Generalization of the dot product; used to define Hilbert spaces
Geometric interpretation of the angle between two vectors defined using an inner product
Scalar product spaces, over any field, have "scalar products" that are symmetrical and linear in the first argument. Hermitian product spaces are restricted to the field of complex numbers and have "Hermitian products" that are conjugate-symmetrical and linear in the first argument. Inner product spaces may be defined over any field, having "inner products" that are linear in the first argument, conjugate-symmetrical, and positive-definite. Unlike inner products, scalar products and Hermitian products need not be positive-definite.
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.^{[1]} If an inner product space H is not a Hilbert space, it can be extended by completion to a Hilbert space ${\overline {H}}.$ This means that $H$ is a linear subspace of ${\overline {H}},$ the inner product of $H$ is the restriction of that of ${\overline {H}},$ and $H$ is dense in ${\overline {H}}$ for the topology defined by the norm.^{[1]}^{[5]}
called an inner product (linear in its first argument) that satisfies the following conditions (1), (2), and (3) for all vectors $x,y,z\in V$ and all scalars $s\in \mathbb {F}$:^{[1]}^{[6]}^{[7]}
Each of the above two properties imply $\langle \mathbf {0} ,x\rangle =0$ for every vector $x.$^{[proof 1]}
This map $\langle \cdot ,\cdot \rangle$ is called a sesquilinear form if condition (1) holds and if $\langle \cdot ,\cdot \rangle$ is also antilinear (sometimes called, conjugate linear) in its second argument,^{[1]} which by definition means that the following always hold:
(note the complex conjugation${\overline {s}}$ of the scalar $s$ in the conjugate homogeneity property above). Every inner product is a special type of sesquilinear form.
Hermitian symmetry or conjugate symmetry:^{[note 2]}
Conditions (1) and (2) are the defining properties of a Hermitian form, which is a special type of sesquilinear form.^{[1]} A complex sesquilinear form is Hermitian if and only if $\langle x,x\rangle$ is real for all $x.$^{[1]}^{[proof 2]}
This condition (Hermitian symmetry) implies that $\langle x,x\rangle$ is a real number for all $x.$^{[proof 3]}
If $\mathbb {F} =\mathbb {R}$ then this condition holds if and only if $\langle \,\cdot ,\cdot \,\rangle$ is a symmetric map, meaning that $\langle x,y\rangle =\langle y,x\rangle$ for all $x,y.$ If $\langle \,\cdot ,\cdot \,\rangle$ is symmetric and condition (1) holds then $\langle \,\cdot ,\cdot \,\rangle$ is a bilinear map (although there exist bilinear maps that are not symmetric). If $\mathbb {F} =\mathbb {C}$ and $\langle \,\cdot ,\cdot \,\rangle$ takes a non-real scalar as a value then it can not be both symmetric and conjugate symmetric.
$\langle x,x\rangle >0\quad {\text{ if }}x\neq \mathbf {0}$
(Positive definite)
The above three conditions are the defining properties of an inner product, which is why an inner product is sometimes (equivalently) defined as being a positive-definite Hermitian form.
An inner product can equivalently be defined as a positive-definite sesquilinear form.^{[1]}^{[note 3]}
If $\mathbb {F} =\mathbb {R}$ (respectively, if $\mathbb {F} =\mathbb {C}$) then an inner product is called a real inner product (respectively, a complex inner product).
Assuming $\langle \mathbf {0} ,\mathbf {0} \rangle =0$ holds, then condition (3) will hold if and only if both conditions (4) and (5) below hold:^{[1]}^{[5]}
If the assignment $V\to \mathbb {C}$ given by $x\mapsto {\sqrt {\langle x,x\rangle }}$ defines a seminorm on $V,$ then this seminorm will be a norm if and only if condition (4) is satisfied.
$\langle x,x\rangle \geq 0\quad {\text{ for every }}x$
(Positive semi-definiteness)
This condition holds if only if the assignment $V\to \mathbb {C}$ defined by $x\mapsto {\sqrt {\langle x,x\rangle }}$ is well-defined and valued in $[0,\infty ).$ If this condition does not hold then this assignment does not define a seminorm (nor a norm) on $V$ (because by definition, seminorms and norms must be non-negative real-valued).
Conditions (1) through (5) are satisfied by every inner product.
This seminorm will be a norm if and only if condition (4) is satisfied.
