For the scalar product or dot product of coordinate vectors, see
dot product.
In linear algebra, an inner product space is a vector space with an additional structure called an inner product. This additional structure associates each pair of vectors in the space with a scalar quantity known as the inner product of the vectors. Inner products allow the rigorous introduction of intuitive geometrical notions such as the length of a vector or the angle between two vectors. They also provide the means of defining orthogonality between vectors (zero inner product). Inner product spaces generalize Euclidean spaces (in which the inner product is the dot product, also known as the scalar product) to vector spaces of any (possibly infinite) dimension, and are studied in functional analysis.
An inner product naturally induces an associated norm, thus an inner product space is also a normed vector space. A complete space with an inner product is called a Hilbert space. An incomplete space with an inner product is called a preHilbert space, since its completion with respect to the norm, induced by the inner product, becomes a Hilbert space. Inner product spaces over the field of complex numbers are sometimes referred to as unitary spaces.
Definition
In this article, the field of scalars denoted $F$ is either
the field of real numbers $\backslash R$ or the field of complex numbers $\backslash mathbb\{C\}$.
Formally, an inner product space is a vector space V over the field $F$ together with an inner product, i.e., with a map
 $\backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle:\; V\; \backslash times\; V\; \backslash rightarrow\; F$
that satisfies the following three axioms for all vectors $x,y,z\; \backslash in\; V$ and all scalars $a\; \backslash in\; F$:^{[1]}^{[2]}
 $\backslash langle\; x,y\backslash rangle\; =\backslash overline\{\backslash langle\; y,x\backslash rangle\}.$
Note that when $F\; =\; \backslash R$, there is symmetry.
 Linearity in the first argument:
 $\backslash langle\; ax,y\backslash rangle=\; a\; \backslash langle\; x,y\backslash rangle.$
 $\backslash langle\; x+y,z\backslash rangle=\; \backslash langle\; x,z\backslash rangle+\; \backslash langle\; y,z\backslash rangle.$
 $\backslash langle\; x,x\backslash rangle\; \backslash geq\; 0$ with equality only for $x\; =\; 0.$
Alternative definition and remarks
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 product $\backslash langle\; x,y\backslash rangle$ as $\backslash langle\; yx\backslash rangle$ (the braket notation of quantum mechanics), respectively $y^\backslash dagger\; x$ (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 dual vectors or linear functionals of the dual space V^{∗}, with conjugacy associated with duality. This reverse order is now occasionally followed in the more abstract literature,^{[3]} taking $\backslash langle\; x,y\backslash rangle$ to be conjugate linear in x rather than y. A few instead find a middle ground by recognizing both $\backslash langle\; ,\; \backslash rangle$ and $\backslash langle\; \; \backslash rangle$ as distinct notations differing only in which argument is conjugate linear.
There are various technical reasons why it is necessary to restrict the basefield to $\backslash R$ and $\backslash C$ in the definition. Briefly, the basefield has to contain an ordered subfield (in order for nonnegativity to make sense) 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 $\backslash R$ or $\backslash mathbb\{C\}$ will suffice for this purpose, e.g., the algebraic numbers, but when it is a proper subfield (i.e., neither $\backslash R$ nor $\backslash mathbb\{C\}$) even finitedimensional inner product spaces will fail to be metrically complete. In contrast all finitedimensional inner product spaces over $\backslash R$ or $\backslash mathbb\{C\}$, such as those used in quantum computation, are automatically metrically complete and hence Hilbert spaces.
In some cases we need to consider nonnegative semidefinite sesquilinear forms. This means that $\backslash langle\; x,x\backslash rangle$ is only required to be nonnegative. We show how to treat these below.
