Inner product on complex vector spaces

Inner Product Spaces: Complex inner product spaces

Inner product on complex vector spaces

In a complex vector space, the definition of inner product is slightly different from the real case:

Let $V$ be a complex vector space. An inner product on $V$ is a function that assigns to every pair of vectors $\vec{a}$ , $\vec{b}$ from $V$ a complex number $\dotprod{\vec{a}}{\vec{b}}$ in such a way that the following three conditions are satisfied:

linearity in the first argument: $\dotprod{\vec{a}}{\vec{b}}$ is linear in $\vec{a}$ ,
Hermiticity: $\dotprod{\vec{a}}{\vec{b}}=\overline{\dotprod{\vec{b}}{\vec{a}}}$ for all $\vec{a},\vec{b}\in V$ , where $\overline{z}$ indicates the complex conjugate of a complex number $z$ ,
positive definiteness: $\dotprod{\vec{a}}{\vec{a}}\geq 0$ for all $\vec{a}\in V$ ; moreover, if $\dotprod{\vec{a}}{\vec{a}} = 0$ then $\vec{a}=\vec{0}$ .

A complex vector space with an inner product is often referred to as a (complex) inner product space.

The length of a vector $\vec{v}$ is given by $\sqrt{\dotprod{\vec{v}}{\vec{v}}}$ (as in the real case).

Also, the distance between $\vec{v}$ and $\vec{w}$ is given by the same formula as in the real case: $\norm{ \vec{v}-\vec{w}}$ .

A complex inner product is also known as a Hermitian inner product (after the mathematician Hermite). Because of Hermiticity, $\dotprod{\vec{a}}{\vec{a}} = \overline{\dotprod{\vec{a}}{\vec{a}}}$ , so $\dotprod{\vec{a}}{\vec{a}}$ is real for all $\vec{a}\in V$ . Therefore the inequality $\dotprod{\vec{a}}{\vec{a}}\geq 0$ of the positive definiteness makes sense.

Each complex-valued map $f:\mathbb{C}\to\mathbb{C}$ with two arguments $a$ and $b$ from the $1$ -dimensional vector space $\mathbb{C}$ which is linear in the first argument, and Hermitian (that is, satisfies the Hermiticity rule), is of the form $f(a,b) = \alpha\cdot a\cdot \overline{b\,}\phantom{xx}\text{ for a real number }\alpha$ This map is an inner product on $\mathbb{C}$ if and only if $\alpha\gt0$ .

On $\mathbb{C}^n$ we can define an inner product in many ways. The most common is the standard inner product of $\vec{a}= \rv{a_1,a_2,\ldots,a_n}$ and $\vec{b}=\rv{b_1,b_2,\ldots,b_n}$ :
$\dotprod{\vec{a}}{\vec{b}}=a_1\cdot \overline{{b}_1}+a_2\cdot \overline{{b}_2}+ \cdots +a_n \cdot\overline{{b}_n}$ If we restrict this inner product to $\mathbb{R}^n$ , then the complex conjugation is no longer needed and we get the standard inner product on $\mathbb{R}^n$ .

We verify that the standard inner product satisfies the conditions on the inner product. Linearity in the first argument and positive definiteness can be proved just as in the real case. We establish Hermiticity by the following chain of equalities for $\vec{a},\vec{b}\in V$ :

$\begin{array}{rcl} \dotprod{\vec{a}}{\vec{b}}&=&a_1\cdot \overline{{b}_1}+a_2\cdot \overline{{b}_2}+ \cdots +a_n \cdot\overline{{b}_n}\\&&\phantom{xx}\color{blue}{\text{definition of standard inner product}}\\& = &\overline{\overline{a_1}\cdot{{b}_1}}+\overline{\overline{a_2}\cdot {{b}_2}}+ \cdots +\overline{\overline{a_n }\cdot{{b}_n}}\\&&\phantom{xx}\color{blue}{\text{multiplicativity of complex conjugation}}\\ & = &\overline{ \overline{a_1}\cdot {b_1}+\overline{a_2}\cdot {b_2}+ \cdots +\overline{a_n }\cdot{b_n} }\\&&\phantom{xx}\color{blue}{\text{additivity of complex conjugation}}\\ & = &\overline{{b_1}\cdot\overline{a_1}+{b_2}\cdot\overline{a_2}+ \cdots +{b_n}\cdot\overline{a_n } }\\&&\phantom{xx}\color{blue}{\text{commutativity of complex multiplication}}\\ &=&\overline{\dotprod{\vec{b}}{\vec{a}}}\\&&\phantom{xx}\color{blue}{\text{definition of standard inner product}}\end{array}$