The constant $0$ map is a positive semi-definite Hermitian form. Indeed, the constant $0$ map is bilinear, sesquilinear, symmetric, conjugate symmetric, and nonnegative-definite; but it is positive definite if and only if $V=\{\mathbf {0} \}.$
Conjugate symmetry versus bilinearity and symmetry:
Reasons for why complex inner products are required to be conjugate symmetric instead of bilinear or symmetric are now given.
Assuming that $\mathbb {F} =\mathbb {C}$ and that $x\in V$ satisfies $\langle x,x\rangle \neq 0,$ if $\langle \,\cdot ,\cdot \,\rangle$ is bilinear then whenever $c\in \mathbb {C}$ is a scalar such that $c^{2}\not \in \mathbb {R}$ then it is impossible for both$\langle x,x\rangle$ and $\langle cx,cx\rangle =c^{2}\langle x,x\rangle$ to be real numbers;
^{[proof 4]} consequently, in this case the bilinear map $\langle \,\cdot ,\cdot \,\rangle$ cannot be nonnegative-definite and thus the assignment $v\mapsto {\sqrt {\langle v,v\rangle }}$ will not define a seminorm (nor a norm) on $V.$
If rather than being bilinear, the map $\langle \,\cdot ,\cdot \,\rangle$ is instead sesquilinear (which will be true when conditions (1) and (2) are satisfied), then for any scalar $c\neq 0,$$\langle x,x\rangle$ is a real number if and only if $\langle cx,cx\rangle =c{\overline {c}}\langle x,x\rangle =|c|^{2}\langle x,x\rangle$ is a real number.
This shows that when $\mathbb {F} =\mathbb {C}$ then it is possible for a sesquilinear map to be nonnegative-definite (and thus to induce a seminorm or norm) but this is never possible for a non-zero bilinear map. This is one reason for requiring inner products to be conjugate symmetric instead of bilinear. If $\langle \,\cdot ,\cdot \,\rangle$ satisfies (1) and is symmetric (instead of conjugate symmetric) then it is necessarily bilinear, which is why complex inner products (unlike real inner products) are not required to be symmetric.
In particular, if $V=\mathbb {C}$ then defining $\langle \,\cdot ,\cdot \,\rangle$ by $\langle x,\,y\rangle :=x{\overline {y}}$ (with complex conjugation) will induce a norm on $V$ but defining it by $\langle x,\,y\rangle :=xy$ (without complex conjugation) will not.
Alternative definitions, notations and remarks
A common special case of the inner product, the scalar product or dot product, is written with a centered dot $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 $\langle x,y\rangle$ as $\langle y\ |\ x\rangle$ (the bra–ket notation of quantum mechanics), respectively $y^{\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 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,^{[8]} taking $\langle x,y\rangle$ to be conjugate linear in $x$ rather than $y.$ A few instead find a middle ground by recognizing both $\langle \,\cdot \,,\,\cdot \,\rangle$ and $\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 $\mathbb {R}$ and $\mathbb {C}$ in the definition. Briefly, the base field has to contain an orderedsubfield in order for non-negativity to make sense,^{[9]} 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 for conjugation. More generally, any quadratically closed subfield of $\mathbb {R}$ or $\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 $\mathbb {R}$ nor $\mathbb {C}$), even finite-dimensional inner product spaces will fail to be metrically complete. In contrast, all finite-dimensional inner product spaces over $\mathbb {R}$ or $\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 $\langle x,x\rangle$ is only required to be non-negative. Treatment for these cases are illustrated below.
Motivation for the definition
Inner products are generalizations of the dot product from the real vector space $\mathbb {R} ^{n}$ to an arbitrary vector space $V$ over a real or complex field $\mathbb {F} ,$ where it will be assumed in this section that $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 $\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, $\langle \,\cdot ,\cdot \,\rangle$ should have the following properties:
This generalizes the following property of the dot product, which is often stated as "the dot product distributes over addition": $\;(x+y)\cdot z\,=\,x\cdot z+y\cdot z\;$ and $\;x\cdot (y+z)\,=\,x\cdot y+x\cdot z.$
There exists a norm$\|\,\cdot \,\|$ on $V$ such that $\langle x,\,x\rangle =\|x\|^{2}$ for every vector $x.$
For the dot product on $\mathbb {R} ^{n},$ this norm is the Euclidean/$L^{2}$-norm$\|\,\cdot \,\|_{2}$ because $x\cdot x=x_{1}^{2}+\cdots +x_{n}^{2}=\|x\|_{2}^{2}$ holds for all $x=\left(x_{1},\ldots ,x_{n}\right)\in \mathbb {R} ^{n}.$
Properties (A) and (B) imply that the norm $\|\,\cdot \,\|$ satisfies the parallelogram law$\|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}.$^{[proof 5]}
Consequently, the polarization identity can be used with $\|\,\cdot \,\|$ to define a unique inner product $\langle \,\cdot ,\cdot \,\rangle _{P}$ on $V$ such that $\langle x,\,x\rangle _{P}:=\|x\|^{2}$ for all $x$^{[10]} (see this footnote^{[note 4]} for its definition; note that $\langle \,\cdot ,\cdot \,\rangle _{P}$ might be different from $\langle \,\cdot ,\cdot \,\rangle$).