Elementary properties
Notice that conjugate symmetry implies that $\backslash langle\; x,x\; \backslash rangle$ is real for all $x$, since we have $\backslash langle\; x,x\; \backslash rangle\; =\; \backslash overline\{\backslash langle\; x,x\; \backslash rangle\}.$
Moreover, sesquilinearity (see below) implies that $\backslash langle\; x,x\; \backslash rangle=\; 1\backslash langle\; x,x\backslash rangle\; =\; \backslash overline\{1\}\backslash langle\; x,x\backslash rangle\; =\; \backslash langle\; x,x\backslash rangle.$
Conjugate symmetry and linearity in the first variable gives
 $\backslash langle\; x,\; a\; y\; \backslash rangle\; =\; \backslash overline\{\backslash langle\; a\; y,\; x\; \backslash rangle\}\; =\; \backslash overline\{a\}\; \backslash overline\{\backslash langle\; y,\; x\; \backslash rangle\}\; =\; \backslash overline\{a\}\; \backslash langle\; x,\; y\; \backslash rangle$
 $\backslash langle\; x,\; y\; +\; z\; \backslash rangle\; =\; \backslash overline\{\backslash langle\; y\; +\; z,\; x\; \backslash rangle\}\; =\; \backslash overline\{\backslash langle\; y,\; x\; \backslash rangle\}\; +\; \backslash overline\{\backslash langle\; z,\; x\; \backslash rangle\}\; =\; \backslash langle\; x,\; y\; \backslash rangle\; +\; \backslash langle\; x,\; z\; \backslash rangle,$
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 positivedefinite Hermitian form.
In the case of $F\; =\; \backslash R$, conjugatesymmetry reduces to symmetry, and sesquilinear reduces to bilinear.
So, an inner product on a real vector space is a positivedefinite symmetric bilinear form.
From the linearity property it is derived that $x\; =\; 0$ implies $\backslash langle\; x,x\; \backslash rangle\; =\; 0,$ while from the positivedefiniteness axiom we obtain the converse, $\backslash langle\; x,x\; \backslash rangle\; =\; 0$ implies $x\; =\; 0.$
Combining these two, we have the property that $\backslash langle\; x,x\; \backslash rangle\; =\; 0$ if and only if $x\; =\; 0.$
Combining the linearity of the inner product in its first argument and the conjugate symmetry gives the following important generalization of the familiar square expansion:
 $\backslash langle\; x\; +\; y,x\; +\; y\backslash rangle\; =\; \backslash langle\; x,x\backslash rangle\; +\; \backslash langle\; x,y\backslash rangle\; +\; \backslash langle\; y,x\backslash rangle\; +\; \backslash langle\; y,y\backslash rangle.$
Assuming that the underlying field is $\backslash R$, the inner product becomes symmetric, and we obtain
 $\backslash langle\; x\; +\; y,x\; +\; y\backslash rangle\; =\backslash langle\; x,x\backslash rangle\; +\; 2\backslash langle\; x,y\backslash rangle\; +\; \backslash langle\; y,y\backslash rangle,$
or similarly,
 $\backslash langle\; x\; \; y,x\; \; y\backslash rangle\; =\backslash langle\; x,x\backslash rangle\; \; 2\backslash langle\; x,y\backslash rangle\; +\; \backslash langle\; y,y\backslash rangle.$
The property of an inner product space $V$ that
 $\backslash langle\; x+y,z\backslash rangle=\; \backslash langle\; x,z\backslash rangle+\; \backslash langle\; y,z\backslash rangle$ and $\backslash langle\; x,y+z\backslash rangle\; =\; \backslash langle\; x,y\backslash rangle\; +\; \backslash langle\; x,z\backslash rangle$
is also known as additivity.
Examples
 A simple example is the real numbers with the standard multiplication as the inner product
 $\backslash langle\; x,y\backslash rangle:=\; x\; y.$
 More generally, the real nspace $\backslash mathbb\{R\}$^{n} with the dot product is an inner product space, an example of a Euclidean nspace.
 $\backslash langle\; (x\_1,\backslash ldots,\; x\_n),(y\_1,\backslash ldots,\; y\_n)\backslash rangle:=\; x^\backslash mathsf\{T\}\; y\; =\; \backslash sum\_\{i=1\}^\{n\}\; x\_i\; y\_i\; =\; x\_1\; y\_1\; +\; \backslash cdots\; +\; x\_n\; y\_n,$
 where x^{T} is the transpose of x.
 The general form of an inner product on $\backslash mathbb\{C\}$^{n} is known as the Hermitian form and is given by
 $\backslash langle\; \backslash mathbf\{x\},\backslash mathbf\{y\}\backslash rangle:=\; \backslash mathbf\{y\}^\backslash dagger\backslash mathbf\{M\}\backslash mathbf\{x\}\; =\; \backslash overline\{\backslash mathbf\{x\}^\backslash dagger\backslash mathbf\{M\}\backslash mathbf\{y\}\},$
 where M is any Hermitian positivedefinite 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. Up to an orthogonal transformation it is a weightedsum version of the dot product, with positive weights.
