Editing Riesz representation theorem (section)

==== Example in finite dimensions using matrix transformations ====

Consider the special case of <math>H = \Complex^n</math> (where <math>n > 0</math> is an integer) with the standard inner product 
<math display=block>\langle z \mid w \rangle := \overline{\,\vec{z}\,\,}^{\operatorname{T}} \vec{w} \qquad \text{ for all } \; w, z \in H</math> 
where <math>w \text{ and } z</math> are represented as [[Column matrix|column matrices]] <math>\vec{w} := \begin{bmatrix}w_1 \\ \vdots \\ w_n\end{bmatrix}</math> and <math>\vec{z} := \begin{bmatrix}z_1 \\ \vdots \\ z_n\end{bmatrix}</math> with respect to the standard orthonormal basis <math>e_1, \ldots, e_n</math> on <math>H</math> (here, <math>e_i</math> is <math>1</math> at its <math>i</math><sup>th</sup> coordinate and <math>0</math> everywhere else; as usual, <math>H^*</math> will now be associated with the [[dual basis]]) and where <math>\overline{\,\vec{z}\,}^{\operatorname{T}} := \left[\overline{z_1}, \ldots, \overline{z_n}\right]</math> denotes the [[conjugate transpose]] of <math>\vec{z}.</math> 
Let <math>\varphi \in H^*</math> be any linear functional and let <math>\varphi_1, \ldots, \varphi_n \in \Complex</math> be the unique scalars such that 
<math display=block>\varphi\left(w_1, \ldots, w_n\right) = \varphi_1 w_1 + \cdots + \varphi_n w_n \qquad \text{ for all } \; w := \left(w_1, \ldots, w_n\right) \in H,</math> 
where it can be shown that <math>\varphi_i = \varphi\left(e_i\right)</math> for all <math>i = 1, \ldots, n.</math> 
Then the Riesz representation of <math>\varphi</math> is the vector 
<math display=block>f_{\varphi} ~:=~ \overline{\varphi_1} e_1 + \cdots + \overline{\varphi_n} e_n
~=~ \left(\overline{\varphi_1}, \ldots, \overline{\varphi_n}\right) 
\in H.</math>
To see why, identify every vector <math>w = \left(w_1, \ldots, w_n\right)</math> in <math>H</math> with the column matrix 
<math>\vec{w} := \begin{bmatrix}w_1 \\ \vdots \\ w_n\end{bmatrix}</math> 
so that <math>f_{\varphi}</math> is identified with <math>\vec{f_{\varphi}} := \begin{bmatrix}\overline{\varphi_1} \\ \vdots \\ \overline{\varphi_n}\end{bmatrix} =  \begin{bmatrix}\overline{\varphi\left(e_1\right)} \\ \vdots \\ \overline{\varphi\left(e_n\right)}\end{bmatrix}.</math> 
As usual, also identify the linear functional <math>\varphi</math> with its [[transformation matrix]], which is the [[row matrix]] <math>\vec{\varphi} := \left[\varphi_1, \ldots, \varphi_n\right]</math> so that <math>\vec{f_{\varphi}} := \overline{\,\vec{\varphi}\,\,}^{\operatorname{T}}</math> and the function <math>\varphi</math> is the assignment <math>\vec{w} \mapsto \vec{\varphi} \, \vec{w},</math> where the right hand side is [[matrix multiplication]]. Then for all <math>w = \left(w_1, \ldots, w_n\right) \in H,</math>
<math display=block>\varphi(w) 
= \varphi_1 w_1 + \cdots + \varphi_n w_n 
= \left[\varphi_1, \ldots, \varphi_n\right] 
\begin{bmatrix}w_1 \\ \vdots \\ w_n\end{bmatrix} 
= \overline{\begin{bmatrix}\overline{\varphi_1} \\ \vdots \\ \overline{\varphi_n}\end{bmatrix}}^{\operatorname{T}} 
\vec{w} 
= \overline{\,\vec{f_{\varphi}}\,\,}^{\operatorname{T}} \vec{w} 
= \left\langle \,\,f_{\varphi}\, \mid \,w\, \right\rangle,
</math>
which shows that <math>f_{\varphi}</math> satisfies the defining condition of the Riesz representation of <math>\varphi.</math> 
The bijective antilinear isometry <math>\Phi : H \to H^*</math> defined in the corollary to the Riesz representation theorem is the assignment that sends <math>z = \left(z_1, \ldots, z_n\right) \in H</math> to the linear functional <math>\Phi(z) \in H^*</math> on <math>H</math> defined by 
<math display=block>w = \left(w_1, \ldots, w_n\right) ~\mapsto~ \langle \,z\, \mid \,w\,\rangle = \overline{z_1} w_1 + \cdots + \overline{z_n} w_n,</math> 
where under the identification of vectors in <math>H</math> with column matrices and vector in <math>H^*</math> with row matrices, <math>\Phi</math> is just the assignment 
<math display=block>\vec{z} = \begin{bmatrix}z_1 \\ \vdots \\ z_n\end{bmatrix} ~\mapsto~ \overline{\,\vec{z}\,}^{\operatorname{T}} = \left[\overline{z_1}, \ldots, \overline{z_n}\right].</math> 
As described in the corollary, <math>\Phi</math>'s inverse <math>\Phi^{-1} : H^* \to H</math> is the antilinear isometry <math>\varphi \mapsto f_{\varphi},</math> which was just shown above to be: 
<math display=block>\varphi ~\mapsto~ f_{\varphi} ~:=~ \left(\overline{\varphi\left(e_1\right)}, \ldots, \overline{\varphi\left(e_n\right)}\right);</math> 
where in terms of matrices, <math>\Phi^{-1}</math> is the assignment 
<math display=block>\vec{\varphi} = \left[\varphi_1, \ldots, \varphi_n\right] ~\mapsto~ \overline{\,\vec{\varphi}\,\,}^{\operatorname{T}} = \begin{bmatrix}\overline{\varphi_1} \\ \vdots \\ \overline{\varphi_n}\end{bmatrix}.</math> 
Thus in terms of matrices, each of <math>\Phi : H \to H^*</math> and <math>\Phi^{-1} : H^* \to H</math> is just the operation of [[Conjugate transpose|conjugate transposition]] <math>\vec{v} \mapsto \overline{\,\vec{v}\,}^{\operatorname{T}}</math> (although between different spaces of matrices: if <math>H</math> is identified with the space of all column (respectively, row) matrices then <math>H^*</math> is identified with the space of all row (respectively, column matrices). 

This example used the standard inner product, which is the map <math>\langle z \mid w \rangle := \overline{\,\vec{z}\,\,}^{\operatorname{T}} \vec{w},</math> but if a different inner product is used, such as <math>\langle z \mid w \rangle_M := \overline{\,\vec{z}\,\,}^{\operatorname{T}} \, M \, \vec{w} \,</math> where <math>M</math> is any [[Hermitian matrix|Hermitian]] [[positive-definite matrix]], or if a different orthonormal basis is used then the transformation matrices, and thus also the above formulas, will be different.