Let $V$ be the set of complex-valued continuous functions defined on a real interval $\ivcc{a}{b}$ . Here, complex-valued means that the values of $f$ lie in $\mathbb{C}$ and continuous means that each of the two compositions $\Re\circ f$ (the real part of $f$ ) and $\Im\circ f$ (the imaginary part of $f$ ) is a continuous function $\ivcc{a}{b}\to\mathbb{C}$ . Then $V$ with the usual pointwise addition and scalar multiplication is a vector space. Take $f,g\in V$ and define
$\dotprod{f}{g}= \int_a^b f(x)\cdot \overline{g(x)}\,\dd x$ The verification that this defines an inner product on $V$ is easy, except for the second part of positive definiteness. Here is a proof of it. Take $f\in V$ and suppose that $f(\alpha )\neq 0$ for a number $\alpha$ in $\ivcc{a}{b}$ . Then there is an interval $\ivcc{h}{h+\delta}$ of positive length $\delta$ around $\alpha$ within $\ivcc{a}{b}$ such that $|f(x)|\gt\frac12|f(\alpha )|\gt0$ for all $x$ in that interval. Then
$\dotprod{f}{f}= \int_a^b|f(x)|^2\dd x\geq \int_h^{h+\delta}|f(x)|^2\dd x\gt \int_h^{h+\delta}\left(\frac12|f(\alpha )|\right)^2\dd x\geq \frac14\delta\cdot |f(\alpha )|^2\gt 0$ So if $\dotprod{f}{f} = \ 0$ , then $f(x) = 0$ for all $x\in\ivcc{a}{b}$ .

An important example is the subspace of the inner product space $V$ of complex functions defined on $\ivcc{0}{2\pi}$ spanned by the functions $e_n$ for $n\in\mathbb{Z}$ defined by
$e_n(x)=\ee^{\ii nx}$ If $n\neq m$ then we have $e_n\perp e_m$ because $\begin{array}{rcl}\dotprod{e_n}{e_m}&=&\displaystyle \int_0^{2\pi}\ee^{\ii nx}\overline{\ee^{\ii mx}}\dd x\\&&\color{blue}{e_n(x)=\ee^{\ii nx}\text{ and definition inner product of functions}}\\&=& \displaystyle\int_0^{2\pi}\ee^{\ii(n-m)x}\dd x\\&&\color{blue}{\overline{\ee^{\ii mx}}=\ee^{-\ii mx}\text{ and }\ee^a\ee^b=\ee^{a+b}}\\&=&\displaystyle\left[\frac{1}{\ii(n-m)}\ee^{\ii(n-m)x}\right]_0^{2\pi}\\&&\color{blue}{\text{primitive of }\ee^{ax}\text{ is }\frac{1}{a}\ee^{ax}}\\&=&\displaystyle \frac{1}{\ii(n-m)}\left(\ee^{\ii(n-m)2\pi}-\ee^0\right)\\&&\color{blue}{\text{integration boundaries substituted}}\\&=&\displaystyle \frac{1}{\ii(n-m)}\left(\cos\left((n-m)2\pi\right)+\ii\cdot\sin\left((n-m)2\pi\right)-1\right)\\&&\color{blue}{\text{Euler's formula and }\ee^0=1}\\&=&\displaystyle \frac{1}{\ii(n-m)}\left(1+\ii\cdot 0-1\right)\\&&\color{blue}{\cos(2\pi k)=1\text{ and }\sin(2\pi k)=0\text{ for all }k\in\mathbb{N}}\\&=&0\end{array}$ Also,
$\begin{array}{rcll} \norm{e_n}^2 &=& \dotprod{e_n}{e_n}&\color{blue}{\text{definition norm}}\\&=&\displaystyle \int_0^{2\pi}\ee^{\ii nx}\overline{\ee^{\ii nx}}\dd x & \color{blue}{e_n(x)=\ee^{\ii nx}\text{ and definition inner product of functions}}\\ &=&\displaystyle \int_0^{2\pi}\ee^{\ii nx}\ee^{-\ii nx}\dd x & \color{blue}{\text{complex conjugation}}\\ &=&\displaystyle \int_0^{2\pi}\dd x & \color{blue}{\ee^a\ee^{-a}=\ee^{a-a}=\ee^0=1}\\ &=&2\pi&\color{blue}{\text{integration carried out}} \end{array}$ so
$\norm{e_n}=\sqrt{2\pi}\text{ for all }n\in\mathbb{Z}$ These functions are at the core of the complex Fourier theory.