 The article on Hilbert space 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 which induces an incomplete metric occurs with the space C([a, b]) of continuous complex valued functions on the interval [a, b]. The inner product is
 $\backslash langle\; f\; ,\; g\; \backslash rangle:=\; \backslash int\_a^b\; f(t)\; \backslash overline\{g(t)\}\; \backslash ,\; dt.$
 This space is not complete; consider for example, for the interval Template:Closedclosed the sequence of continuous "step" functions { f_{k} }_{k} where
 f_{k}(t) is 0 for t in the subinterval Template:Closedclosed
 f_{k}(t) is 1 for t in the subinterval Template:Closedclosed
 f_{k} is affine in Template:Openopen. That is, f_{k}(t) = kt.
 This sequence is a Cauchy sequence for the norm induced by the preceding inner product, which does not converge to a continuous function.
 $\backslash langle\; X,\; Y\; \backslash rangle:=\; \backslash operatorname\{E\}(X\; Y)$
 is an inner product. 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 square real matrices, $\backslash langle\; A,\; B\; \backslash rangle:=\; \backslash mathrm\{tr\}(AB^\backslash mathsf\{T\})$ with transpose as conjugation $\backslash big(\backslash langle\; A,\; B\; \backslash rangle\; =\; \backslash langle\; B^\backslash mathsf\{T\},\; A^\backslash mathsf\{T\}\; \backslash rangle\; \backslash big)$ is an inner product.
Norms on inner product spaces
A linear space with a norm such as:
 $\backslash x\backslash \_p\; =\; \backslash left[\; \backslash sum\_\{i=1\}^\{\backslash infty\}\; \backslash xi\_i^p\; \backslash right]\; ^\{1/p\}\; \backslash \; x\; =\; \backslash \{\backslash xi\_i\backslash \}\; \backslash in\; \backslash mathit\; \{l\}^p\; \backslash \; ,$
where p ≠ 2 is a normed space but not an inner product space, because this norm does not satisfy the parallelogram equality required of a norm to have an inner product associated with it.^{[4]}^{[5]}
However, inner product spaces have a naturally defined norm based upon the inner product of the space itself that does satisfy the parallelogram equality:
 $\backslash x\backslash \; =\backslash sqrt\{\backslash langle\; x,\; x\backslash rangle\}.$
This is well defined by the nonnegativity axiom of the definition of inner product space. The norm is thought of as the length of the vector x.
Directly from the axioms, we can prove the following:
 $\backslash langle\; x,y\backslash rangle\; \backslash leq\; \backslash x\backslash \; \backslash cdot\; \backslash y\backslash $
 with equality if and only if x and y are linearly dependent. This is one of the most important inequalities in mathematics. It is also known in the Russian mathematical literature as the Cauchy–Bunyakowski–Schwarz inequality.
 Because of its importance, its short proof should be noted.
 It is trivial to prove the inequality true in the case y = 0. Thus we assume $\backslash langle\; y,\; y\; \backslash rangle$ is nonzero, giving us the following:
 $\backslash lambda\; =\; \backslash langle\; y\; ,\; y\; \backslash rangle^\{1\}\; \backslash langle\; x,\; y\backslash rangle$
 $0\; \backslash leq\; \backslash langle\; x\; \backslash lambda\; y,\; x\; \backslash lambda\; y\; \backslash rangle\; =\; \backslash langle\; x,\; x\backslash rangle\; \; \backslash langle\; y\; ,\; y\; \backslash rangle^\{1\}\; \; \backslash langle\; x,y\backslash rangle^2.$
 The complete proof can be obtained by multiplying out this result.
 Orthogonality: The geometric interpretation of the inner product in terms of angle and length, motivates much of the geometric terminology we use in regard to these spaces. Indeed, an immediate consequence of the Cauchy–Schwarz inequality is that it justifies defining the angle between two nonzero vectors x and y in the case $F\; =\; \backslash mathbb\{R\}$ by the identity
 $\backslash operatorname\{angle\}(x,y)\; =\; \backslash arccos\; \backslash frac\{\backslash langle\; x,\; y\; \backslash rangle\}\{\backslash x\backslash \; \backslash cdot\; \backslash y\backslash \}.$
 We assume the value of the angle is chosen to be in the interval Template:Closedclosed. This is in analogy to the situation in twodimensional Euclidean space.
