Linear Algebra Essentials: PART 1

1. An introduction

Theorem 1

Let $V$ be an inner product space over $F$ and let $g\,:\,V\rightarrow F$ be a linear transformation. Then there exists a unique vector $y \in V$ such that $g(x) = \langle x,y \rangle$ for all $x \in V$ .

Definition

Let $T$ be a linear operator on an inner product space $V$ . The unique linear operator $T^*$ which is such that $\langle T(x), y\rangle = \langle x, T^*(y)\rangle$ is called the adjoint of $T$ .

Theorem 2

The adjoint of any linear operator exists.

Theorem 3

Let $V$ be a finite-dimensional inner product space, and let $\beta$ be an orthonormal basis for $V$ . Then $[T^*]_\beta = [T]_\beta^*$ .

Corollary: $L_{A^*} = (L_A)^*$

2. Proofs to the Theorems

PROOF OF THEOREM 1 (Every linear operator is an inner product operator)

Let $\beta=\{u_1,u_2, . . . ,u_n\}$ be a base for $V$ . Let A

$y = \sum\limits_{i=1}^n \overline{g(u_i)}u_i$ .

Then let $h(x) := \langle x,y\rangle$ . For $i \in [n]$ we have: $h(u_i) = \langle u_i, \sum\limits_{j=1}^n \overline{g(u_j)}u_j \rangle = \sum\limits_{j=1}^n g(u_j) \langle u_i, u_j \rangle = g(u_i)$ . Since $h$ and $g$ agree on every element of $\beta$ , they are equal linear operators. The uniqueness of $y$ follows easily through a contradiction argument.

PROOF OF THEOREM 2 (existence and uniqueness of the adjoint linear operator)

Let $y \in V$ . Define $g(x) = \langle T(x), y \rangle$ . It is easy to see that $g$ is linear. So, by Theorem 1, there exists a $y^* \in V$ such that $g(x) = \langle x, y^* \rangle$ . Defining $T^*(y) = y^*$ gives us the adjoint. Linearity and uniqueness follow easily.