Inner product space

In the following properties, which result almost immediately from the definition of an inner product, x, y and z are arbitrary vectors, and a and b are arbitrary scalars.

• ${\displaystyle \langle \mathbf {0} ,x\rangle =\langle x,\mathbf {0} \rangle =0.}$
• ${\displaystyle \langle x,x\rangle }$ is real and nonnegative.
• ${\displaystyle \langle x,x\rangle =0}$ if and only if ${\displaystyle x=\mathbf {0} .}$
• ${\displaystyle \langle x,ay+bz\rangle ={\overline {a}}\langle x,y\rangle +{\overline {b}}\langle x,z\rangle .}$
  This implies that an inner product is a sesquilinear form.
• ${\displaystyle \langle x+y,x+y\rangle =\langle x,x\rangle +2\operatorname {Re} (\langle x,y\rangle )+\langle y,y\rangle ,}$ where ${\displaystyle \operatorname {Re} }$
  denotes the real part of its argument.

Over ${\displaystyle \mathbb {R} }$, conjugate-symmetry reduces to symmetry, and sesquilinearity reduces to bilinearity. Hence an inner product on a real vector space is a positive-definite symmetric bilinear form. The binomial expansion of a square becomes

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

Some authors, especially in physics and matrix algebra, prefer to define inner products and sesquilinear forms with linearity in the second argument rather than the first. Then the first argument becomes conjugate linear, rather than the second.

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 an inner product space with arithmetic multiplication as its inner product:

$${\displaystyle \langle x,y\rangle$$ :=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 an inner product space with the 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, 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

$${\displaystyle \langle X,Y\rangle =\mathbb {E} [XY]}$$

is an inner product.[8][9][10] In this case, ${\displaystyle \langle X,X\rangle =0}$ if and only if ${\displaystyle \mathbb {P} [X=0]=1}$ (that is, ${\displaystyle X=0}$ almost surely), where ${\displaystyle \mathbb {P} }$ 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

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

With this norm, every inner product space becomes a normed vector space.

So, every general property of normed vector spaces applies to inner product spaces. In particular, one has the following properties:

Absolute homogeneity

$${\displaystyle \|ax\|=|a|\,\|x\|}$$

for every ${\displaystyle x\in V}$ and ${\displaystyle a\in F}$ (this results from ${\displaystyle \langle ax,ax\rangle =a{\overline {a}}\langle x,x\rangle }$).

Triangle inequality

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

for ${\displaystyle x,y\in V.}$

These two properties show that one has indeed a norm.

Cauchy–Schwarz inequality

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

for every ${\displaystyle x,y\in V,}$ with equality if and only if ${\displaystyle x}$ and ${\displaystyle y}$ are linearly dependent.

Parallelogram law

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

for every ${\displaystyle x,y\in V.}$

The parallelogram law is a necessary and sufficient condition for a norm to be defined by an inner product.

Polarization identity

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

for every ${\displaystyle x,y\in V.}$ The inner product can be retrieved from the norm by the polarization identity, since its imaginary part is the real part of ${\displaystyle \langle x,iy\rangle .}$

Ptolemy's inequality

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

for every ${\displaystyle x,y,z\in V.}$ Ptolemy's inequality is a necessary and sufficient condition for a seminorm to be the norm defined by an inner product.[11]

Orthogonality

Two vectors ${\displaystyle x}$ and ${\displaystyle y}$ are said to be orthogonal, often written ${\displaystyle x\perp y,}$ if their inner product is zero, that is, if ${\displaystyle \langle x,y\rangle =0.}$
This happens if and only if ${\displaystyle \|x\|\leq \|x+sy\|}$ for all scalars ${\displaystyle s,}$[12] and if and only if the real-valued function ${\displaystyle f(s):=\|x+sy\|^{2}-\|x\|^{2}}$ 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 non positive).
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 ${\displaystyle x\in V.}$[12]

Orthogonal complement

The orthogonal complement of a subset ${\displaystyle C\subseteq V}$ is the set ${\displaystyle C^{\bot }}$ of the vectors that are orthogonal to all elements of C; that is,

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

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

If ${\displaystyle x}$ and ${\displaystyle y}$ are orthogonal, then

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

This may be proved by expressing the squared norms in terms of the inner products, using additivity for expanding the right-hand side of the equation.

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 pairwise orthogonal, then

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

Angle

When ${\displaystyle \langle x,y\rangle }$ is a real number then the Cauchy–Schwarz inequality implies that ${\textstyle {\frac {\langle x,y\rangle }{\|x\|\,\|y\|}}\in [-1,1],}$ and thus that

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

is a real number. This allows defining the (non oriented) angle of two vectors in modern definitions of Euclidean geometry in terms of linear algebra. This is also used in data analysis, under the name "cosine similarity", for comparing two vectors of data.

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. Moreover, this real part defines an inner product on ${\displaystyle V,}$ considered as a real vector space. There is thus a one-to-one correspondence between complex inner products on a complex vector space ${\displaystyle V,}$ and real inner products on ${\displaystyle V.}$

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 ${\displaystyle \langle x,ix\rangle _{\mathbb {R} }=0.}$

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

If ${\displaystyle V}$ is a vector space and ${\displaystyle \langle \,\cdot \,,\,\cdot \,\rangle }$ a semi-definite sesquilinear form, then the function:

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

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

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

Alternatively, one may require that the pairing be a nondegenerate form, meaning that for all non-zero ${\displaystyle x\neq 0}$ there exists some ${\displaystyle y}$ such that ${\displaystyle \langle x,y\rangle \neq 0,}$ though ${\displaystyle y}$ need not equal ${\displaystyle x}$; in other words, the induced map to the dual space ${\displaystyle V\to V^{*}}$ 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 ${\displaystyle V\to V^{*}}$) and thus hold more generally.