The complex dot product is not bilinear. It is linear in the first argument, but semi-linear or half-linear in the second argument; that is, $\dotprod{\vec{a}}{(\lambda\vec{b}+\mu\vec{c})} =\overline{\lambda} \cdot (\dotprod{\vec{a}}{\vec{b}})+\overline{\mu} \cdot (\dotprod{\vec{a}}{\vec{c}})$ . Here is a derivation of this property:

$\begin{array}{rcl}\dotprod{\vec{a}}{(\lambda\vec{b}+\mu\vec{c})} &=& \overline{\dotprod{(\lambda \vec{b}+\mu\vec{c})}{ \vec{a}}} \\&&\phantom{xx}\color{blue}{\text{Hermitian}}\\ &=& \overline{\lambda \cdot(\dotprod{\vec{b}}{\vec{a}})+\mu \cdot(\dotprod{\vec{c}}{\vec{a}})}\\&&\phantom{xx}\color{blue}{\text{linearity of inner product in first argument}}\\&=&\overline{\lambda}\cdot \overline{\dotprod{\vec{b}}{\vec{a}}}+\overline{\mu}\cdot \overline{\dotprod{\vec{c}}{\vec{a}}}\\&&\phantom{xx}\color{blue}{\text{properties of complex conjugation}}\\ & =& \overline{\lambda} \cdot (\dotprod{\vec{a}}{\vec{b}})+\overline{\mu} \cdot (\dotprod{\vec{a}}{\vec{c}})\\&&\phantom{xx}\color{blue}{\text{Hermitian}}\end{array}$

We conclude that the complex dot product is linear in the first argument and non-linear but half-linear (namely, up to a complex conjugation) in the second argument. Therefore, the complex inner product is referred to as one-and-a-half-linear.

AngleThe concept of angle in the complex case is a lot more complicated than in the real case. We will not deal with it in this course.

The Cauchy-Schwarz inequality known for real inner product spaces reads as follows in the complex case.

In a complex inner product space $V$ we have, for all vectors $\vec{a},\vec{b}\in V$ ,
$|\dotprod{\vec{a}}{\vec{b}}|\leq\norm{\vec{a}}\cdot\norm{\vec{b}}$

This inequality is an equality if and only if $\vec{a}$ and $\vec{b}$ are linearly dependent.

The proof of the Cauchy-Schwarz inequality in the complex case is slightly more complicated than in the real case.

If $\vec{a} =\vec{0}$ , then it follows from the linearity of the inner product in the first argument that $\dotprod{\vec{a}}{\vec{b}} = \dotprod{(0\cdot\vec{a})}{\vec{b}}= 0\cdot(\dotprod{\vec{a}}{\vec{b}})= 0$ for all $\vec{b}$ . In particular, $\dotprod{\vec{a}}{\vec{a}}= 0$ , so $\norm{\vec{a}}=0$ . Hence the Cauchy-Schwarz inequality holds with the equality sign. This proves the rule in case $\vec{a}=\vec{0}$ because $\vec{0}$ and $\vec{b}$ are linearly dependent.

