structural:summarization, quality:corruption
Real and complex parts of inner products

Suppose that ⟨ ⋅, ⋅ ⟩ {\displaystyle \langle \cdot,\cdot \rangle } is an inner product on V {\displaystyle V} (so it is antilinear in its second argument). The polarization identity shows that the real part of the inner product is Re ⟨ x, y ⟩ = 1 4 (‖ x + y ‖ 2 − ‖ x − y ‖ 2). {\displaystyle \operatorname {Re} \langle x,y\rangle ={\frac {1}{4}}\left(\|x+y\|^{2}-\|x-y\|^{2}\right).}

If V {\displaystyle V} is a real vector space then ⟨ x, y ⟩ = Re ⟨ x, y ⟩ = 1 4 (‖ x + y ‖ 2 − ‖ x − y ‖ 2) {\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 0. {\displaystyle 0.}

Assume for the rest of this section that V {\displaystyle V} is a complex vector space.The polarization identity for complex vector spaces shows that

The map defined by ⟨ x ∣ y ⟩ = ⟨ y, x ⟩ {\displaystyle \langle x\mid y\rangle =\langle y,x\rangle } for all x, y ∈ V {\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 ⟨ x ∣ y ⟩ {\displaystyle \langle x\mid y\rangle } and ⟨ x, y ⟩ {\displaystyle \langle x,y\rangle } are equal to Re ⟨ x, y ⟩ {\displaystyle \operatorname {Re} \langle x,y\rangle } but the inner products differ in their complex part:

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 V, {\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 V, {\displaystyle V,} and real inner products on V. {\displaystyle V.}

For example, suppose that V = C n {\displaystyle V=\mathbb {C} ^{n}} for some integer n > 0. {\displaystyle n>0.} When V {\displaystyle V} is considered as a real vector space in the usual way (meaning that it is identified with the 2 n − {\displaystyle 2n-} dimensional real vector space R 2 n, {\displaystyle \mathbb {R} ^{2n},} with each (a 1 + i b 1, …, a n + i b n) ∈ C n {\displaystyle \left(a_{1}+ib_{1},\ldots,a_{n}+ib_{n}\right)\in \mathbb {C} ^{n}} identified with (a 1, b 1, …, a n, b n) ∈ R 2 n {\displaystyle \left(a_{1},b_{1},\ldots,a_{n},b_{n}\right)\in \mathbb {R} ^{2n}}), then the dot product x ⋅ y = (x 1, …, x 2 n) ⋅ (y 1, …, y 2 n):= x 1 y 1 + ⋯ + x 2 n y 2 n {\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 V = C n {\displaystyle V=\mathbb {C} ^{n}} induced by the dot product is the map that sends c = (c 1, …, c n), d = (d 1, …, d n) ∈ C n {\displaystyle c=\left(c_{1},\ldots,c_{n}\right),d=\left(d_{1},\ldots,d_{n}\right)\in \mathbb {C} ^{n}} to ⟨ c, d ⟩:= c 1 d 1 ¯ + ⋯ + c n d n ¯ {\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 V R {\displaystyle V_{\mathbb {R} }} denote V {\displaystyle V} considered as a vector space over the real numbers rather than complex numbers.The real part of the complex inner product ⟨ x, y ⟩ {\displaystyle \langle x,y\rangle } is the map ⟨ x, y ⟩ R = Re ⟨ x, y ⟩ : V R × V R → R, {\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 V R. {\displaystyle V_{\mathbb {R} }.} Every inner product on a real vector space is a bilinear and symmetric map.

