Inner product space

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 ${\displaystyle \langle y\ |\ x\rangle }$ (the bra–ket notation of quantum mechanics), respectively ${\displaystyle y^{\dagger }x}$ (dot product as a case of the convention of forming the matrix product ${\displaystyle AB,}$ as the dot products of rows of ${\displaystyle A}$ with columns of ${\displaystyle B}$). Here, the kets and columns are identified with the vectors of ${\displaystyle V,}$ and the bras and rows with the linear functionals (covectors) of the dual space ${\displaystyle V^{*},}$ with conjugacy associated with duality. This reverse order is now occasionally followed in the more abstract literature,[8] taking ${\displaystyle \langle x,y\rangle }$ to be conjugate linear in ${\displaystyle x}$ rather than ${\displaystyle y.}$ A few instead find a middle ground by recognizing both ${\displaystyle \langle \,\cdot \,,\,\cdot \,\rangle }$ and ${\displaystyle \langle \,\cdot \ |\ \cdot \,\rangle }$ 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,[9] and therefore has to have characteristic equal to ${\displaystyle 0}$ (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 for conjugation. More generally, any quadratically closed subfield of ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$ will suffice for this purpose (for example, algebraic numbers, constructible numbers). However, in the cases where it is a proper subfield (that is, 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.

Inner products are generalizations of the dot product from the real vector space ${\displaystyle \mathbb {R} ^{n}}$ to an arbitrary vector space ${\displaystyle V}$ over a real or complex field ${\displaystyle \mathbb {F} ,}$ where it will be assumed in this section that ${\displaystyle V\neq \{0\}.}$ There are many different ways to explain how the definition of an inner product arises from generalizing the dot product, and this section gives just one of these. Specifically, this section will detail how the properties (A) and (B) listed below motivate the defining properties of inner products (that is, properties (1), (2), and (3) mentioned above).

A generalization of the dot product should be a scalar-valued map of the form ${\displaystyle \langle \,\cdot ,\cdot \,\rangle :V\times V\to \mathbb {F} .}$ It should also, at a minimum, distribute over addition (that is, be additive in each argument) and induce a norm. That is, ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ should have the following properties:

1. Additive in its 1st argument and additive in its 2nd argument.
   • This generalizes the following property of the dot product, which is often stated as "the dot product distributes over addition": ${\displaystyle \;(x+y)\cdot z\,=\,x\cdot z+y\cdot z\;}$ and ${\displaystyle \;x\cdot (y+z)\,=\,x\cdot y+x\cdot z.}$
2. There exists a norm ${\displaystyle \|\,\cdot \,\|}$ on ${\displaystyle V}$ such that ${\displaystyle \langle x,\,x\rangle =\|x\|^{2}}$ for every vector ${\displaystyle x.}$
   • For the dot product on ${\displaystyle \mathbb {R} ^{n},}$ this norm is the Euclidean/${\displaystyle L^{2}}$-norm ${\displaystyle \|\,\cdot \,\|_{2}}$ because ${\displaystyle x\cdot x=x_{1}^{2}+\cdots +x_{n}^{2}=\|x\|_{2}^{2}}$ holds for all ${\displaystyle x=\left(x_{1},\ldots ,x_{n}\right)\in \mathbb {R} ^{n}.}$

Properties (A) and (B) imply that the norm ${\displaystyle \|\,\cdot \,\|}$ satisfies the parallelogram law ${\displaystyle \|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}.}$[proof 5] Consequently, the polarization identity can be used with ${\displaystyle \|\,\cdot \,\|}$ to define a unique inner product ${\displaystyle \langle \,\cdot ,\cdot \,\rangle _{P}}$ on ${\displaystyle V}$ such that ${\displaystyle \langle x,\,x\rangle _{P}:=\|x\|^{2}}$ for all ${\displaystyle x}$[10] (see this footnote[note 4] for its definition; note that ${\displaystyle \langle \,\cdot ,\cdot \,\rangle _{P}}$ might be different from ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$). Thus merely requiring that ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ satisfy (A) and (B) has already led to the appearance of an inner product (although this inner product ${\displaystyle \langle \,\cdot ,\cdot \,\rangle _{P}}$ will no longer be discussed). In addition, (A) and (B) imply that ${\displaystyle \langle x,y\rangle +\langle y,x\rangle ={\tfrac {1}{2}}\left[\|x+y\|^{2}-\|x-y\|^{2}\right]}$ so if ${\displaystyle \mathbb {F} =\mathbb {C} }$ then ${\displaystyle \langle x,ix\rangle =-\langle ix,x\rangle }$[proof 6] and ${\displaystyle \operatorname {im} \langle x,y\rangle =-\operatorname {im} \langle y,x\rangle ,}$[proof 5] where ${\displaystyle \operatorname {im} }$ denotes the imaginary part of a scalar; this indicates that ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ should not be expected to be a symmetric map in the complex case and that Hermitian symmetry holds if and only if the real parts of ${\displaystyle \langle x,y\rangle }$ and ${\displaystyle \langle y,x\rangle }$ are always equal.

