kevinlui's site

October 11

Announcements

• Section 2.1, 2.2 due tomorrow

• Section 2.3, 3.1 due next Thursday

• Worksheet 2 due Friday

• Midterm next week

• Worksheet 3 will be posted on Friday, it'll have some practice exam problems

3.1 Linear transformation

One-to-one and Onto linear transformation

Definition: Let $T:\mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation. Then

• $T$ is one-to-one if for every vector $w\in \mathbb{R}^n$, there exists at most one vector $v\in \mathbb{R}^m$ such that $T(v)=w$.

• $T$ is onto if for every vector $w\in\mathbb{R}^n$, there is exists at least one vector $v\in \mathbb{R}^n$ such that $T(v)=w$.

A linear transformation $T$ is one-to-one if $T(u)=T(v)$ implies $u=v$. In other words, if $u\neq v$, then $T(u)\neq T(v)$. (Two-to-two!)

Talk about the general idea of one-to-one and onto.

Theorem: Let $T$ be a linear transformation $T$ is one-to-one if $T(u)=0$ implies $u=0$.

Example: Let $T$ be the linear transformation defined by $T(x)=Ax$, where $$\begin{bmatrix} 4 & -1 \\ -2 & 2 \\ 0 & 3 \end{bmatrix}$$ Is $T$ one-to-one? Onto?

Let $T$ be the linear transformation defined by $T(x)=Ax$, where $$\begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 3 & 0 \end{bmatrix}$$ Is $T$ one-to-one? Onto?

Theorem: Let $T:\mathbb{R}^m \to \mathbb{R}^n$ be a linear transformation. Let $A$ be the matrix so that $T(x)=Ax$. Then

• $T$ is one-to-one if the columns of $A$ are linearly independent.

• $T$ is onto if the columns of $A$ span $\mathbb{R}^n$

In particular, the dimension of $A$ can sometimes implies that $T$ cannot be one-to-one and onto.

Theorem: Let $S=\{a_1,\ldots,a_n\}$ with $a_i\in \mathbb{R}^n$, $A=[a_i]$, and $T(x)=Ax$. (So $A$ is square). Then the following are equivalent:

• $S$ spans $\mathbb{R}^n$

• $S$ is linearly independent

• $Ax=b$ has a unique solution for all $b\in \mathbb{R}^n$

• $T(xs)=b$ has a unique solution for all $b\in \mathbb{R}^n$

• $T$ is onto

• $T$ is one-to-one.

Geometry of linear transformations from R^2 to R^2

Lines go to lines (or points)! Why? $T((1-s)u+sv)=(1-s)T(u)+sT(v)$.

The columns of the matrix tells you where the standard basis goes.

Let's see what happens to the square $\{(x,y):0\leq x,y\leq 1\}$ under the following transforms $$\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix}$$ $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$ $$\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$