Week 2 - Linear Transformations and Matrices #

Opening Remarks #

A linear transformation is a vector function that has the following two properties:
- Transforming a scaled vector is the same as scaling the transformed vector: $$L(\alpha x) = \alpha L(x)$$
- Transforming the sum of two vectors is the same as summing the two transformed vectors: $$L(x + y) = L(x) + L(y)$$
Lemma: $L: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation if and only if(iff) for all $u,v \in \mathbb{R}^n$ and $\alpha, \beta \in \mathbb{R}^n$ $$L(\alpha u + \beta v) = \alpha L(u) + \beta L(v)$$
Lemma: Let $v_0, v_1, \ldots, v_{k-1} \in \mathbb{R}^n$ and let $L: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation. Then
- $$L(v_0 + v_1 + \ldots + v_{k-1}) = L(v_0) + L(v_1) + \ldots + L(v_{k-1})$$

What is the Principle of Mathematical Induction(week induction)(数学归纳法)?
- if one can show that:
  - (Base case) a property holds for $k = k_b$; and
  - (Inductive step) if it holds for $k = K$, where $K \ge k_b$ , then it is also holds for $k = K +1$,
- then one can conclude that the property holds for all integers $k \ge k_b$ . Often $k_b = 0$ or $k_b = 1$.
Example: To proof: $\displaystyle\sum_{i=0}^{n-1}{i} = n(n-1)/2$
- Base case: $n = 1$. For this case, we must show that $\displaystyle\sum_{i=0}^{i-1}{i} = 1(1-1)/2$
  - $\displaystyle\sum_{i=0}^{i-1}{i} = 0 = 1(1-1)/2$
  - So this proves the base case.
- Inductive step: Inductive Hypothesis (IH): Assume that the result is true for $n = k$ where $k \ge 1$: $\displaystyle\sum_{i=0}^{k-1}{i} = k(k-1)/2$
  - We need to show that the result is then also true for $n=k+1$: $\displaystyle\sum_{i=0}^{(k+1)-1}{i} = (k+1)((k+1)-1)/2$
  - Assume that $k \ge 1$, Then
    - $$\begin{aligned}\sum_{i=0}^{(k+1)-1}{i} &= \sum_{i=0}^{k-1}{i} + k \\ &= k(k-1)/2 + k = k(k+1)/2 \\ &= (k+1)((k+1)-1)/2 \end{aligned}$$
  - This proves the inductive step.

$\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}$ is a 3 by 2 matrix: m = 3 rows and n = 2 columns.

The Big Idea. The linear transformation L is completely described by the vectors
- $a_0 ,a_1 ,…,a_{n-1}$, where $a_j = L(e_j)$
because for any vector x, $L(x) = \sum^{n-1}_{j=0} x_j a_j$.
e.g. $Ax = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a combination of the columns. $Ax = x_1 \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix}$

Let $L: \mathbb{R}^n \to \mathbb{R}^m$ be defined by $L(x) = Ax$ where $A \in \mathbb{R}^{m \times n}$. Then L is a linear transformation.
Alternatively, A vector function $f: \mathbb{R}^n \to \mathbb{R}^m$ is a linear transformation if and only if it can be represented by an $m \times n$ matrix, which is a very special two dimensional array of numbers (elements).
The set of all real valued $m \times n$ matrices is denoted by $\mathbb{R}^{m \times n}$