Orthonormal systems in complex vector spaces

Inner Product Spaces: Complex inner product spaces

Orthonormal systems in complex vector spaces

The inner product on a complex vector space is not symmetric, but perpendicularity is.

We say that the vector $\vec{a}$ is perpendicular to the vector $\vec{b}$ in a complex inner product space $V$ if $\dotprod{\vec{a}}{\vec{b}}=0$ .

This relation is symmetric; that is, if $\dotprod{\vec{a}}{\vec{b}}=0$ , then $\dotprod{\vec{b}}{\vec{a}}=0$ .

The vector $\rv{1,\complexi}$ is perpendicular to $\rv{1,-\complexi}$ in the inner product space $\mathbb{C}^2$ , since

$\dotprod{\rv{1,\complexi}}{\rv{1,-\complexi}} = 1\cdot \overline{1}+\complexi\cdot \overline{-\complexi} = 1+\complexi^2 = 0$

If $\vec{a}$ is perpendicular to $\vec{b}$ , then $\dotprod{\vec{a}}{\vec{b}}=0$ , so, by Hermiticity of the inner product,

$\dotprod{\vec{b}}{\vec{a}}=\overline{\dotprod{\vec{a}}{\vec{b}}} = \overline{0} = 0$

We conclude that $\vec{b}$ is perpendicular to $\vec{a}$ .

Instead of saying that they are perpendicular to each other, we also speak of orthogonality of $\vec{a}$ and $\vec{b}$ . This term is also used when it comes to more vectors, as we shall see below.

The following notions regarding mutual perpendicularity of vectors, are direct extensions of the real case.

Let $\basis{\vec{v}_1,\ldots ,\vec{v}_n}$ be a system of vectors of a complex inner product space $V$ .

The system is called orthogonal if for $1\leq i, j\leq n$ with $i\neq j$ we have
$\dotprod{\vec{v}_i}{\vec{v}_j}=0$
The system is called orthonormal if for $1\leq i, j\leq n$ $\dotprod{\vec{v}_i}{\vec{v}_j}=\left\{\,\begin{array}{l} 0\ \text{if}\ i\neq j\\ 1\ \text{if}\ i=j\ \end{array}\right.$

If, in addition, the system $\basis{\vec{v}_1,\ldots ,\vec{v}_n}$ is a basis for $V$ , we call it an orthonormal basis of $V$ .

Standard basis

The standard basis of $\mathbb{C}^n$ is orthonormal.

The column vectors of the matrix $Q = \frac{1}{3}\matrix{2&\ii&2\,\ii\\ 2\,\ii&2&1\\ \ii&-2&2 }$ form an orthonormal basis for $\mathbb{C}^3$ (provided with the standard inner product). This statement is equivalent to $Q^\top\, \overline{Q} =I_3$

where $\overline{Q}$ is the matrix of which each entry is the complex conjugate of the corresponding entry of $Q$ . In particular, $Q$ is invertible with inverse $Q^* = \overline{Q^\top}$ .

The properties of complex orthonormal systems are just as good as those of real orthonormal systems:

Properties of orthonormal systems

Let $V$ be a complex inner product space.

Orthonormal systems in $V$ are independent.
If $\basis{\vec{a}_1,\ldots ,\vec{a}_n}$ is an orthonormal basis of $V$ , then the coordinates of $\vec{x}$ with respect to this basis are successively $\dotprod{\vec{x}}{\vec{a}_1}, \ldots , \dotprod{\vec{x}}{\vec{a}_n}$ : $\vec{x}=(\dotprod{\vec{x}}{\vec{a}_1})\,\vec{a}_1+\cdots+(\dotprod{\vec{x}}{\vec{a}_n})\,\vec{a}_n$
The length of $\vec{x}$ is equal to the length of the coordinate vector of $\vec{x}$ relative to the standard inner product:
$\norm{\vec{x}}^2 ={\left|\dotprod{\vec{x}}{\vec{a}_1}\right|}^2 + \cdots + {\left|\dotprod{\vec{x}}{\vec{a}_n}\right|}^2$
Write the vectors as linear combinations of the basis vectors: $\vec{x}=\sum_{i=1}^n x_i\vec{a}_i\quad\text{ and }\quad \vec{y}=\sum_{i=1}^n y_i\vec{a}_i$ Then the inner product $\dotprod{\vec{x}}{\vec{y}}$ can be expressed as $\dotprod{\vec{x}}{\vec{y}}=\sum_{i=1}^n x_i\cdot\overline{y_i}$

The proofs of these properties are very similar to those of the real case.

Let the vector $\vec{x}^{\,*}=\rv{1,\complexi}$ be given in the inner product space $\mathbb{C}^2$ (with standard inner product).

Enter a vector of length one in the span of $\vec{x}$ .

$\vec{x}=$ ${\rv{{{1}\over{\sqrt{2}}},{{\complexi}\over{\sqrt{2}}}}}$

We first calculate the norm $\norm{\vec{x}^{\,*}}$ :
$\begin{array}{rcl} \norm{\vec{x}^{\,*}} &=&\sqrt{\dotprod{\vec{x}^{\,*}}{\vec{x}^{\,*}}}\\ &&\phantom{xx}\color{blue}{\text{definition of norm}}\\ &=&\sqrt{\dotprod{\rv{1,\complexi}}{\rv{1,\complexi}}}\\ &&\phantom{xx}\color{blue}{\vec{x}^{\,*}\text{ substituted}}\\ &=&\sqrt{(1)\cdot (\overline{1})+(\complexi)\cdot (\overline{\complexi})}\\ &&\phantom{xx}\color{blue}{\text{definition of complex inner product}}\\ &=&\sqrt{(1)\cdot (1)+(\complexi)\cdot (-\complexi)}\\ &&\phantom{xx}\color{blue}{\text{conjugated}}\\ &=&\sqrt{2}\\ &&\phantom{xx}\color{blue}{\text{simplified}} \end{array}$ Next we divide the vector by the norm in order to get a normalized vector: $\vec{x}=\frac{1}{\sqrt{2}}\,\rv{1,\complexi}=\rv{{{1}\over{\sqrt{2}}},{{\complexi}\over{\sqrt{2}}}}$ Note that the answers $-\vec{x}$ , $\ii \cdot \vec{x}$ , and $-\ii\cdot \vec{x}$ are also correct.

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