cdot, \cdot \rangle$ 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
:$\begin \langle x, \ y \rangle &= \frac \left(\, x + y\, ^2 - \, x - y\, ^2 + i\, x + iy\, ^2 - i\, x - iy\, ^2 \right) \\ &= \operatorname \langle x, y \rangle + i \operatorname \langle x, i y \rangle. \\ \end$

The map defined by $\langle x \mid y \rangle = \langle y, x \rangle$ for all $x, y \in V$ satisfies the axioms of the inner product except that it is antilinear in its , rather than its second, argument. The real part of both $\langle x \mid y \rangle$ and $\langle x, y \rangle$ are equal to $\operatorname \langle x, y \rangle$ but the inner products differ in their complex part:
:$\begin \langle x \mid y \rangle &= \frac \left(\, x + y\, ^2 - \, x - y\, ^2 - i\, x + iy\, ^2 + i\, x - iy\, ^2 \right) \\ &= \operatorname \langle x, y \rangle - i \operatorname \langle x, i y \rangle. \\ \end$

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

These formulas show that every complex inner product is completely determined by its real part. There is thus a one-to-one correspondence between complex inner products and real inner products. For example, suppose that $V := \Complex^n$ for some integer $n > 0.$ When $V$ is considered as a real vector space in the usual way (meaning that it is identified with the $2 n-$dimensional real vector space $\R^,$ with each $\left(a_1 + i b_1, \ldots, a_n + i b_n\right) \in \Complex^n$ identified with $\left(a_1, b_1, \ldots, a_n, b_n\right) \in \R^$), then the dot product $x \,\cdot\, y = \left(x_1, \ldots, x_\right) \, \cdot \, \left(y_1, \ldots, y_\right) := x_1 y_1 + \cdots + x_ y_$ defines a real inner product on this space. The unique complex inner product $\langle \,\cdot, \cdot\, \rangle$ on $V = \C^n$ induced by the dot product is the map that sends $c = \left(c_1, \ldots, c_n\right), d = \left(d_1, \ldots, d_n\right) \in \Complex^n$ to $\langle c, d \rangle := c_1 \overline + \cdots + c_n \overline$ (because the real part of this map $\langle \,\cdot, \cdot\, \rangle$ is equal to the dot product).

Real vs. complex inner products

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

For example, if $V = \Complex$ with inner product $\langle x, y \rangle = x \overline,$ where $V$ is a vector space over the field $\Complex,$ then $V_ = \R^2$ is a vector space over $\R$ and $\langle x, y \rangle_$ is the dot product $x \cdot y,$ where $x = a + i b \in V = \Complex$ is identified with the point $(a, b) \in V_ = \R^2$ (and similarly for $y$); thus the standard inner product $\langle x, y \rangle = x \overline,$ on $\Complex$ is an "extension" the dot product . Also, had $\langle x, y \rangle$ been instead defined to be the $\langle x, y \rangle = x y$ (rather than the usual $\langle x, y \rangle = x \overline$) then its real part $\langle x, y \rangle_$ would be the dot product; furthermore, without the complex conjugate, if $x \in \C$ but $x \not\in \R$ then $\langle x, x \rangle = x x = x^2 \not\in [0, \infty)$ so the assignment $x \mapsto \sqrt$ would not define a norm.

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 $\langle x, y \rangle = 0$ then $\langle x, y \rangle_ = 0,$ but the next example shows that the converse is in general true.
Given any $x \in V,$ the vector $i x$ (which is the vector $x$ rotated by 90°) belongs to $V$ and so also belongs to $V_$ (although scalar multiplication of $x$ by $i = \sqrt$ is not defined in $V_,$ the vector in $V$ denoted by $i x$ is nevertheless still also an element of $V_$). For the complex inner product, $\langle x, ix \rangle = -i \, x\, ^2,$ whereas for the real inner product the value is always $\langle x, ix \rangle_ = 0.$

