Linear Functionals Part II

 This is a continuation of the previous post: Linear Functionals Part I

Let us take a look at a Lemma...

LEMMA: Let $A=[a_{ij}]\in {\mathbb{F}}^{m\times n}$, that is either the real or complex scalar fields. Then the adjoint $A^{\ast }$ satisfies...

$$⟨Ax,y⟩=⟨x,A^{\ast }y⟩\mathrm{\forall }x\in {\mathbb{F}}^n,y\in {\mathbb{F}}^m$$

PROOF: Let us write this out...

$$\begin{array}{r}Ax=\left(\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \cdots & a_{1n}\\ a_{21}& a_{22}& a_{23}& & a_{2n}\\ ⋮& ⋮& ⋮& & ⋮\\ a_{m1}& a_{m2}& a_{m3}& \cdots & a_{mn}\end{array}\right)\end{array}\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\\ ⋮\\ x_n\end{array}\right)=\left(\begin{array}{c}\sum _{j=1}^n{a}_{1j}{x}_j\\ \sum _{j=1}^n{a}_{2j}{x}_j\\ \sum _{j=1}^n{a}_{3j}{x}_j\\ ⋮\\ \sum _{j=1}^n{a}_{mj}{x}_j\end{array}\right)$$

then (we will look at the real case, then $A^{\ast }=A^T$ that is the tranpose.

$$⟨Ax,y⟩=\sum _{i=1}^m\sum _{j=1}^n{a}_{ij}{x}_j{y}_i=\sum _{j=1}^n{x}_j\sum _{i=1}^m{a}_{ij}{y}_i=⟨x,\left(\begin{array}{c}\sum _{i=1}^m{a}_{i1}{y}_i\\ \sum _{i=1}^m{a}_{i2}{y}_i\\ \sum _{i=1}^m{a}_{i3}{y}_i\\ ⋮\\ \sum _{i=1}^m{a}_{in}{y}_i\end{array}\right)⟩$$

and

$$=⟨x,\left(\begin{array}{ccccc}{a}_{11}& a_{21}& a_{31}& \cdots & a_{m1}\\ a_{12}& a_{22}& a_{32}& \cdots & a_{m2}\\ ⋮& ⋮& ⋮& & ⋮\\ a_{1n}& a_{2n}& a_{3n}& \cdots & a_{mn}\end{array}\right)\left(\begin{array}{c}{y}_1\\ y_2\\ y_2\\ ⋮\\ y_m\end{array}\right)⟩=⟨x,A^{T}y⟩$$