 In the case $F\; =\; \backslash mathbb\{C\}$, the angle in the interval Template:Closedclosed is typically defined by
 $\backslash operatorname\{angle\}(x,y)\; =\; \backslash arccos\; \backslash frac\{\backslash langle\; x,\; y\; \backslash rangle\}\{\backslash x\backslash \; \backslash cdot\; \backslash y\backslash \}.$
 Correspondingly, we will say that nonzero vectors x and y of V are orthogonal if and only if their inner product is zero.
 $\backslash r\; \backslash cdot\; x\backslash \; =\; r\; \backslash cdot\; \backslash \; x\backslash .$
 The homogeneity property is completely trivial to prove.
 $\backslash x\; +\; y\backslash \; \backslash leq\; \backslash x\; \backslash \; +\; \backslash y\backslash .$
 The last two properties show the function defined is indeed a norm.
 Because of the triangle inequality and because of axiom 2, we see that · is a norm which turns V into a normed vector space and hence also into a metric space. The most important inner product spaces are the ones which are complete with respect to this metric; they are called Hilbert spaces. Every inner product V space is a dense subspace of some Hilbert space. This Hilbert space is essentially uniquely determined by V and is constructed by completing V.
 $\backslash x\backslash ^2\; +\; \backslash y\backslash ^2\; =\; \backslash x+y\backslash ^2.$
 The proof of the identity requires only expressing the definition of norm in terms of the inner product and multiplying out, using the property of additivity of each component.
 The name Pythagorean theorem arises from the geometric interpretation of this result as an analogue of the theorem in synthetic geometry. Note that the proof of the Pythagorean theorem in synthetic geometry is considerably more elaborate because of the paucity of underlying structure. In this sense, the synthetic Pythagorean theorem, if correctly demonstrated, is deeper than the version given above.
 An induction on the Pythagorean theorem yields:
 If x_{1}, ..., x_{n} are orthogonal vectors, that is, $\backslash langle\; x\_j,x\_k\backslash rangle=0$ for distinct indices j, k, then
 $\backslash sum\_\{i=1\}^n\; \backslash x\_i\backslash ^2\; =\; \backslash left\backslash \backslash sum\_\{i=1\}^n\; x\_i\; \backslash right\backslash ^2.$
 In view of the CauchySchwarz inequality, we also note that $\backslash langle\backslash cdot,\backslash cdot\backslash rangle$ is continuous from V × V to F. This allows us to extend Pythagoras' theorem to infinitely many summands:
 Parseval's identity: Suppose V is a complete inner product space. If {x_{k}} are mutually orthogonal vectors in V then
 $\backslash sum\_\{i=1\}^\backslash infty\backslash x\_i\backslash ^2\; =\; \backslash left\backslash \backslash sum\_\{i=1\}^\backslash infty\; x\_i\backslash right\backslash ^2,$
 provided the infinite series on the left is convergent. Completeness of the space is needed to ensure that the sequence of partial sums
 $S\_k\; =\; \backslash sum\_\{i=1\}^k\; x\_i,$
 which is easily shown to be a Cauchy sequence, is convergent.
 $\backslash x\; +\; y\backslash ^2\; +\; \backslash x\; \; y\backslash ^2\; =\; 2\backslash x\backslash ^2\; +\; 2\backslash y\backslash ^2.$
The Parallelogram law is, in fact, a necessary and sufficient condition for the existence of a scalar
product corresponding to a given norm. If it holds, the scalar product is defined by the
polarization identity:
 $\backslash x\; +\; y\backslash ^2\; =\; \backslash x\backslash ^2\; +\; \backslash y\backslash ^2\; +\; 2\; \backslash real\; \backslash langle\; x\; ,\; y\; \backslash rangle.$
 which is a form of the law of cosines.
Orthonormal sequences
Let V be a finite dimensional inner product space of dimension n. Recall that every 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 $\backslash textstyle\; \{\backslash \{e\_1,\backslash ldots,e\_n\backslash \}\}$ is orthonormal if $\backslash textstyle\; \backslash langle\; e\_i,\; e\_j\backslash rangle=0$ if $\backslash textstyle\; i\backslash neq\; j$ and $\backslash textstyle\; \backslash langle\; e\_i,\; e\_i\backslash rangle\; =\; e\_i\; =\; 1$ for each i.