Now assume that $\vec{a}\neq\vec{0}$ and let $\vec{b}\in V$ . Write $\varphi$ for the complex argument of $\dotprod{\vec{a}}{\vec{b}}$ and $\vec{a}^{\,*}$ for $\ee^{-\ii\varphi}\cdot\vec{a}$ . Then we have
$\dotprod{\vec{a}^{\,*}}{\vec{b}}=\ee^{-\ii\varphi}\cdot(\dotprod{\vec{a}}{\vec{b}})\in\mathbb{R}$ For $\lambda \in \mathbb{R}$ we define
$f(\lambda )= \dotprod{(\lambda\vec{a}^{\,*} +\vec{b})}{(\lambda \vec{a}^{\,*}+\vec{b})}$ Caclulations with $f(\lambda)$ show that it is a quadratic function in $\lambda$ : $\begin{array}{rcll} f(\lambda ) &=&\dotprod{(\lambda \vec{a}^{\,*}+\vec{b})}{(\lambda\vec{a}^{\,*}+\vec{b})}&\color{blue}{\text{definition }f}\\ &=&\lambda^2(\dotprod{\vec{a}^{\,*}}{\vec{a}^{\,*}})+\lambda(\dotprod{\vec{a}^{\,*}}{\vec{b}})+\lambda (\dotprod{\vec{b}}{\vec{a}^{\,*}})+\dotprod{\vec{b}}{\vec{b}}&\color{blue}{\text{brackets expanded}}\\&=&\lambda^2(\dotprod{\vec{a}^{\,*}}{\vec{a}^{\,*}})+\lambda(\dotprod{\vec{a}^{\,*}}{\vec{b}})+\lambda (\overline{\dotprod{\vec{a}^{\,*}}{\vec{b}}})+\dotprod{\vec{b}}{\vec{b}}&\color{blue}{\text{Hermiticity}}\\&=&\lambda^2(\dotprod{\vec{a}^{\,*}}{\vec{a}^{\,*}})+\lambda(\dotprod{\vec{a}^{\,*}}{\vec{b}})+\lambda (\dotprod{\vec{a}^{\,*}}{\vec{b}})+\dotprod{\vec{b}}{\vec{b}}&\color{blue}{\dotprod{\vec{a}^{\,*}}{\vec{b}}\in\mathbb{R}}\\&=&\norm{\vec{a}^{\,*}}^2\lambda^2+2(\dotprod{\vec{a}^{\,*} }{\vec{b}})\lambda+\norm{\vec{b}}^2&\color{blue}{\text{definition norm}} \end{array}$ In particular, $f(\lambda)$ is a quadratic polynomial in $\lambda$ with real coefficients (recall from above that $\dotprod{\vec{a}^{\,*}}{\vec{b}}$ is real) and satisfies $f(\lambda)\geq 0$ for each $\lambda \in\mathbb{R}$ . Since this real function is never negative, the discriminant must be less than or equal to zero:
$4(\dotprod{\vec{a}^{\,*}}{\vec{b}})^2-4 \norm{\vec{a}^{\,*}}^2\cdot\norm{\vec{b}}^2\leq0$ After dividing by $4$ and moving the second term to the right, we find $(\dotprod{\vec{a}^{\,*}}{\vec{b}})^2\leq \norm{\vec{a}^{\,*}}^2\cdot\norm{\vec{b}}^2$ . Because the values at both sides are never negative, this is equivalent to
$|\dotprod{\vec{a}^{\,*}}{\vec{b}}|\leq\norm{\vec{a}^{\,*}}\cdot\norm{\vec{b}}$ where $|x|$ is the absolute value of a complex number $x$ . We conclude
$\begin{array}{rclcl}|\dotprod{\vec{a}}{\vec{b}}|&=& |\dotprod{(\ee^{\ii\varphi}\cdot\vec{a}^{\,*})}{\vec{b}}|&\phantom{xx}&\color{blue}{\vec{a}^{\,*}=\ee^{-\ii\varphi}\cdot\vec{a}}\\&=& |\ee^{\ii\varphi}\cdot(\dotprod{\vec{a}^{\,*}}{\vec{b}})|&\phantom{xx}&\color{blue}{\text{linearity of inner product in first argument}}\\& =&|\ee^{\ii\varphi}|\cdot |\dotprod{\vec{a}^{\,*}}{\vec{b}}|&\phantom{xx}&\color{blue}{\text{multiplicativity of the absolute value}}\\ &=& |\dotprod{\vec{a}^{\,*}}{\vec{b}}|&\phantom{xx}&\color{blue}{|\ee^{\ii\varphi}|=1}\\ &\leq &\norm{\vec{a}^{\,*}}\cdot\norm{\vec{b}}&\phantom{xx}&\color{blue}{\text{just proved above}}\\ &=&\norm{\ee^{-\ii\varphi}\vec{a}}\cdot\norm{\vec{b}}&\phantom{xx}&\color{blue}{\vec{a}^{\,*}=\ee^{-\ii\varphi}\vec{a}}\\&=&\norm{\vec{a}}\cdot\norm{\vec{b}}&\phantom{xx}&\color{blue}{\text{multiplicativity of the norm and }|\ee^{-\ii\varphi}|=1\text{ as above} }\end{array}$

We end by establishing that the Cauchy-Schwarz inequality is an equality if and only if $\vec{a}$ and $\vec{b}$ are linearly dependent. The two vectors $\vec{a}$ and $\vec{b}$ are linearly dependent if and only if at least one of the two is the zero vector or both are nonzero and there is a complex scalar $\mu$ such that $\vec{b} = \mu \,\vec{a}$ .

