Linear Functionals Part I

 Let us look at linear functionals briefly, these are the foundation of frame theory...

DEFINITION: Let $\mathcal{V}$ be a finite-dimensional vector space. A linear functional is linear transformation $\phi : \mathcal{V}\longrightarrow \mathbb{F}$, that is a scalar-valued function that takes as input a vector.

Now let us take a look at the Riesz Representation Theorem, this will require some preliminaries...

First, let $A=[a_{ij}]$ be an $m\times n$ matrix over the field $\mathbb{F}$, the ADJOINT matrix $A^*$ of $A$ is the matrix $B=[b_{ij}]\in\mathbb{F}^{n\times m}$ such that $b_{ij}=\overline{a_{ji}}\ \forall \ \ i,j\in\mathbb{N}$. Simply put, the adjoint is the conjugate transpose of the matrix...

PROPOSITION Let $A,B$ be two matrices over the field $\mathbb{F}$, and let $\alpha,\beta\in\mathbb{F}$. Assume that $A,B\in\mathbb{F}^{m\times n}$, that is the same sizes. Then we have the following properties...

i) $(\alpha A+\beta B)^*=\overline{a}A^*+\overline{b}B^*$

ii) $(A^*)^*=A$

iii) $(AB)^*=B^*A^*$

iv) $(A^{-1})^*=(A^*)^{-1}$

PROOF: Let us go step by step here

i) $(\alpha A+\beta B)^*=(\alpha [a_{ij}]+\beta [b_{ij}])^*=([\alpha a_{ij}]+[\beta b_{ij}])^*=[\alpha a_{ij}+\beta b_{ij}]^*=[c_{ij}]^*=[\overline{c_{ji}}]=[\overline{\alpha a_{ji}+\beta b_{ji}}]$

$=[\overline{\alpha}\overline{a_{ji}}+\overline{\beta}\overline{b_{ji}}]=[\overline{\alpha}\overline{a_{ji}}]+[\overline{\beta}\overline{b_{ji}}]=\overline{\alpha}[\overline{a_{ji}}]+\overline{\beta}[\overline{b_{ji}}]=\overline{\alpha}A^*+\overline{\beta}B^*\ \ \ \square$

ii) $(A^*)^*=([a_{ij}]^*)^*=([\overline{a_{ji}}])^*=[c_{ij}]^*=[\overline{c_{ji}}]=[\overline{\overline{a_{ij}} }]=[a_{ij}]=A\ \ \ \square$

iii) This one needs careful attention $A\in\mathbb{F}^{m\times n}, B\in\mathbb{F}^{n\times p}$...

$$AB=\begin{pmatrix}\leftarrow & a_{1:} &\rightarrow\\ \leftarrow & a_{2:} & \rightarrow \\ & \vdots & \\ \leftarrow & a_{m:} & \rightarrow\end{pmatrix}\begin{pmatrix}\uparrow & \uparrow & & \uparrow \\ b_{:1} & b_{:2} & \cdots & b_{:p} \\ \downarrow & \downarrow & & \downarrow \end{pmatrix}=\begin{pmatrix}\langle a_{1:},b_{:1}\rangle & \langle a_{1:},b_{:2}\rangle & \cdots & \langle a_{1:},b_{:p}\rangle \\ \langle a_{2:},b_{:1}\rangle & \langle a_{2:},b_{:2}\rangle& \cdots & \langle a_{2:},b_{:p}\rangle \\ \vdots & \vdots & & \vdots \\ \langle a_{m:},b_{:1}\rangle & \langle a_{m:},b_{:2}\rangle & \cdots & \langle a_{m:},b_{:p}\rangle \end{pmatrix}$$

let

$$AB=[\langle a_{i:},b_{:j}\rangle]_{i,j=1}^{m,p}\Longrightarrow (AB)^*=[\langle a_{i:},b_{:j}\rangle]^*=[c_{ij}]^*=[\overline{c_{ji}}]=[\overline{\langle b_{:j},a_{i:} \rangle}]=[\langle \overline{b_{:j}},\overline{a_{i:}}\rangle]$$

then

$$=\begin{pmatrix}\langle \overline{b_{:1}},\overline{a_{1:}}\rangle & \langle \overline{b_{:1}},\overline{a_{2:}}\rangle & \cdots & \langle \overline{b_{:1}},\overline{a_{m:}}\rangle \\ \langle \overline{b_{:2}},\overline{a_{1:}}\rangle & \langle \overline{b_{:2}},\overline{a_{2:}}\rangle & \cdots & \langle \overline{b_{:2}},\overline{a_{m:}} \rangle \\ \vdots & \vdots & & \vdots \\ \langle \overline{b_{:p}},\overline{a_{1:}} \rangle & \langle \overline{b_{:p}},\overline{a_{2:}} \rangle & \cdots & \langle \overline{b_{:p}},\overline{a_{m:}}\rangle \end{pmatrix}=\underbrace{\begin{pmatrix}\leftarrow & \overline{b_{:1}}^T & \rightarrow \\ \leftarrow & \overline{b_{:2}}^T & \rightarrow \\ & \vdots & \\ \leftarrow & \overline{b_{:p}}^T& \rightarrow\end{pmatrix}}_{[p\times n]}\underbrace{\begin{pmatrix}\uparrow & \uparrow & & \uparrow \\ \overline{a_{1:}}^T & \overline{a_{2:}}^T & \cdots & \overline{a_{m:}}^T\\  \downarrow & \downarrow & & \downarrow\end{pmatrix}}_{[n\times m]}=B^*A^*\ \ \ \square$$

iv) This one is easy, we use iii) to help, assume invertibility...

$$\begin{align*} AA^{-1}=I &\Longleftrightarrow (AA^{-1})^* =I^*=I\\ &\Longleftrightarrow A^*(A^{-1})^*=I\\ &\Longleftrightarrow (A^*)^{-1}A^*(A^{-1})^*=(A^*)^{-1}I\\ &\Longleftrightarrow (A^{-1})^*=(A^*)^{-1}\ \ \ \square\end{align*}$$