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 : ℝ^m → ℝⁿ be a linear transformation. Then

• T is one-to-one if for every vector w ∈ ℝⁿ, there exists at most one vector v ∈ ℝ^m such that T(v) = w.
• T is onto if for every vector w ∈ ℝⁿ, there is exists at least one vector v ∈ ℝⁿ 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 ≠ v, then T(u) ≠ 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 : ℝ^m → ℝⁿ 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 ℝⁿ

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

Theorem: Let S = {a₁, …, a_n} with a_i ∈ ℝⁿ, A = [a_i], and T(x) = Ax. (So A is square). Then the following are equivalent:

• S spans ℝⁿ
• S is linearly independent
• Ax = b has a unique solution for all b ∈ ℝⁿ
• T(xs) = b has a unique solution for all b ∈ ℝⁿ
• 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 ≤ x, y ≤ 1} under the following transforms
$$ \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} $$

$$ \begin{bmatrix} 1 & 2 \\ 0 & 2 \end{bmatrix} $$

$$ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$

$$ \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix} $$