Thus merely requiring that $\langle \,\cdot ,\cdot \,\rangle$ satisfy (A) and (B) has already led to the appearance of an inner product (although this inner product $\langle \,\cdot ,\cdot \,\rangle _{P}$ will no longer be discussed).
In addition, properties (A) and (B) imply that the sum $\langle x,y\rangle +\langle y,x\rangle$ is always a real number^{[proof 5]} and that if $\mathbb {F} =\mathbb {C}$ then $\langle x,ix\rangle =-\langle ix,x\rangle$ for all $x,$^{[proof 6]} which indicates that $\langle \,\cdot ,\cdot \,\rangle$ should not be expected to be a symmetric map in the complex case.
For these equalities to hold for all real$q\in \mathbb {R}$ (that is, for $\langle \,\cdot ,\cdot \,\rangle$ to be real homogeneous in each argument), it suffices for there to exist some norm-induced topology on $V$ (or even just some vector topology) that makes the map $\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 $\langle \,\cdot ,\cdot \,\rangle$ be real homogeneous in each argument, which if $\mathbb {F} =\mathbb {R}$ is the same as being homogeneous (and thus linear) in each argument.
Given real homogeneity in both arguments, if $\mathbb {F} =\mathbb {C}$ then it is reasonable to demand that $\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 $\langle ix,y\rangle =i\langle x,y\rangle$ holds for all $x,y.$
Thus when $\mathbb {F} =\mathbb {C} ,$ property (B) makes it impossible for $\langle \,\cdot ,\cdot \,\rangle$ to be homogeneous in both arguments.
Moreover, the formula $\langle x,\,cx\rangle ={\overline {c}}\langle x,\,x\rangle$ suggests^{[note 5]} that $\langle \,\cdot ,\cdot \,\rangle$ should be conjugate homogeneous in its second argument.
The above derivation of the formula $\langle x,\,cx\rangle ={\overline {c}}\langle x,\,x\rangle$ shows that the appearance of the complex conjugation${\overline {c}}$ of the scalar $c$ may be viewed as a consequence of the fact that (1) absolute homogeneity$\|cx\|=|c|\,\|x\|$ introduces an absolute value around $c$ while homogeneity does not, and (2) the equation $|c|^{2}=cz$ is solved by the complex conjugate $z={\overline {c}}.$
Said differently, the complex conjugate arises from solving $\langle x,\,cx\rangle =z\langle x,\,x\rangle$ for $z,$ where (1) absolute homogeneity leads to the factor $|c|^{2}$ in $|c|^{2}\langle x,\,x\rangle =cz\langle x,\,x\rangle ,$ and (2) this newest equation happens to be solved by $z={\overline {c}}.$
Property (B) implies that the map $\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, $\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 $\langle x,x\rangle =\|x\|^{2}$ is real for all $x,$ when $\mathbb {F} =\mathbb {C}$ then $\langle \,\cdot ,\cdot \,\rangle$ is a Hermitian form if and only if it is a sesquilinear form.^{[1]}
Because $\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 $\langle \,\cdot ,\cdot \,\rangle$ should be a sesquilinear form, which has already been discussed.
For every vector $x,$conjugate symmetry guarantees $\langle x,x\rangle ={\overline {\langle x,x\rangle }},$ which implies that $\langle x,x\rangle$ is a real number. It also guarantees that for all vectors $x$ and $y,$
where $\operatorname {Re} z$ denotes the real part of a scalar $z.$
Conjugate symmetry and linearity in the first variable imply^{[proof 8]}conjugate linearity, also known as antilinearity, in the second argument; explicitly, this means that for any vectors $x,y,z$ and any scalar $s,$
This shows that every inner product is also a sesquilinear form and that inner products are additive in each argument, meaning that for all vectors $x,y,z:$
${\begin{alignedat}{4}&{\text{Additivity in the 1st argument: }}&&\quad \langle x+y,z\rangle &&=\langle x,z\rangle +\langle y,z\rangle \\&{\text{Additivity in the 2nd argument: }}&&\quad \langle x,y+z\rangle &&=\langle x,y\rangle +\langle x,z\rangle \end{alignedat}}$
Additivity in each argument implies the following important generalization of the familiar square expansion:
In the case of $\mathbb {F} =\mathbb {R} ,$conjugate symmetry$\langle x,y\rangle ={\overline {\langle y,x\rangle }}$ reduces to symmetry $\langle x,y\rangle =\langle y,x\rangle$ and so sesquilinearity reduces to bilinearity. Hence an inner product on a real vector space is a positive-definite symmetric bilinear form. That is, when $\mathbb {F} =\mathbb {R}$ then
Every inner product is an L-semi-inner product although not all L-semi-inner products are inner products.