Some motivation for why ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ should be homogeneous in one argument and conjugate homogeneous in the other will now be given. Additivity in each argument guarantees that ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ is homogeneous over the rational numbers ${\displaystyle \mathbb {Q} }$ in each argument, meaning that for all vectors ${\displaystyle x,y,}$ and all rational ${\displaystyle q\in \mathbb {Q} }$

$${\displaystyle \langle qx,y\rangle ~=~q\,\langle x,y\rangle \qquad {\text{ and }}\qquad \langle x,qy\rangle ~=~q\,\langle x,y\rangle .}$$

For these equalities to hold for all real ${\displaystyle q\in \mathbb {R} }$ (that is, for ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ to be real homogeneous in each argument), it suffices for there to exist some norm-induced topology on ${\displaystyle V}$ (or even just some vector topology) that makes the map ${\displaystyle \langle \,\cdot ,\cdot \,\rangle :V\times V\to \mathbb {F} }$ continuous (or even just separately continuous),[proof 7] which is a relatively mild condition. These observations suggest that it is not too much to ask that ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ be real homogeneous in each argument, which if ${\displaystyle \mathbb {F} =\mathbb {R} }$ is the same as being homogeneous (and thus linear) in each argument. Given real homogeneity in both arguments, if ${\displaystyle \mathbb {F} =\mathbb {C} }$ then it is reasonable to demand that ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ be (fully) homogeneous in at least one argument, say in its first argument, which will be true if and only if ${\displaystyle \langle ix,y\rangle =i\langle x,y\rangle }$ holds for all ${\displaystyle x,y.}$

Property (B) and homogeneity in the 1st argument imply that for any scalar ${\displaystyle c}$ and vector ${\displaystyle x,}$

$${\displaystyle c\langle x,\,cx\rangle =\langle cx,\,cx\rangle =\|cx\|^{2}=|c|^{2}\|x\|^{2}=|c|^{2}\langle x,\,x\rangle =c{\overline {c}}\langle x,\,x\rangle }$$

where solving for ${\displaystyle \langle x,\,cx\rangle }$ gives:

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

Thus when ${\displaystyle \mathbb {F} =\mathbb {C} ,}$ property (B) makes it impossible for ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ to be homogeneous in both arguments. Moreover, the formula ${\displaystyle \langle x,\,cx\rangle ={\overline {c}}\langle x,\,x\rangle }$ suggests[note 5] that ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ should be conjugate homogeneous in its second argument. The above derivation of the formula ${\displaystyle \langle x,\,cx\rangle ={\overline {c}}\langle x,\,x\rangle }$ shows that the appearance of the complex conjugation ${\displaystyle {\overline {c}}}$ of the scalar ${\displaystyle c}$ may be viewed as a consequence of the fact that (1) absolute homogeneity ${\displaystyle \|cx\|=|c|\,\|x\|}$ introduces an absolute value around ${\displaystyle c}$ while homogeneity does not, and (2) the equation ${\displaystyle |c|^{2}=cz}$ is solved by the complex conjugate ${\displaystyle z={\overline {c}}.}$ Said differently, the complex conjugate arises from solving ${\displaystyle \langle x,\,cx\rangle =z\langle x,\,x\rangle }$ for ${\displaystyle z,}$ where (1) absolute homogeneity leads to the factor ${\displaystyle |c|^{2}}$ in ${\displaystyle |c|^{2}\langle x,\,x\rangle =cz\langle x,\,x\rangle ,}$ and (2) this newest equation happens to be solved by ${\displaystyle z={\overline {c}}.}$

Property (B) implies that the map ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ must be positive definite (condition (3) above), which explains why the definition of an inner product requires positive definiteness. Because it is positive definite, ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ is an inner product if and only if it satisfies (1) and (2) above (that is, linearity in the first argument and Hermitian symmetry), which are the two defining properties of a Hermitian form. Since ${\displaystyle \langle x,x\rangle =\|x\|^{2}}$ is real for all ${\displaystyle x,}$ when ${\displaystyle \mathbb {F} =\mathbb {C} }$ then ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ is a Hermitian form if and only if it is a sesquilinear form.[1] Because ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ is assumed to be additive in each argument, it is a sesquilinear form if and only if it is homogeneous in one argument and conjugate homogeneous in the other. Thus to motivate why inner products should satisfy properties (1) and (2) above, it is enough to give motivation for why ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ should be a sesquilinear form, which has already been discussed.

