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.
Perpendicular 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 following notions regarding mutual perpendicularity of vectors, are direct extensions of the real case.
Orthogonal and orthonormal systems 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#.
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 \[\dotprod{\vec{x}}{\vec{y}}=\sum_{i=1}^n x_i\cdot\overline{y_i}\]
We first calculate the norm #\norm{\vec{x}^{\,*}}#:
\[\begin{array}{rcl}
\norm{\vec{x}^{\,*}} &=&\sqrt{\dotprod{\vec{x}^{\,*}}{\vec{x}^{\,*}}}\\
&&\phantom{xx}\color{blue}{\text{definition of norm}}\\
&=&\sqrt{\dotprod{\rv{2 \complexi-1,2 \complexi+1}}{\rv{2 \complexi-1,2 \complexi+1}}}\\
&&\phantom{xx}\color{blue}{\vec{x}^{\,*}\text{ substituted}}\\
&=&\sqrt{(2 \complexi-1)\cdot (\overline{2 \complexi-1})+(2 \complexi+1)\cdot (\overline{2 \complexi+1})}\\
&&\phantom{xx}\color{blue}{\text{definition of complex inner product}}\\
&=&\sqrt{(2 \complexi-1)\cdot (-2\cdot \complexi-1)+(2 \complexi+1)\cdot (1-2\cdot \complexi)}\\
&&\phantom{xx}\color{blue}{\text{conjugated}}\\
&=&\sqrt{10}\\
&&\phantom{xx}\color{blue}{\text{simplified}}
\end{array}\] Next we divide the vector by the norm in order to get a normalized vector: \[\vec{x}=\frac{1}{\sqrt{10}}\,\rv{2 \complexi-1,2 \complexi+1}=\rv{{{2 \complexi-1}\over{\sqrt{10}}},{{2 \complexi+1}\over{\sqrt{10}}}}\] Note that the answers #-\vec{x}#, #\ii \cdot \vec{x}#, and #-\ii\cdot \vec{x}# are also correct.
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