One can define an inner product on every finite vector space with a basis, by taking the dot-product of the unique coordinate vectors in the baseis. Conversely, the inner product of a vector space with an orthogonal basis, can always be expressed as a dot-product, using Parseval's identity.
Some examples
Real and complex numbers
Among the simplest examples of inner product spaces are $\mathbb {R}$ and $\mathbb {C} .$
The real numbers$\mathbb {R}$ are a vector space over $\mathbb {R}$ that becomes a real inner product space when endowed with standard multiplication as its real inner product:^{[3]}
$\langle x,y\rangle :=xy\quad {\text{ for }}x,y\in \mathbb {R} .$
The complex numbers$\mathbb {C}$ are a vector space over $\mathbb {C}$ that becomes a complex inner product space when endowed with the complex inner product
$\langle x,y\rangle :=x{\overline {y}}\quad {\text{ for }}x,y\in \mathbb {C} .$
Unlike with the real numbers, the assignment $(x,y)\mapsto xy$ does not define a complex inner product on $\mathbb {C} .$
where $x^{\operatorname {T} }$ is the transpose of $x.$
A function $\langle \,\cdot ,\cdot \,\rangle :\mathbb {R} ^{n}\times \mathbb {R} ^{n}\to \mathbb {R}$ is an inner product on $\mathbb {R} ^{n}$ if and only if there exists a symmetricpositive-definite matrix$\mathbf {M}$ such that $\langle x,y\rangle =x^{\operatorname {T} }\mathbf {M} y$ for all $x,y\in \mathbb {R} ^{n}.$ If $\mathbf {M}$ is the identity matrix then $\langle x,y\rangle =x^{\operatorname {T} }\mathbf {M} y$ is the dot product. For another example, if $n=2$ and $\mathbf {M} ={\begin{bmatrix}a&b\\b&d\end{bmatrix}}$ is positive-definite (which happens if and only if $\det \mathbf {M} =ad-b^{2}>0$ and one/both diagonal elements are positive) then for any $x:=\left[x_{1},x_{2}\right]^{\operatorname {T} },y:=\left[y_{1},y_{2}\right]^{\operatorname {T} }\in \mathbb {R} ^{2},$
where $M$ is any Hermitianpositive-definite matrix and $y^{\dagger }$ 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([a,b])$ of continuous complex valued functions $f$ and $g$ on the interval $[a,b].$ The inner product is
is an inner product.^{[11]}^{[12]}^{[13]} 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 {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,
Finally, since for $A$ nonzero, $\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.
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 canonical norm, that is defined by^{[3]}
with equality if and only if $x$ and $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.
When $\langle x,y\rangle$ is a real number then the Cauchy–Schwarz inequality guarantees that ${\frac {\langle x,y\rangle }{\|x\|\,\|y\|}}\in [-1,1]$ lies in the domain of the inverse trigonometric function$\arccos :[-1,1]\to [0,\pi ]$ and so the (non oriented) angle between $x$ and $y$ can be defined as:
Two vectors $x$ and $y$ are called orthogonal, written $x\perp y,$ if their inner product is zero: $\langle x,y\rangle =0.$ This happens if and only if $\|x\|\leq \|x+sy\|$ for all scalars $s.$^{[14]} Moreover, for $y\neq 0,$ the scalar $s_{0}:=-{\frac {\overline {\langle x,y\rangle }}{\|y\|^{2}}}$ minimizes $f(s):=\left\|x+sy\right\|^{2}-\|x\|^{2}$ with value $f\left(s_{0}\right)=-{\frac {|\langle x,y\rangle |^{2}}{\|y\|^{2}}}.$
For a complex − but not real − inner product space $H,$ a linear operator $T:V\to V$ is identically $0$ if and only if $x\perp Tx$ for every $x\in V.$^{[14]}
The orthogonal complement of a subset $C\subseteq V$ is the set $C^{\bot }$ of all vectors $y\in V$ such that $y$ and $c$ are orthogonal for all $c\in C$; that is, it is the set
$C^{\bot }:=\left\{\,y\in V:\langle y,c\rangle =0{\text{ for all }}c\in C\,\right\}.$
This set $C^{\bot }$ is always a closed vector subspace of $V$ and if the closure$\operatorname {cl} _{V}C$ of $C$ in $V$ is a vector subspace then $\operatorname {cl} _{V}C=\left(C^{\bot }\right)^{\bot }.$
Whenever $x,y\in V$ and $\langle x,y\rangle =0$ then
$\|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.
