### Inner Product Spaces: Complex inner product spaces

### Orthogonal complements in complex inner product spaces

The theory of orthogonality in the complex case has many similarities with the real case.

Perpendicularity

Let #V# be a complex inner product space.

- Suppose that #W# is a linear subspace of #V#. The
**orthogonal complement**of #W# in #V# is the set

\[

W^\perp =\left\{\vec{x}\in V\mid \dotprod{\vec{x}}{\vec{w}}=0\ \text{ for all }\ \vec{w}\in W\right\}

\] This is a linear subspace of #V# which only has #\vec{0}# in common with #W#. - If #\vec{x}# is a vector of #V# and #W# a finite-dimensional linear subspace of #V#, then there is a unique vector #\vec{y}# in #W# such that #\vec{x}-\vec{y}# is perpendicular to #W#. We call this vector the
**orthogonal projection**of #\vec{x}# on #W# and often denote it by #P_W(\vec{x})#.

We briefly mention some of the results and properties regarding the orthogonal projection and perpendicular vectors. The definition of an orthonormal basis is similar to that in the real case; later we will go into this in greater detail.

Let #V# be a complex inner product space and let #W# be a subspace #W# of #V#. Suppose that #\basis{\vec{a}_1, \ldots ,\vec{a}_k}# is an orthonormal basis of #W# for a natural number #k#. Then the following statements hold.

- Two arbitrary vectors #\vec{a}# and #\vec{b}# are perpendicular to one another if and only if \[\norm{\vec{a}+\vec{b}}^2 = \norm{\vec{a}}^2+\norm{\vec{b}}^2\]
- If the vectors #\vec{a}_1 \ldots ,\vec{a}_k# are mutually perpendicular, that is to say: #\dotprod{\vec{a}_i}{\vec{a}_j }=0# if #i\neq j#, then

\[\norm{\vec{a}_1+\cdots + \vec{a}_k }^2 =\norm{\vec{a}_1}^2 +

\cdots + \norm{\vec{a}_k}^2\] - #\vec{x}-P_W(\vec{x})# is perpendicular to each vector from #W#.
- The orthogonal projection #P_W(\vec{x})# is given by #(\dotprod{\vec{x}}{\vec{a}_1})\vec{a}_1 + \cdots +(\dotprod{\vec{x}}{\vec{a}_k})\vec{a}_k#.
- #\norm{\vec{x}-P_W(\vec{x})}=\min_{\vec{z}\in W} \norm{\vec{x}-\vec{z}}#, that is to say, the distance from #\vec{x}# to a vector from #W# is minimal for #P_W(\vec{x})#.
- The perpendicular projection is the unique vector for which this minimum occurs.
- #\norm{P_W(\vec{x})}\leq\norm{\vec{x}}# with equality if and only if #\vec{x}=P_W(\vec{x})#.
- We have #P_W(\vec{x})=\vec{x}# if and only #\vec{x}# belongs to #W#.

Finally, we give a thinned-out version of the dimension formula as discussed *previously* for the real case.

Dimension formula

Let #V# be a finite-dimensional complex inner product space. For each linear subspace #W# of #V# we have \[ \dim{V}=\dim{W}+\dim{W^{\perp}}\]

In order to obtain an orthonormal basis for #W#, we normalize the vector # \rv{ 1 , \complexi } #: \[\frac{1}{\norm{\rv{ 1 , \complexi } }} \cdot \rv{ 1 , \complexi } = \frac{1}{\sqrt{2}}\cdot \rv{ 1 , \complexi } =\rv{ {{1}\over{\sqrt{2}}} , {{\complexi}\over{\sqrt{2}}} } \]

Next, we calculate the inner product of #\vec{x} # with this vector:\[ \begin{array}{rcl}\displaystyle\dotprod{\vec{x}}{\rv{ {{1}\over{\sqrt{2}}} , {{\complexi}\over{\sqrt{2}}} } }&=&\displaystyle\dotprod{\left[ 3 , 2\, \complexi \right] }{ \rv{ {{1}\over{\sqrt{2}}} , {{\complexi}\over{\sqrt{2}}} } }\\

&&\phantom{xx}\color{blue}{\text{vector }\vec{x}\text{ substituted}}\\

&=&\displaystyle (3)\cdot\overline{{{1}\over{\sqrt{2}}}}+(2\cdot \complexi)\cdot\overline{{{\complexi}\over{\sqrt{2}}}}\\

&&\phantom{xx}\color{blue}{\text{definition of complex inner product}}\\

&=&\displaystyle {{5}\over{\sqrt{2}}}\\

&&\phantom{xx}\color{blue}{\text{simplified}}

\end{array}\]

We now obtain the orthogonal projection by taking this inner product as the coefficient of the normalized basis vector of #W#: \[P_W(\vec{x})={{5}\over{\sqrt{2}}} \cdot {\rv{ {{1}\over{\sqrt{2}}} , {{\complexi}\over{\sqrt{2}}} } }=\left[ {{5}\over{2}} , {{5\, \complexi}\over{2}} \right] \]

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