If at least one of the two is the zero vector, both sides of the Cauchy-Schwarz inequality are zero, and so the inequality is an equality. Conversely, if both sides of the inequality are zero, then in particular the right-hand side equals zero. This means that $\norm{\vec{a}}=0$ or $\norm{\vec{b}}=0$ , which implies $\vec{a} =\vec{0}$ or $\vec{b} = \vec{0}$ , and so $\vec{a}$ and $\vec{b}$ are linearly dependent. This establishes that the right-hand side of the inequality is zero if and only if at least one of the two vectors is equal to $\vec{0}$ .

Therefore, for the remainder of the proof we will assume that $\vec{a}$ and $\vec{b}$ are nonzero vectors. If they are linearly dependent, then there is a complex scalar $\mu$ such that $\vec{b} = \mu \vec{a}$ , and so $|\dotprod{\vec{a}}{\vec{b}}| =|\overline{\mu}|\cdot\dotprod{\vec{a}}{\vec{a}}=|{\mu}|\cdot \norm{\vec{a} }\cdot\norm{\vec{a}}=\norm{\vec{a}}\cdot\left(|{\mu}|\cdot\norm{\vec{a}}\right)=\norm{\vec{a}}\cdot\norm{\vec{b}}$ so equality holds in the Cauchy-Schwartz inequality. Conversely, if equality holds, then the inequality sign in the above deduction of the inequality must be an equality sign: $|\dotprod{\vec{a}^{\,*}}{\vec{b}}|=\norm{\vec{a}^{\,*}}\cdot\norm{\vec{b}}$ . This means that the discriminant of the real quadratic function $f(\lambda)$ is equal to zero, so $f(\lambda)$ has a zero. Let $\lambda =\mu$ be a solution of $f(\lambda) = 0$ . Then $\dotprod{(\mu\vec{a}^{\,*} +\vec{b})}{(\mu \vec{a}^{\,*}+\vec{b})}=f(\mu) =0$ , and so $\vec{b}=-\mu\vec{a}^{\,*}=-\mu\cdot\ee^{-\ii\varphi}\cdot\vec{a}$ . We have shown that $\vec{a}$ and $\vec{b}$ are linearly dependent. This concludes the proof of the statement about equality in the Cauchy-Schwartz inequality.

An interesting consequence of the Cauchy-Schwarz inequality is the following fact: let $\rv{a_1,\ldots ,a_n}$ and $\rv{b_1,\ldots ,b_n}$ be sequences of complex numbers. Then the following inequality holds:

$\left|\,\sum_{i=1}^n\ a_i\overline{b_i}\,\right|^2\leq\left(\,\sum_{i=1}^n|a_i|^2\right)\cdot \left(\,\sum_{i=1}^n|b_i|^2\right)$

Consider the inner product space $\mathbb{C}^2$ with standard inproduct.

Determine the inner product of the vectors
$\vec{x}=\rv{2\cdot \complexi+5,3\cdot \complexi-2}\quad\text{ and } \quad \vec{y}=\rv{2\cdot \complexi+4,6-3\cdot \complexi}$

$10\cdot \complexi+3$

$\begin{array}{rcl} \dotprod{\vec{x}}{\vec{y}}&=&\dotprod{\rv{2\cdot \complexi+5,3\cdot \complexi-2}}{\rv{2\cdot \complexi+4,6-3\cdot \complexi}}\\ &&\phantom{xx}\color{blue}{\vec{x}\text{ and }\vec{y}\text{ written out }}\\ &=&(2\cdot \complexi+5)\cdot( \overline{2\cdot \complexi+4})+(3\cdot \complexi-2)\cdot (\overline{6-3\cdot \complexi})\\ &&\phantom{xx}\color{blue}{\text{definition of standard inner product}}\\ &=&(2\cdot \complexi+5)\cdot(4-2\cdot \complexi)+(3\cdot \complexi-2)\cdot (3\cdot \complexi+6)\\ &&\phantom{xx}\color{blue}{\text{complex conjugation written out}}\\ &=&10\cdot \complexi+3\\ &&\phantom{xx}\color{blue}{\text{simplified}} \end{array}$

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