An induction on the Pythagorean theorem yields: if $x_{1},\ldots ,x_{n}$ are orthogonal vectors (meaning that $\left\langle x_{j},x_{k}\right\rangle =0$ for distinct indices $j\neq k$) then
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]}
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
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 first, rather than its second, argument. The real part of both $\langle x\mid y\rangle$ and $\langle x,y\rangle$ are equal to $\operatorname {Re} \langle x,y\rangle$ but the inner products differ in their complex 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:=\mathbb {C} ^{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 $2n-$dimensional real vector space $\mathbb {R} ^{2n},$ with each $\left(a_{1}+ib_{1},\ldots ,a_{n}+ib_{n}\right)\in \mathbb {C} ^{n}$ identified with $\left(a_{1},b_{1},\ldots ,a_{n},b_{n}\right)\in \mathbb {R} ^{2n}$), then the dot product$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 $\langle \,\cdot ,\cdot \,\rangle$ on $V=\mathbb {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 \mathbb {C} ^{n}$ to $\langle c,d\rangle :=c_{1}{\overline {d_{1}}}+\cdots +c_{n}{\overline {d_{n}}}$ (because the real part of this map $\langle \,\cdot ,\cdot \,\rangle$ is equal to the dot product).
Real vs. complex inner products
Let $V_{\mathbb {R} }$ 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 _{\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 $V_{\mathbb {R} }.$ Every inner product on a real vector space is a bilinear and symmetric map.
For example, if $V=\mathbb {C}$ with inner product $\langle x,y\rangle =x{\overline {y}},$ where $V$ is a vector space over the field $\mathbb {C} ,$ then $V_{\mathbb {R} }=\mathbb {R} ^{2}$ is a vector space over $\mathbb {R}$ and $\langle x,y\rangle _{\mathbb {R} }$ is the dot product$x\cdot y,$ where $x=a+ib\in V=\mathbb {C}$ is identified with the point $(a,b)\in V_{\mathbb {R} }=\mathbb {R} ^{2}$ (and similarly for $y$); thus the standard inner product $\langle x,y\rangle =x{\overline {y}},$ on $\mathbb {C}$ is an "extension" the dot product . Also, had $\langle x,y\rangle$ been instead defined to be the symmetric map$\langle x,y\rangle =xy$ (rather than the usual conjugate symmetric map$\langle x,y\rangle =x{\overline {y}}$) then its real part $\langle x,y\rangle _{\mathbb {R} }$ would not be the dot product; furthermore, without the complex conjugate, if $x\in \mathbb {C}$ but $x\not \in \mathbb {R}$ then $\langle x,x\rangle =xx=x^{2}\not \in [0,\infty )$ so the assignment $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 $\langle x,y\rangle =0$ then $\langle x,y\rangle _{\mathbb {R} }=0,$ but the next example shows that the converse is in general not true.
Given any $x\in V,$ the vector $ix$ (which is the vector $x$ rotated by 90°) belongs to $V$ and so also belongs to $V_{\mathbb {R} }$ (although scalar multiplication of $x$ by $i={\sqrt {-1}}$ is not defined in $V_{\mathbb {R} },$ the vector in $V$ denoted by $ix$ is nevertheless still also an element of $V_{\mathbb {R} }$). 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 _{\mathbb {R} }=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,Ax\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=\mathbb {C}$ has the inner product $\langle x,y\rangle :=x{\overline {y}}$ mentioned above. Then the map $A:V\to V$ defined by $Ax=ix$ is a linear map (linear for both $V$ and $V_{\mathbb {R} }$) that denotes rotation by $90^{\circ }$ in the plane. Because $x$ and $Ax$ perpendicular vectors and $\langle x,Ax\rangle _{\mathbb {R} }$ is just the dot product, $\langle x,Ax\rangle _{\mathbb {R} }=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 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 $\{e_{1},\ldots ,e_{n}\}$ 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\{e_{a}\right\}_{a\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). Say that $E$ is an orthonormal basis for $V$ if it is a basis and
$\left\langle e_{a},e_{b}\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.