Among the simplest examples of inner product spaces are ${\displaystyle \mathbb {R} }$ and ${\displaystyle \mathbb {C} .}$ The real numbers ${\displaystyle \mathbb {R} }$ are a vector space over ${\displaystyle \mathbb {R} }$ that becomes a real inner product space when endowed with standard multiplication as its real inner product:[3]

$${\displaystyle \langle x,y\rangle$$ :=xy\quad {\text{ for }}x,y\in \mathbb {R} .}

The complex numbers ${\displaystyle \mathbb {C} }$ are a vector space over ${\displaystyle \mathbb {C} }$ that becomes a complex inner product space when endowed with the complex inner product

$${\displaystyle \langle x,y\rangle$$ :=x{\overline {y}}\quad {\text{ for }}x,y\in \mathbb {C} .}

Unlike with the real numbers, the assignment ${\displaystyle (x,y)\mapsto xy}$ does not define a complex inner product on ${\displaystyle \mathbb {C} .}$

More generally, the real ${\displaystyle n}$-space ${\displaystyle \mathbb {R} ^{n}}$ with the dot product is an inner product space,[3] 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 ${\displaystyle x^{\operatorname {T} }}$ is the transpose of ${\displaystyle x.}$

A function ${\displaystyle \langle \,\cdot ,\cdot \,\rangle$ :\mathbb {R} ^{n}\times \mathbb {R} ^{n}\to \mathbb {R} } is an inner product on ${\displaystyle \mathbb {R} ^{n}}$ if and only if there exists a symmetric positive-definite matrix ${\displaystyle \mathbf {M} }$ such that ${\displaystyle \langle x,y\rangle =x^{\operatorname {T} }\mathbf {M} y}$ for all ${\displaystyle x,y\in \mathbb {R} ^{n}.}$ If ${\displaystyle \mathbf {M} }$ is the identity matrix then ${\displaystyle \langle x,y\rangle =x^{\operatorname {T} }\mathbf {M} y}$ is the dot product. For another example, if ${\displaystyle n=2}$ and ${\displaystyle \mathbf {M} ={\begin{bmatrix}a&b\\b&d\end{bmatrix}}}$ is positive-definite (which happens if and only if ${\displaystyle \det \mathbf {M} =ad-b^{2}>0}$ and one/both diagonal elements are positive) then for any ${\displaystyle x:=\left[x_{1},x_{2}\right]^{\operatorname {T} },y:=\left[y_{1},y_{2}\right]^{\operatorname {T} }\in \mathbb {R} ^{2},}$

$${\displaystyle \langle x,y\rangle$$ :=x^{\operatorname {T} }\mathbf {M} y=\left[x_{1},x_{2}\right]{\begin{bmatrix}a&b\\b&d\end{bmatrix}}{\begin{bmatrix}y_{1}\\y_{2}\end{bmatrix}}=ax_{1}y_{1}+bx_{1}y_{2}+bx_{2}y_{1}+dx_{2}y_{2}.}

As mentioned earlier, every inner product on ${\displaystyle \mathbb {R} ^{2}}$ is of this form (where ${\displaystyle b\in \mathbb {R} ,a>0}$ and ${\displaystyle d>0}$ satisfy ${\displaystyle ad>b^{2}}$).

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 ${\displaystyle M}$ is any Hermitian positive-definite matrix and ${\displaystyle y^{\dagger }}$ is the conjugate transpose of ${\displaystyle y.}$ 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, ${\displaystyle \{f_{k}\}_{k},}$ 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 ${\displaystyle X}$ and ${\displaystyle Y,}$ the expected value of their product

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

is an inner product.[11][12][13] In this case, ${\displaystyle \langle X,X\rangle =0}$ if and only if ${\displaystyle \Pr(X=0)=1}$ (that is, ${\displaystyle X=0}$ almost surely), where ${\displaystyle \Pr }$ denotes the probability of the event. This definition of expectation as inner product can be extended to random vectors as well.

