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 $\varphi :\mathcal{V}⟶\mathbb{F}$, 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 $A^{\ast }$ of $A$ is the matrix $B=[b_{ij}]\in {\mathbb{F}}^{n\times m}$ such that $b_{ij}=\overline{a_{ji}}\mathrm{\forall }i,j\in \mathbb{N}$. 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) $(\alpha A+\beta B{)}^{\ast }=\bar{a}{A}^{\ast }+\bar{b}{B}^{\ast }$

ii) $(A^{\ast }{)}^{\ast }=A$

iii) $(AB{)}^{\ast }=B^{\ast }{A}^{\ast }$

iv) $(A^{-1}{)}^{\ast }=(A^{\ast }{)}^{-1}$

PROOF: Let us go step by step here

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

$=[\bar{\alpha }\overline{a_{ji}}+\bar{\beta }\overline{b_{ji}}]=[\bar{\alpha }\overline{a_{ji}}]+[\bar{\beta }\overline{b_{ji}}]=\bar{\alpha }[\overline{a_{ji}}]+\bar{\beta }[\overline{b_{ji}}]=\bar{\alpha }{A}^{\ast }+\bar{\beta }{B}^{\ast }◻$

ii) $(A^{\ast }{)}^{\ast }=([a_{ij}{]}^{\ast }{)}^{\ast }=([\overline{a_{ji}}]{)}^{\ast }=[c_{ij}{]}^{\ast }=[\overline{c_{ji}}]=[\overline{\stackrel{―}{a_{ij}}}]=[a_{ij}]=A◻$

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

$$AB=\left(\begin{array}{ccc}\leftarrow & a_{1:}& \to \\ \leftarrow & a_{2:}& \to \\ & ⋮& \\ \leftarrow & a_{m:}& \to \end{array}\right)\left(\begin{array}{cccc}\uparrow & \uparrow & & \uparrow \\ b_{:1}& b_{:2}& \cdots & b_{:p}\\ \downarrow & \downarrow & & \downarrow \end{array}\right)=\left(\begin{array}{cccc}⟨a_{1:},b_{:1}⟩& ⟨a_{1:},b_{:2}⟩& \cdots & ⟨a_{1:},b_{:p}⟩\\ ⟨a_{2:},b_{:1}⟩& ⟨a_{2:},b_{:2}⟩& \cdots & ⟨a_{2:},b_{:p}⟩\\ ⋮& ⋮& & ⋮\\ ⟨a_{m:},b_{:1}⟩& ⟨a_{m:},b_{:2}⟩& \cdots & ⟨a_{m:},b_{:p}⟩\end{array}\right)$$

let

$$AB=[⟨a_{i:},b_{:j}⟩{]}_{i,j=1}^{m,p}⟹(AB{)}^{\ast }=[⟨a_{i:},b_{:j}⟩{]}^{\ast }=[c_{ij}{]}^{\ast }=[\overline{c_{ji}}]=[\overline{⟨b_{:j},a_{i:}⟩}]=[⟨\overline{b_{:j}},\overline{a_{i:}}⟩]$$

then

$$=\left(\begin{array}{cccc}⟨\overline{b_{:1}},\overline{a_{1:}}⟩& ⟨\overline{b_{:1}},\overline{a_{2:}}⟩& \cdots & ⟨\overline{b_{:1}},\overline{a_{m:}}⟩\\ ⟨\overline{b_{:2}},\overline{a_{1:}}⟩& ⟨\overline{b_{:2}},\overline{a_{2:}}⟩& \cdots & ⟨\overline{b_{:2}},\overline{a_{m:}}⟩\\ ⋮& ⋮& & ⋮\\ ⟨\overline{b_{:p}},\overline{a_{1:}}⟩& ⟨\overline{b_{:p}},\overline{a_{2:}}⟩& \cdots & ⟨\overline{b_{:p}},\overline{a_{m:}}⟩\end{array}\right)=\underset{[p\times n]}{\underbrace{\left(\begin{array}{ccc}\leftarrow & {\overline{b_{:1}}}^T& \to \\ \leftarrow & {\overline{b_{:2}}}^T& \to \\ & ⋮& \\ \leftarrow & {\overline{b_{:p}}}^T& \to \end{array}\right)}}\underset{[n\times m]}{\underbrace{\left(\begin{array}{cccc}\uparrow & \uparrow & & \uparrow \\ {\overline{a_{1:}}}^T& {\overline{a_{2:}}}^T& \cdots & {\overline{a_{m:}}}^T\\ \downarrow & \downarrow & & \downarrow \end{array}\right)}}=B^{\ast }{A}^{\ast }◻$$

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

$$\begin{array}{rl}A{A}^{-1}=I& ⟺(A{A}^{-1}{)}^{\ast }=I^{\ast }=I\\ & ⟺A^{\ast }(A^{-1}{)}^{\ast }=I\\ & ⟺(A^{\ast }{)}^{-1}{A}^{\ast }(A^{-1}{)}^{\ast }=(A^{\ast }{)}^{-1}I\\ & ⟺(A^{-1}{)}^{\ast }=(A^{\ast }{)}^{-1}◻\end{array}$$