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).^{[citation needed]}
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 the converse need not hold in general. If $G$ is a dense subspace of an inner product space $V,$ then any orthonormal basis for $G$ is automatically an orthonormal basis for $V.$ Thus, it suffices to construct an inner product space $V$ with a dense subspace $G$ whose dimension is strictly smaller than that of $V.$
Let $K$ be a Hilbert space of dimension $\aleph _{0}.$ (for instance, $K=\ell ^{2}(\mathbb {N} )$). Let $E$ be an orthonormal basis of $K,$ so $|E|=\aleph _{0}.$ Extend $E$ to a Hamel basis$E\cup F$ for $K,$where $E\cap F=\varnothing .$ 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=\ell ^{2}(\mathbb {R} )$). Let $B$ be an orthonormal basis for $L$ and let $\varphi :F\to B$ be a bijection. Then there is a linear transformation $T:K\to L$ such that $Tf=\varphi (f)$ for $f\in F,$ and $Te=0$ for $e\in E.$
Let $V=K\oplus L$ and let $G=\{(k,Tk):k\in K\}$ be the graph of $T.$ Let ${\overline {G}}$ be the closure of $G$ in $V$; we will show ${\overline {G}}=V.$ Since for any $e\in E$ we have $(e,0)\in G,$ it follows that $K\oplus 0\subseteq {\overline {G}}.$
Next, if $b\in B,$ then $b=Tf$ for some $f\in F\subseteq K,$ so $(f,b)\in G\subseteq {\overline {G}}$; since $(f,0)\in {\overline {G}}$ as well, we also have $(0,b)\in {\overline {G}}.$ It follows that $0\oplus L\subseteq {\overline {G}},$ so ${\overline {G}}=V,$ and $G$ is dense in $V.$
Finally, $\{(e,0):e\in E\}$ is a maximal orthonormal set in $G$; if
for all $e\in E$ then $k=0,$ so $(k,Tk)=(0,0)$ is the zero vector in $G.$ Hence the dimension of $G$ is $|E|=\aleph _{0},$ whereas it is clear that the dimension of $V$ is $c.$ This completes the proof.
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 {e^{ikt}}{\sqrt {2\pi }}}$
is an orthonormal basis of the space $C[-\pi ,\pi ]$ with the $L^{2}$ inner product. The mapping
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 $[-\pi ,\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:V\to W$ between inner product spaces $V$ and $W$ are of relevance:
Continuous linear maps: $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 $\{\|Ax\|:\|x\|\leq 1\},$ where $x$ ranges over the closed unit ball of $V,$ is bounded.
Symmetric linear operators: $A:V\to W$ is linear and $\langle Ax,y\rangle =\langle x,Ay\rangle$ for all $x,y\in V.$
Isometries: $A:V\to W$ satisfies $\|Ax\|=\|x\|$ for all $x\in V.$ A linear isometry (resp. an antilinear isometry) 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 real 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 morphisms between inner product spaces, and morphisms of real inner product spaces are orthogonal transformations (compare with orthogonal matrix).
Isometrical isomorphisms: $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 {\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 $\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 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\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$covector 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 trace 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.
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 (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 outer product (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).
^By combining the linear in the first argument property with the conjugate symmetry property you get conjugate-linear in the second argument: ${\textstyle \langle x,by\rangle =\langle x,y\rangle {\overline {b}}.}$ 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 ${\textstyle M}$ such that ${\textstyle \langle x,y\rangle =x^{*}My.}$ (Here, ${\textstyle x^{*}}$ is the conjugate transpose of ${\textstyle x.}$)
^A line over an expression or symbol, such as ${\overline {s}}$ or ${\overline {\langle y,x\rangle }},$ denotes complex conjugation. A scalar $s$ is real if and only if $s={\overline {s}}.$
^This is because condition (1) (that is, linearity in the first argument) and positive definiteness implies that $\langle x,x\rangle$ is always a real number. And as mentioned before, a sesquilinear form is Hermitian if and only if $\langle x,x\rangle$ is real for all $x.$
^Let $R(x,y):={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right).$ If $\mathbb {F} =\mathbb {R}$ then let $\langle x,\,y\rangle _{P}:=R(x,y)$ while if $\mathbb {F} =\mathbb {C}$ then let $\langle x,\,y\rangle _{P}:=R(x,y)+iR(x,iy).$ See the polarization identity article for more details.