The inner product for complex square matrices of the same size is the Frobenius inner product ${\displaystyle \langle A,B\rangle$ :=\operatorname {tr} \left(AB^{\textsf {H}}\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,

$${\displaystyle \langle A,B\rangle =\operatorname {tr} \left(AB^{\textsf {H}}\right)={\overline {\operatorname {tr} \left(BA^{\textsf {H}}\right)}}={\overline {\left\langle B,A\right\rangle }}}$$

Finally, since for ${\displaystyle A}$ nonzero, ${\displaystyle \langle A,A\rangle =\sum _{ij}\left|A_{ij}\right|^{2}>0}$, we get that the Frobenius inner product is positive definite too, and so is an inner product.

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

Every inner product space induces a norm, called its canonical norm, that is defined by[3]

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

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

$${\displaystyle d(x,y)=\|y-x\|.}$$

The axioms of the inner product guarantee that the map above forms a norm, which will have the following properties.

Homogeneity

For a vector ${\displaystyle x\in V}$ and a scalar ${\displaystyle s}$

$${\displaystyle \|sx\|=|s|\,\|x\|.}$$

Triangle inequality

For vectors ${\displaystyle x,y\in V}$

$${\displaystyle \|x+y\|\leq \|x\|+\|y\|.}$$

These two properties show that one has indeed a norm.

Cauchy–Schwarz inequality

For vectors ${\displaystyle x,y\in V}$

$${\displaystyle |\langle x,y\rangle |\leq \|x\|\,\|y\|}$$

with equality if and only if ${\displaystyle x}$ and ${\displaystyle y}$ are linearly dependent. In the Russian mathematical literature, this inequality is also known as the Cauchy–Bunyakovsky inequality or the Cauchy–Bunyakovsky–Schwarz inequality.

Cosine similarity

When ${\displaystyle \langle x,y\rangle }$ is a real number then the Cauchy–Schwarz inequality guarantees that ${\displaystyle {\frac {\langle x,y\rangle }{\|x\|\,\|y\|}}\in [-1,1]}$ lies in the domain of the inverse trigonometric function ${\displaystyle \arccos$ :[-1,1]\to [0,\pi ]} and so the (non oriented) angle between ${\displaystyle x}$ and ${\displaystyle y}$ can be defined as:

$${\displaystyle \angle (x,y)=\arccos {\frac {\langle x,y\rangle }{\|x\|\,\|y\|}},}$$

where

$${\displaystyle 0\leq \;\angle (x,y)\;\leq \pi .}$$

Polarization identity

The inner product can be retrieved from the norm by the polarization identity

$${\displaystyle \|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}+2\operatorname {Re} \langle x,y\rangle ,}$$

which is a form of the law of cosines.

Orthogonality

Two vectors ${\displaystyle x}$ and ${\displaystyle y}$ are called orthogonal, written ${\displaystyle x\perp y,}$ if their inner product is zero: ${\displaystyle \langle x,y\rangle =0.}$ This happens if and only if ${\displaystyle \|x\|\leq \|x+sy\|}$ for all scalars ${\displaystyle s,}$[14] or said differently, if and only if the real-valued map ${\displaystyle f(s):=\|x+sy\|^{2}-\|x\|^{2}}$ is non-negative. This is a consequence of the fact that if ${\displaystyle y\neq 0}$ then the scalar ${\displaystyle s_{0}:=-{\tfrac {\overline {\langle x,y\rangle }}{\|y\|^{2}}}}$ minimizes ${\displaystyle f}$ with value ${\displaystyle f\left(s_{0}\right)=-{\tfrac {|\langle x,y\rangle |^{2}}{\|y\|^{2}}},}$ which is always ${\displaystyle \leq 0.}$ For a complex − but not real − inner product space ${\displaystyle H,}$ a linear operator ${\displaystyle T:V\to V}$ is identically ${\displaystyle 0}$ if and only if ${\displaystyle x\perp Tx}$ for every ${\displaystyle x\in V.}$[14]

Orthogonal complement

The orthogonal complement of a subset ${\displaystyle C\subseteq V}$ is the set ${\displaystyle C^{\bot }}$ of all vectors ${\displaystyle y\in V}$ such that ${\displaystyle y}$ and ${\displaystyle c}$ are orthogonal for all ${\displaystyle c\in C}$; that is, it is the set

$${\displaystyle C^{\bot }:=\{\,y\in V:\langle y,c\rangle =0{\text{ for all }}c\in C\,\}.}$$

This set ${\displaystyle C^{\bot }}$ is always a closed vector subspace of ${\displaystyle V}$ and if the closure ${\displaystyle \operatorname {cl} _{V}C}$ of ${\displaystyle C}$ in ${\displaystyle V}$ is a vector subspace then ${\displaystyle \operatorname {cl} _{V}C=\left(C^{\bot }\right)^{\bot }.}$

Pythagorean theorem

Whenever ${\displaystyle x,y\in V}$ and ${\displaystyle \langle x,y\rangle =0}$ then

$${\displaystyle \|x\|^{2}+\|y\|^{2}=\|x+y\|^{2}.}$$

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 in Euclidean geometry.