If $\langle \,\cdot, \cdot\, \rangle$ is a complex inner product and $A : V \to V$ is a continuous linear operator that satisfies $\langle x, A x \rangle = 0$ for all $x \in V,$ then $A = 0.$ This statement is no longer true if $\langle \,\cdot, \cdot\, \rangle$ is instead a real inner product, as this next example shows.
Suppose that $V = \Complex$ has the inner product $\langle x, y \rangle := x \overline$ mentioned above. Then the map $A : V \to V$ defined by $A x = ix$ is a linear map (linear for both $V$ and $V_$) that denotes rotation by $90^$ in the plane. Because $x$ and $A x$ perpendicular vectors and $\langle x, Ax \rangle_$ is just the dot product, $\langle x, Ax \rangle_ = 0$ for all vectors $x;$ nevertheless, this rotation map $A$ is certainly not identically $0.$ In contrast, using the complex inner product gives $\langle x, Ax \rangle = -i \, x\, ^2,$ which (as expected) is not identically zero.

Orthonormal sequences

Let $V$ be a finite dimensional inner product space of dimension $n.$ Recall that every Basis (linear algebra), basis of $V$ consists of exactly $n$ linearly independent vectors. Using the Gram–Schmidt process we may start with an arbitrary basis and transform it into an orthonormal basis. That is, into a basis in which all the elements are orthogonal and have unit norm. In symbols, a basis $\$ is orthonormal if $\langle e_i, e_j \rangle = 0$ for every $i \neq j$ and $\langle e_i, e_i \rangle = \, e_a\, ^2 = 1$ for each index $i.$

This definition of orthonormal basis generalizes to the case of infinite-dimensional inner product spaces in the following way. Let $V$ be any inner product space. Then a collection
$E = \left\_$
is a for $V$ if the subspace of $V$ generated by finite linear combinations of elements of $E$ is dense in $V$ (in the norm induced by the inner product). Say that $E$ is an for $V$ if it is a basis and
$\left\langle e_, e_ \right\rangle = 0$
if $a \neq b$ and $\langle e_a, e_a \rangle = \, e_a\, ^2 = 1$ for all $a, b \in A.$

Using an infinite-dimensional analog of the Gram-Schmidt process one may show:

Theorem. Any separable inner product space has an orthonormal basis.

Using the Hausdorff maximal principle and the fact that in a complete inner product space orthogonal projection onto linear subspaces is well-defined, one may also show that

Theorem. Any complete inner product space has an orthonormal basis.

The two previous theorems raise the question of whether all inner product spaces have an orthonormal basis. The answer, it turns out is negative. This is a non-trivial result, and is proved below. The following proof is taken from Halmos's ''A Hilbert Space Problem Book'' (see the references).

:

Parseval's identity leads immediately to the following theorem:

Theorem. Let $V$ be a separable inner product space and $\left\_k$ an orthonormal basis of $V.$ Then the map
$x \mapsto \bigl\_$
is an isometric linear map $V \mapsto \ell^2$ with a dense image.

This theorem can be regarded as an abstract form of Fourier series, in which an arbitrary orthonormal basis plays the role of the sequence of trigonometric polynomials. Note that the underlying index set can be taken to be any countable set (and in fact any set whatsoever, provided $\ell^2$ is defined appropriately, as is explained in the article Hilbert space). In particular, we obtain the following result in the theory of Fourier series:

Theorem. Let $V$ be the inner product space C \pi, \pi Then the sequence (indexed on set of all integers) of continuous functions
$e_k(t) = \frac$
is an orthonormal basis of the space C \pi, \pi/math> with the $L^2$ inner product. The mapping
$f \mapsto \frac \left\_$
is an isometric linear map with dense image.

Orthogonality of the sequence $\_k$ follows immediately from the fact that if $k \neq j,$ then
$\int_^\pi e^ \, \mathrmt = 0.$

Normality of the sequence is by design, that is, the coefficients are so chosen so that the norm comes out to 1. Finally the fact that the sequence has a dense algebraic span, in the , follows from the fact that the sequence has a dense algebraic span, this time in the space of continuous periodic functions on \pi, \pi/math> with the uniform norm. This is the content of the Weierstrass theorem on the uniform density of trigonometric polynomials.

