October 11

Announcements

• Section 2.3, 3.1 due next Thursday
• Write down name if you did worksheet 2
• Midterm next week
• Worksheet 3 will be posted tonight, it’ll have some practice exam problems

3.1 Linear transformation

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. Once you know this, you should know everything.

Let’s see what happens to the square {(x, y) : 0 ≤ x, y ≤ 1} under the following transforms
$$ \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} $$

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

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

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

Piecing things together

Theorem: Let S = {v₁, …, v_n}. Let A be the matrix with the elements of S as columns. Let B be an echelon matrix equivalent to A. Let T be a linear transform with T(x) = Ax. Then the following are equivalent

• The set S is linearly independent.
• The linear equation x₁v₁ + … + x_nv_n = 0 has only the trivial solution.
• Every columns of B has a pivot. (computationally useful)
• For any b ∈ ℝⁿ, the equation x₁v₁ + … + x_nv_n = b has a unique solution.
• The homogenous equation Ax = 0 has only the trivial solution.
• For any b ∈ ℝⁿ, the equation Ax = b has at most one solution.
• For any b ∈ ℝⁿ, b can be expressed as a linear combination of elements in S in at most one way.
• The zero vector can be expressed as a linear combination of elements in S in only one way.
• T is a one-to-one linear transformation.
• The only solution to T(x) = 0 is x = 0. If T(x) = 0, then x = 0.
• There is at most one solution to T(x) = b.

Theorem: Let S = {v₁, …, v_n} be a set of vectors in ℝ^m. Let A be the matrix with the elements of S as columns. Let B be an echelon matrix equivalent to A. Let T be a linear transform with T(x) = Ax. Then the following are equivalent

• The set S spans ℝ^m.
• The linear equation x₁v₁ + … + x_nv_n = b always has a solution.
• Every row of B has a pivot. (computationally useful)
• For any b ∈ ℝⁿ, the equation Ax = b has at least one solution.
• For any b ∈ ℝⁿ, b can be expressed as a linear combination of elements in S in at least one way.
• T is a onto linear transformation.
• There is always a solution to T(x) = b.

Examples:

Kristin DeVleming exam: Let u₁ = (4, 4, 2) and u₂ = (8, 5,  − 3). Let v = (26, 17,  − 8). Write v as a linear combination of u₁, u₂. Write a vector w that is not in the span of u₁, u₂.

Josh Swanson exam: Are the following sets spanning?

• {(1, 2, 3), ( − 1,  − 1, 2), ( − 1, 0, 7)}
• {(1,  − 1, 1), (0, 1, 2), ( − 2, 0, 2), (1, 3, 1)}.