Lecture 07 - Linear Transformation of a Vector space

#math

Definition

Let $X, Y$ be vector spaces. A mapping

T : X \to Y

is linear if for all $x, y \in X$ and $a, b$ scalars we have:

T (a x + b y) = a T (x) + b T (y)

Such a linear transformation is uniquely determined on all of $X$ by its action on the basis vectors of $X$ . That is, if $S = {x_{α}}_{α \in A}$ is a basis of $X$ and $y_{α} = T x_{α}$ then for $x = \sum_{k = 1}^{m} λ_{k} x_{α_{k}}$ we have:

T x = \sum_{k = 1}^{m} λ_{k} y_{α_{k}}

If $X$ and $Y$ are finite dimensional spaces and $T : X \overset{linear}{\to} Y$ where we have $B_{X} = {x_{1}, . . ., x_{m}}$ and $B_{Y} = {y_{1}, . . ., y_{n}}$ are basis vectors for their respective spaces then for $1 \leq j \leq m, \exists α_{k_{j}}$ such that:

T x_{j} = \sum_{k = 1}^{n} α_{k_{j}} y_{k}

Definition

If $T$ is a linear transformation between two vector spaces $X$ and $Y$ and $T$ is bijective then $T$ is called an isomorphism between $X$ and $Y$ . We say $X$ and $Y$ are isomorphic.

If $T$ is an isomorphism and $B_{X}$ is a basis of $X$ then $T (B)$ is a basis of $Y$ . That is both spaces have the same finite dimension or are both infinite dimensional.

Definition

Let $T : X \overset{linear}{\to} Y$ . Then we define the following terms:

Kernel of $T$ or null space of $T$ is:

K e r (T) = n u l l (T) = {x \in X : T (x) = 0}

Range of $T$ is:

R (T) = {y \in Y : y = T x for some x \in X}

Recall that $T$ is an isomorphism if

K e r (T) = {0} & R (T) = Y

Contraction Mapping Theorem

One of the most important theorems used in applied mathematics which concerns fixed points of a mapping of $X$ into $X$

Definition

A mapping $T : X \to X$ is a contraction if it is Lipschitz continuous with Lipschitz constant $ρ < 1$ . That is $\exists ρ \in [0, 1]$ such that:

d (T x, T y) \leq ρ d (x, y)

If $ρ = 1$ , then $T$ is called non-expansive.

Theorem

(Contraction Mapping Theorem) If $X$ is a complete metric space and $T$ is a contraction on $X$ then $\exists! x \in X$ such that $T x = x$ .

Proof

(Proof of uniqueness)
Assume that we have two different fixed points $x_{1}$ and $x_{2}$ then:

\begin{aligned} d (x_{1}, x_{2}) & = d (T x_{1}, T x_{2}) \\ \leq ρ d (x_{1}, x_{2}) \\ ⟹ 0 & < (1 - ρ) d (x_{1}, x_{2}) \leq 0 \end{aligned}

Its positive as $ρ < 1$ thus $d (x_{1}, x_{2}) = 0$ i.e. $x_{1} = x_{2}$
(Existence of fixed point $x$ )
Fix any point $x_{1} i n X$ and define the iteration $x_{n + 1} = T x_{n}$ . As $X$ is complete, we want to show ${x_{n}}_{n = 1}^{\infty}$ is a Cauchy sequence in $X$ . Notice that:

d (x_{n}, x_{n + 1}) \leq ρ^{n - 1} d (x_{1}, x_{2})

Then let $m > n \geq N$ we have:

\begin{aligned} d (x_{m}, x_{n}) & \leq \sum_{j = n + 1}^{m} d (x_{j}, x_{j - 1}) \\ \leq \sum_{j = n + 1}^{m} ρ^{j} d (x_{1}, x_{2}) \\ = ρ^{n} \frac{(1 - ρ^{m - n})}{1 - ρ} d (x_{1}, x_{2}) \end{aligned}

Now as $m, n \to \infty$ we have $d (x_{m}, x_{n}) \to 0$ , and as $X$ is complete $\exists x \in X$ such that $lim_{n \to \infty} x_{n} = x$ .

Example

Consider the ODE

\frac{d u}{d t} = f (t, u) u (t_{0}) = u_{0}

f \in C ([a, b] \times R), t_{0} \in [a, b]

If $u$ is a classical solution of ODE then

\begin{matrix} (**) & u (t) = u_{0} + \int_{t_{0}}^{t} f (s, u (s)) d s \end{matrix}

Conversely if $u \in C ([a, b])$ and satisfies $(* *)$ , then $u^{'}$ exists, so $u^{'}$ is continuous so the ODE holds.

\begin{aligned} T : X = & C ([a, b]) ⟶ X \\ T (u (t)) & = u_{0} + \int_{t_{0}^{t}} f (s, u (s)) d s \end{aligned}

Condition on f:
f Lipschitz w.r.t $u$ :

| f (t, u) - f (t, v) \leq L | u - v |

for some $L$ .

| T u - T v | \leq L \int_{t_{0}}^{t} | u (s) - v (s) | d s \leq L | b - a | d (u, v)

If $L | b - a | \leq 1 ⟺ | b - a | < 1 / L$ then the contraction mapping provides a unique solution.