Operators on inner product spaces

Several types of

linear

maps $A : V \to W$ between inner product spaces $V$ and $W$ are of relevance:
* : $A : V \to W$ is linear and continuous with respect to the metric defined above, or equivalently, $A$ is linear and the set of non-negative reals $\,$ where $x$ ranges over the closed unit ball of $V,$ is bounded.
* : $A : V \to W$ is linear and $\langle Ax, y \rangle = \langle x, Ay \rangle$ for all $x, y \in V.$
* : $A : V \to W$ satisfies $\, A x\, = \, x\,$ for all $x \in V.$ A (resp. an ) is an isometry that is also a linear map (resp. an antilinear map). For inner product spaces, the polarization identity can be used to show that $A$ is an isometry if and only if $\langle Ax, Ay \rangle = \langle x, y \rangle$ for all $x, y \in V.$ All isometries are injective. The Mazur–Ulam theorem establishes that every surjective isometry between two normed spaces is an affine transformation. Consequently, an isometry $A$ between real inner product spaces is a linear map if and only if $A(0) = 0.$ Isometries are

morphism

s between inner product spaces, and morphisms of real inner product spaces are orthogonal transformations (compare with orthogonal matrix).
* : $A : V \to W$ is an isometry which is surjective (and hence bijective). Isometrical isomorphisms are also known as unitary operators (compare with unitary matrix).

From the point of view of inner product space theory, there is no need to distinguish between two spaces which are isometrically isomorphic. The spectral theorem provides a canonical form for symmetric, unitary and more generally normal operators on finite dimensional inner product spaces. A generalization of the spectral theorem holds for continuous normal operators in Hilbert spaces.

Generalizations

Any of the axioms of an inner product may be weakened, yielding generalized notions. The generalizations that are closest to inner products occur where bilinearity and conjugate symmetry are retained, but positive-definiteness is weakened.

Degenerate inner products

If $V$ is a vector space and $\langle \,\cdot\,, \,\cdot\, \rangle$ a semi-definite sesquilinear form, then the function:
$\, x\, = \sqrt$
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 $W = V / \.$ The sesquilinear form $\langle \,\cdot\,, \,\cdot\, \rangle$ 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.

Nondegenerate conjugate symmetric forms

Alternatively, one may require that the pairing be a nondegenerate form, meaning that for all non-zero $x \neq 0$ there exists some $y$ such that $\langle x, y \rangle \neq 0,$ though $y$ need not equal $x$; in other words, the induced map to the dual space $V \to V^*$ 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 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 Dimension (mathematics), dimensions and indices 3 and 1 (assignment of Sign (mathematics), "+" and "−" to them Sign convention#Metric signature, differs depending on conventions).

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

Related products

The term "inner product" is opposed to outer product, which is a slightly more general opposite. Simply, in coordinates, the inner product is the product of a $1 \times n$ with an $n \times 1$ vector, yielding a $1 \times 1$ matrix (a scalar), while the outer product is the product of an $m \times 1$ vector with a $1 \times n$ covector, yielding an $m \times n$ matrix. The outer product is defined for different dimensions, while the inner product requires the same dimension. If the dimensions are the same, then the inner product is the of the outer product (trace only being properly defined for square matrices). In an informal summary: "inner is horizontal times vertical and shrinks down, outer is vertical times horizontal and expands out".

More abstractly, the outer product is the bilinear map $W \times V^* \to \hom(V, W)$ sending a vector and a covector to a rank 1 linear transformation (simple tensor of type (1, 1)), while the inner product is the bilinear evaluation map $V^* \times V \to F$ given by evaluating a covector on a vector; the order of the domain vector spaces here reflects the covector/vector distinction.

The inner product and outer product should not be confused with the interior product and exterior product, which are instead operations on vector fields and differential forms, or more generally on the exterior algebra.

As a further complication, in geometric algebra the inner product and the (Grassmann) product are combined in the geometric product (the Clifford product in a Clifford algebra) – the inner product sends two vectors (1-vectors) to a scalar (a 0-vector), while the exterior product sends two vectors to a bivector (2-vector) – and in this context the exterior product is usually called the (alternatively, ). The inner product is more correctly called a product in this context, as the nondegenerate quadratic form in question need not be positive definite (need not be an inner product).

See also

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Notes

References

Bibliography

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{{HilbertSpace

Normed spaces
Bilinear forms