^If $\langle x,\,cy\rangle$ can be written as $\langle x,\,cy\rangle =f(c,y)\langle x,\,y\rangle$ for some function $f$ (in particular, this assumes that the scalar in front of $\langle x,\,y\rangle$ that results from trying to "pull $c$ out of $\langle x,\,cy\rangle$" does not depend on $x$) then $\langle y,\,cy\rangle ={\overline {c}}\langle y,\,y\rangle$ implies that $f(c,y)={\overline {c}}$ (when $y\neq 0$) and consequently, $\langle x,\,cy\rangle ={\overline {c}}\langle x,\,y\rangle$ will hold for all $x,y,{\text{ and }}c.$
Proofs
^ ^{a}^{b}Homogeneity in the 1st argument implies $\langle \mathbf {0} ,x\rangle =\langle 0x,x\rangle =0\langle x,x\rangle =0.$$\blacksquare$Additivity in the 1st argument implies $\langle \mathbf {0} ,x\rangle =\langle \mathbf {0} +\mathbf {0} ,x\rangle =\langle \mathbf {0} ,x\rangle +\langle \mathbf {0} ,x\rangle$ so adding $-\langle \mathbf {0} ,x\rangle$ to both sides proves $\langle \mathbf {0} ,x\rangle =0.$$\blacksquare$
^Assume that it is a sesquilinear form that satisfies $\langle x,x\rangle \in \mathbb {R}$ for all $x.$ To conclude that $\langle x,y\rangle ={\overline {\langle y,x\rangle }},$ it is necessary and sufficient to show that the real parts of $\langle y,x\rangle$ and $\langle x,y\rangle$ are equal and that their imaginary parts are negatives of each other. For all $x,y,$ because $\langle x+y,x+y\rangle -\langle x,x\rangle -\langle y,y\rangle =\langle y,x\rangle +\langle x,y\rangle$ and the left hand side is real, $\langle y,x\rangle +\langle x,y\rangle$ is also real, which implies that the $0=\operatorname {im} \left[\langle y,x\rangle +\langle x,y\rangle \right]=\left(\operatorname {im} \langle y,x\rangle \right)+\operatorname {im} \langle x,y\rangle .$ Similarly, $\langle iy,x\rangle +\langle x,iy\rangle \in \mathbb {R} .$ But sesquilinearity implies $\langle iy,x\rangle +\langle x,iy\rangle =i(\langle y,x\rangle -\langle x,y\rangle ),$ which is only possible if the real parts of $\langle y,x\rangle$ and $\langle x,y\rangle$ are equal. $\blacksquare$
^A complex number $c$ is a real number if and only if $c={\overline {c}}.$ Using $y=x$ in condition (2) gives $\langle x,x\rangle ={\overline {\langle x,x\rangle }},$ which implies that $\langle x,x\rangle$ is a real number. $\blacksquare$
^Assume that $\langle \,\cdot ,\cdot \,\rangle$ is a bilinear map and that $x\in V$ satisfies $\langle x,x\rangle \neq 0.$ Let $N:\mathbb {F} \to \mathbb {F}$ be defined by $N(c):=\langle cx,cx\rangle$ where bilinearity implies that $N(c)=\langle cx,cx\rangle =c^{2}\langle x,x\rangle =c^{2}N(1)$ holds for all scalars $c.$ Since $N(1)=\langle x,x\rangle \neq 0,$ the scalar $1/N(1)$ is well-defined and so $N(c)=0$ if and only if $c=0.$ If $c\in \mathbb {C}$ is a scalar such that $c^{2}\not \in \mathbb {R}$ (which implies $c\neq 0$ and ${\frac {1}{c^{2}}}\not \in \mathbb {R}$) then $N(1)\in \mathbb {R}$ implies $N(c)=c^{2}N(1)\not \in \mathbb {R}$ and similarly, $N(c)\in \mathbb {R}$ implies $N(1)={\frac {1}{c^{2}}}N(c)\not \in \mathbb {R} ;$ this shows that for such a $c,$ at most one of $N(1){\text{ and }}N(c)$ can be real. $\blacksquare$