Parseval's identity

An induction on the Pythagorean theorem yields: if ${\displaystyle x_{1},\ldots ,x_{n}}$ are orthogonal vectors (meaning that ${\displaystyle \left\langle x_{j},x_{k}\right\rangle =0}$ for distinct indices ${\displaystyle j\neq k}$) then

$${\displaystyle \sum _{i=1}^{n}\|x_{i}\|^{2}=\left\|\sum _{i=1}^{n}x_{i}\right\|^{2}.}$$

Parallelogram law

For all ${\displaystyle x,y\in V,}$

$${\displaystyle \|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}.}$$

The parallelogram law is, in fact, a necessary and sufficient condition for the existence of an inner product corresponding to a given norm.

Ptolemy's inequality

For all ${\displaystyle x,y,z\in V,}$

$${\displaystyle \|x-y\|\|z\|+\|y-z\|\|x\|\geq \|x-z\|\|y\|.}$$

Ptolemy's inequality is, in fact, a necessary and sufficient condition for the existence of an inner product corresponding to a given norm. In detail, Isaac Jacob Schoenberg proved in 1952 that, given any real, seminormed space, if its seminorm is ptolemaic, then the seminorm is the norm associated with an inner product.[15]

Suppose that ${\displaystyle \langle \cdot ,\cdot \rangle }$ is an inner product on ${\displaystyle V}$ (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 ${\displaystyle 0.}$

Assume for the rest of this section that ${\displaystyle V}$ 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.

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 ${\displaystyle V:=\mathbb {C} ^{n}}$ for some integer ${\displaystyle n>0.}$ When ${\displaystyle V}$ is considered as a real vector space in the usual way (meaning that it is identified with the ${\displaystyle 2n-}$dimensional real vector space ${\displaystyle \mathbb {R} ^{2n},}$ with each ${\displaystyle \left(a_{1}+ib_{1},\ldots ,a_{n}+ib_{n}\right)\in \mathbb {C} ^{n}}$ identified with ${\displaystyle \left(a_{1},b_{1},\ldots ,a_{n},b_{n}\right)\in \mathbb {R} ^{2n}}$), then the dot product ${\displaystyle x\,\cdot \,y=\left(x_{1},\ldots ,x_{2n}\right)\,\cdot \,\left(y_{1},\ldots ,y_{2n}\right):=x_{1}y_{1}+\cdots +x_{2n}y_{2n}}$ defines a real inner product on this space. The unique complex inner product ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ on ${\displaystyle V=\mathbb {C} ^{n}}$ induced by the dot product is the map that sends ${\displaystyle c=\left(c_{1},\ldots ,c_{n}\right),d=\left(d_{1},\ldots ,d_{n}\right)\in \mathbb {C} ^{n}}$ to ${\displaystyle \langle c,d\rangle$ :=c_{1}{\overline {d_{1}}}+\cdots +c_{n}{\overline {d_{n}}}} (because the real part of this map ${\displaystyle \langle \,\cdot ,\cdot \,\rangle }$ is equal to the dot product).

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 a bilinear and symmetric map.

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 ${\displaystyle y}$); thus the standard inner product ${\displaystyle \langle x,y\rangle =x{\overline {y}},}$ on ${\displaystyle \mathbb {C} }$ is an "extension" the dot product . 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 conjugate symmetric 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; furthermore, without the complex conjugate, if ${\displaystyle x\in \mathbb {C} }$ but ${\displaystyle x\not \in \mathbb {R} }$ then ${\displaystyle \langle x,x\rangle =xx=x^{2}\not \in [0,\infty )}$ so the assignment ${\displaystyle x\mapsto {\sqrt {\langle x,x\rangle }}}$ 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 ${\displaystyle \langle x,y\rangle =0}$ 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 ${\displaystyle i={\sqrt {-1}}}$ is not defined in ${\displaystyle V_{\mathbb {R} },}$ the vector in ${\displaystyle V}$ denoted by ${\displaystyle ix}$ is nevertheless still also 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 ${\displaystyle \langle x,ix\rangle _{\mathbb {R} }=0.}$