For example, if V = C {\displaystyle V=\mathbb {C} } with inner product ⟨ x, y ⟩ = x y ¯, {\displaystyle \langle x,y\rangle =x{\overline {y}},} where V {\displaystyle V} is a vector space over the field C, {\displaystyle \mathbb {C},} then V R = R 2 {\displaystyle V_{\mathbb {R} }=\mathbb {R} ^{2}} is a vector space over R {\displaystyle \mathbb {R} } and ⟨ x, y ⟩ R {\displaystyle \langle x,y\rangle _{\mathbb {R} }} is the dot product x ⋅ y, {\displaystyle x\cdot y,} where x = a + i b ∈ V = C {\displaystyle x=a+ib\in V=\mathbb {C} } is identified with the point (a, b) ∈ V R = R 2 {\displaystyle (a,b)\in V_{\mathbb {R} }=\mathbb {R} ^{2}} (and similarly for y {\displaystyle y}); thus the standard inner product ⟨ x, y ⟩ = x y ¯, {\displaystyle \langle x,y\rangle =x{\overline {y}},} on C {\displaystyle \mathbb {C} } is an "extension" the dot product. Also, had ⟨ x, y ⟩ {\displaystyle \langle x,y\rangle } been instead defined to be the symmetric map ⟨ x, y ⟩ = x y {\displaystyle \langle x,y\rangle =xy} (rather than the usual conjugate symmetric map ⟨ x, y ⟩ = x y ¯ {\displaystyle \langle x,y\rangle =x{\overline {y}}}) then its real part ⟨ x, y ⟩ R {\displaystyle \langle x,y\rangle _{\mathbb {R} }} would not be the dot product; furthermore, without the complex conjugate, if x ∈ C {\displaystyle x\in \mathbb {C} } but x ∉ R {\displaystyle x\not \in \mathbb {R} } then ⟨ x, x ⟩ = x x = x 2 ∉ [ 0, ∞) {\displaystyle \langle x,x\rangle =xx=x^{2}\not \in [0,\infty)} so the assignment x ↦ ⟨ x, x ⟩ {\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 ⟨ x, y ⟩ = 0 {\displaystyle \langle x,y\rangle =0} then ⟨ x, y ⟩ R = 0, {\displaystyle \langle x,y\rangle _{\mathbb {R} }=0,} but the next example shows that the converse is in general not true.Given any x ∈ V, {\displaystyle x\in V,} the vector i x {\displaystyle ix} (which is the vector x {\displaystyle x} rotated by 90°) belongs to V {\displaystyle V} and so also belongs to V R {\displaystyle V_{\mathbb {R} }} (although scalar multiplication of x {\displaystyle x} by i = − 1 {\displaystyle i={\sqrt {-1}}} is not defined in V R, {\displaystyle V_{\mathbb {R} },} the vector in V {\displaystyle V} denoted by i x {\displaystyle ix} is nevertheless still also an element of V R {\displaystyle V_{\mathbb {R} }}). For the complex inner product, ⟨ x, i x ⟩ = − i ‖ x ‖ 2, {\displaystyle \langle x,ix\rangle =-i\|x\|^{2},} whereas for the real inner product the value is always ⟨ x, i x ⟩ R = 0. {\displaystyle \langle x,ix\rangle _{\mathbb {R} }=0.}

If ⟨ ⋅, ⋅ ⟩ {\displaystyle \langle \,\cdot,\cdot \,\rangle } is a complex inner product and A: V → V {\displaystyle A:V\to V} is a continuous linear operator that satisfies ⟨ x, A x ⟩ = 0 {\displaystyle \langle x,Ax\rangle =0} for all x ∈ V, {\displaystyle x\in V,} then A = 0. {\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 V = C {\displaystyle V=\mathbb {C} } has the inner product ⟨ x, y ⟩:= x y ¯ {\displaystyle \langle x,y\rangle :=x{\overline {y}}} mentioned above. Then the map A: V → V {\displaystyle A:V\to V} defined by A x = i x {\displaystyle Ax=ix} is a linear map (linear for both V {\displaystyle V} and V R {\displaystyle V_{\mathbb {R} }}) that denotes rotation by 90 ∘ {\displaystyle 90^{\circ }} in the plane. Because x {\displaystyle x} and A x {\displaystyle Ax} are perpendicular vectors and ⟨ x, A x ⟩ R {\displaystyle \langle x,Ax\rangle _{\mathbb {R} }} is just the dot product, ⟨ x, A x ⟩ R = 0 {\displaystyle \langle x,Ax\rangle _{\mathbb {R} }=0} for all vectors x; {\displaystyle x;} nevertheless, this rotation map A {\displaystyle A} is certainly not identically 0. {\displaystyle 0.} In contrast, using the complex inner product gives ⟨ x, A x ⟩ = − i ‖ x ‖ 2, {\displaystyle \langle x,Ax\rangle =-i\|x\|^{2},} which (as expected) is not identically zero.