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 $A^*$ satisfies...

$$\langle Ax,y\rangle=\langle x,A^*y\rangle\ \ \forall \ \ x\in\mathbb{F}^n,y\in\mathbb{F}^m$$

PROOF: Let us write this out...

$$\begin{align*} Ax=\begin{pmatrix}a_{11} & a_{12} & a_{13}& \cdots & a_{1n}\\ a_{21} & a_{22} & a_{23} & & a_{2n}\\ \vdots & \vdots & \vdots & & \vdots\\ a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn}\end{pmatrix}\end{align*}\begin{pmatrix}x_1\\ x_2 \\ x_3\\ \vdots \\ x_n\end{pmatrix}=\begin{pmatrix}\sum_{j=1}^n a_{1j}x_j \\ \sum_{j=1}^n a_{2j}x_j \\ \sum_{j=1}^n a_{3j}x_j\\ \vdots \\ \sum_{j=1}^n a_{mj}x_j \end{pmatrix}$$

then (we will look at the real case, then $A^*=A^T$ that is the tranpose.

$$\langle Ax,y\rangle=\sum_{i=1}^m \sum_{j=1}^n a_{ij}x_jy_i=\sum_{j=1}^n x_j\sum_{i=1}^m a_{ij}y_i=\left\langle x, \begin{pmatrix} \sum_{i=1}^m a_{i1}y_i\\ \sum_{i=1}^m a_{i2}y_i \\ \sum_{i=1}^m a_{i3}y_i \\ \vdots \\ \sum_{i=1}^m a_{in}y_i \end{pmatrix}\right\rangle $$

and

$$=\left\langle x, \begin{pmatrix}a_{11} & a_{21} & a_{31} & \cdots & a_{m1}\\ a_{12} & a_{22} & a_{32} & \cdots & a_{m2} \\ \vdots & \vdots & \vdots & & \vdots \\ a_{1n} & a_{2n} & a_{3n} & \cdots & a_{mn}\end{pmatrix}\begin{pmatrix}y_1\\ y_2 \\ y_2 \\ \vdots \\ y_m\end{pmatrix}\right\rangle=\langle x,A^Ty\rangle$$

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