This definition of orthonormal basis generalizes to the case of infinitedimensional inner product spaces in the following way. Let V be any inner product space. Then a collection $\backslash textstyle\; E=\backslash \{e\_\{\backslash alpha\}\backslash \}\_\{\backslash alpha\; \backslash in\; A\}$ is a basis 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). We say that E is an orthonormal basis for V if it is a basis and $\backslash textstyle\; \backslash langle\; e\_\{\backslash alpha\},\; e\_\{\backslash beta\}\backslash rangle=0$ if $\backslash textstyle\; \backslash alpha\; \backslash neq\; \backslash beta$ and $\backslash textstyle\; \backslash langle\; e\_\{\backslash alpha\},\; e\_\{\backslash alpha\}\backslash rangle\; =\; e\_\{\backslash alpha\}\; =\; 1$ for all $\backslash textstyle\; \backslash alpha,\backslash beta\; \backslash in\; A$.
Using an infinitedimensional analog of the GramSchmidt process one may show:
Theorem. Any separable inner product space V 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 welldefined, one may also show that
Theorem. Any complete inner product space V 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 nontrivial result, and is proved below. The following proof is taken from Halmos's A Hilbert Space Problem Book (see the references).
Proof

Recall that the dimension of an inner product space is the cardinality of a maximal orthonormal system that it contains (by Zorn's lemma it contains at least one, and any two have the same cardinality). An orthonormal basis is certainly a maximal orthonormal system, but as we shall see, the converse need not hold. Observe that if G is a dense subspace of an inner product space H, then any orthonormal basis for G is automatically an orthonormal basis for H. Thus, it suffices to construct an inner product space space H with a dense subspace G whose dimension is strictly smaller than that of H.
Let K be a Hilbert space of dimension $\backslash aleph\_0$ (for instance, $K=\backslash ell^2(\backslash mathbb\{N\})$). Let E be an orthonormal basis of K, so $E\; =\; \backslash aleph\_0$. Extend E to a Hamel basis $E\; \backslash cup\; F$ for K, where $E\; \backslash cap\; F\; =\; \backslash emptyset$. Since it is known that the Hamel dimension of K is c, the cardinality of the continuum, it must be that $F\; =\; c$.
Let L be a Hilbert space of dimension c (for instance, $L\; =\; \backslash ell^2(\backslash mathbb\{R\})$). Let B be an orthonormal basis for L, and let $\backslash phi:\; F\; \backslash to\; B$ be a bijection. Then there is a linear transformation $T:\; K\; \backslash to\; L$ such that $Tf\; =\; \backslash phi(f)$ for $f\; \backslash in\; F$, and $Te\; =\; 0$ for $e\; \backslash in\; E$.
Let $H\; =\; K\; \backslash oplus\; L$ and let $G\; =\; \backslash \{(k,Tk):\; k\; \backslash in\; K)\backslash \}$ be the graph of T. Let $\backslash bar\{G\}$ be the closure of G in H; we will show $\backslash bar\{G\}\; =\; H$. Since for any $e\; \backslash in\; E$ we have $(e,0)\; \backslash in\; G$, it follows that $K\; \backslash oplus\; 0\; \backslash subset\; \backslash bar\{G\}$.
Next, if $b\; \backslash in\; B$, then $b\; =\; Tf$ for some $f\; \backslash in\; F\; \backslash subset\; K$, so $(f,b)\; \backslash in\; G\; \backslash subset\; \backslash bar\{G\}$; since $(f,0)\; \backslash in\; \backslash bar\{G\}$ as well, we also have $(0,b)\; \backslash in\; \backslash bar\{G\}$. It follows that $0\; \backslash oplus\; L\; \backslash subset\; \backslash bar\{G\}$, so $\backslash bar\{G\}\; =\; H$, and G is dense in H.
Finally, $\backslash \{(e,0):\; e\; \backslash in\; E\backslash \}$ is a maximal orthonormal set in G; if
 $0\; =\; \backslash langle\; (e,0),\; (k,\; Tk)\; \backslash rangle\; =\; \backslash langle\; e,k\; \backslash rangle\; +\; \backslash langle\; 0,Tk\; \backslash rangle\; =\; \backslash langle\; e,k\; \backslash rangle$
for all $e\; \backslash in\; E$ then certainly $k=0$, so $(k,Tk)=(0,0)$ is the zero vector in G. Hence the dimension of G is $E\; =\; \backslash aleph\_0$, whereas it is clear that the dimension of H is c. This completes the proof.

Parseval's identity leads immediately to the following theorem:
Theorem. Let V be a separable inner product space and {e_{k}}_{k} an orthonormal basis of V.