If $\mathbb {F} =\mathbb {C}$ and $s\in \mathbb {F}$ then pick $c\in \mathbb {C}$ such that $c^{2}={\frac {s}{N(1)}},$ which implies that $N(c)=c^{2}N(1)={\frac {s}{N(1)}}N(1)=s;$ thus $N(\mathbb {C} )=\mathbb {C}$ so $N:\mathbb {C} \to \mathbb {C}$ is surjective. If $\mathbb {F} =\mathbb {R}$ and $R(1)>0$ (resp. $R(1)<0$) then for any $s\geq 0$ (resp. any $s\leq 0$), $N\left({\sqrt {s/N(1)}}\right)=s,$ which shows that $N(\mathbb {R} )=[0,\infty )$ (resp. $N(\mathbb {R} )=(-\infty ,0]$). $\blacksquare$
^ ^{a}^{b}Note that $\|x+y\|^{2}=\langle x+y,x+y\rangle =\langle x,x\rangle +\langle x,y\rangle +\langle y,x\rangle +\langle y,y\rangle$ and $\|x-y\|^{2}=\langle x-y,x-y\rangle =\langle x,x\rangle -\langle x,y\rangle -\langle y,x\rangle +\langle y,y\rangle ,$ which implies that $\|x+y\|^{2}+\|x-y\|^{2}=2\langle x,x\rangle +2\langle y,y\rangle =2\|x\|^{2}+2\|y\|^{2}.$ This proves that $\|\,\cdot \,\|$ satisfies the parallelogram law. This also shows that $\|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}+\langle x,y\rangle +\langle y,x\rangle ,$ which proves that $\langle x,y\rangle +\langle y,x\rangle$ is a real number. $\blacksquare$
^Combining $\|ix\|=|i|\|x\|=\|x\|$ and $2\|x\|^{2}=|1+i|^{2}\,\|x\|^{2}=\|(1+i)x\|^{2}=\langle x+ix,x+ix\rangle =\|x\|^{2}+\langle x,ix\rangle +\langle ix,x\rangle +\|ix\|^{2}$ proves that $0=\langle x,ix\rangle +\langle ix,x\rangle .$$\blacksquare$
^Fix $x,y\in V.$ The equality $\langle qx,y\rangle =q\langle x,y\rangle$ will be discussed first. Define $L,R:\mathbb {R} \to \mathbb {F}$ by $L(q):=\langle qx,y\rangle$ and $R(q):=q\langle x,y\rangle .$ Because $\langle qx,y\rangle =q\langle x,y\rangle$ for all $q\in \mathbb {Q} ,$$L$ and $R$ are equal on a dense subset of $\mathbb {R} .$ Since $\langle x,y\rangle$ is constant, the map $R:\mathbb {R} \to \mathbb {F}$ is continuous (where the Hausdorff space$\mathbb {F} ,$ which is either $\mathbb {R}$ or $\mathbb {C} ,$ has its usual Euclidean topology). Consequently, if $L:\mathbb {R} \to \mathbb {F}$ is also continuous then $L$ and $R$ will necessarily be equal on all of $\mathbb {R} ;$ that is, $\langle qx,y\rangle =q\langle x,y\rangle$ will hold for all real$q\in \mathbb {R} .$ If $f:\mathbb {R} \to V{\text{ and }}g:V\to \mathbb {F}$ are defined by $f(q):=qx$ and $g(v):=\langle v,y\rangle$ then $L=g\circ f.$ So for $L$ to be continuous, it suffices for there to exist some topology $\tau$ on $V$ that makes both $f$ and $g$ continuous (or even just sequentially continuous). The map $f:\mathbb {R} \to (V,\tau )$ will automatically be continuous if $\tau$ is a topological vector space topology, such as a topology induced by a norm. The map $g:(V,\tau )\to \mathbb {F}$ will be continuous if $\langle \,\cdot ,\cdot \,\rangle :V\times V\to \mathbb {F}$ is separately continuous (which will be true if $\langle \,\cdot ,\cdot \,\rangle$ is continuous). The discussion of the equality $\langle x,qy\rangle =q\langle x,y\rangle$ is nearly identical, with the main difference being that $L,f,g$ must be redefined as $L(q):=\langle x,qy\rangle ,$$f(q):=qy,$ and $g(v):=\langle x,v\rangle .$$\blacksquare$
^Let $x,y,z$ be vectors and let $s$ be a scalar. Then $\langle x,sy\rangle ={\overline {\langle sy,x\rangle }}={\overline {s}}{\overline {\langle y,x\rangle }}={\overline {s}}\langle x,y\rangle$ and $\langle x,y+z\rangle ={\overline {\langle y+z,x\rangle }}={\overline {\langle y,x\rangle }}+{\overline {\langle z,x\rangle }}=\langle x,y\rangle +\langle x,z\rangle .$$\blacksquare$
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