Then the map
 $x\; \backslash mapsto\; \backslash \{\backslash langle\; e\_k,\; x\backslash rangle\backslash \}\_\{k\; \backslash in\; \backslash mathbb\{N\}\}$
is an isometric linear map V → ℓ^{ 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 ℓ^{ 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 $C[\backslash pi,\backslash pi]$. Then the sequence (indexed on set of all integers) of continuous functions
 $e\_k(t)\; =\; \{1\; \backslash over\; \backslash sqrt\{2\; \backslash pi\}\}e^\{i\; k\; t\}$
is an orthonormal basis of the space $C[\backslash pi,\backslash pi]$ with the L^{2} inner product. The mapping
 $f\; \backslash mapsto\; \backslash frac\{1\}\{\backslash sqrt\{2\; \backslash pi\}\}\; \backslash left\backslash \{\backslash int\_\{\backslash pi\}^\backslash pi\; f(t)\; e^\{i\; k\; t\}\; \backslash ,\; dt\; \backslash right\backslash \}\_\{k\; \backslash in\; \backslash mathbb\{Z\}\}$
is an isometric linear map with dense image.
Orthogonality of the sequence {e_{k}}_{k} follows immediately from the fact that if k ≠ j, then
 $\backslash int\_\{\backslash pi\}^\backslash pi\; e^\{i\; (jk)\; t\}\; \backslash ,\; dt\; =\; 0.$
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 inner product norm, follows from the fact that the sequence has a dense algebraic span, this time in the space of continuous periodic functions on $[\backslash pi,\backslash pi]$ 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 from an inner product space V to an inner product space W are of relevance:
 Continuous linear maps, i.e., A is linear and continuous with respect to the metric defined above, or equivalently, A is linear and the set of nonnegative reals {Ax}, where x ranges over the closed unit ball of V, is bounded.
 Symmetric linear operators, i.e., A is linear and $\backslash langle\; Ax,\; y\backslash rangle\; =\backslash langle\; x,\; Ay\backslash rangle$ for all x, y in V.
 Isometries, i.e., A is linear and $\backslash langle\; Ax,\; Ay\backslash rangle\; =\backslash langle\; x,\; y\backslash rangle$ for all x, y in V, or equivalently, A is linear and Ax = x for all x in V. All isometries are injective. Isometries are morphisms between inner product spaces, and morphisms of real inner product spaces are orthogonal transformations (compare with orthogonal matrix).
 Isometrical isomorphisms, i.e., A 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 positivedefiniteness is weakened.
Degenerate inner products
If V is a vector space and $\backslash langle\; ,\; \backslash rangle$ a semidefinite sesquilinear form,
then the function ‖x‖ = $\backslash langle\; x,\; x\backslash rangle^\{1/2\}$ 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 seminorm). We can produce an inner product space by considering the
quotient W = V/{ x : ‖x‖ = 0}. The sesquilinear form $\backslash langle\; ,\; \backslash rangle$ 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 semidefinite kernels on arbitrary sets.
Nondegenerate conjugate symmetric forms
Alternatively, one may require that the pairing be a nondegenerate form, meaning that for all nonzero x there exists some y such that $\backslash langle\; x,\; y\backslash rangle\; \backslash neq\; 0,$ though y need not equal x; in other words, the induced map to the dual space $V\; \backslash to\; V^*$ 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 pseudoRiemannian 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 $V\; \backslash to\; V^*$) 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 × n covector with an n × 1 vector, yielding a 1×1 matrix (a scalar), while the outer product is the product of an m×1 vector with a 1×n covector, yielding an m×n matrix. Note that 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 trace of the outer product (trace only being properly defined for square matrices).
On an inner product space, or more generally a vector space with a nondegenerate form (so an isomorphism $V\; \backslash to\; V^*$) vectors can be sent to covectors (in coordinates, via transpose), so one can take the inner product and outer product of two vectors, not simply of a vector and a covector.
In a quip: "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 $W\; \backslash times\; V^*\; \backslash to\; \backslash operatorname\{Hom\}(V,W)$ 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 $V^*\; \backslash times\; V\; \backslash to\; F$ 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 exterior (Grassmann) product are combined in the geometric product (the Clifford product in a Clifford algebra) – the inner product sends two vectors (1vectors) to a scalar (a 0vector), while the exterior product sends two vectors to a bivector (2vector) – and in this context the exterior product is usually called the "outer (alternatively, wedge) product". The inner product is more correctly called a scalar product in this context, as the nondegenerate quadratic form in question need not be positive definite (need not be an inner product).
Notes and inline references
See also
References
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