October 25

Announcements

• Worksheet 4 posted, due Friday
• Webassign 3.2 due Thursday

3.2 Matrix Algebra

Definition: The ith standard basis vector, denoted e_i, for ℝⁿ is the length n vector consisting of all zeros except a one in the ith position. The set of all standard basis vectors for ℝⁿ is call the standard basis. When it matters, we will regard e_i a column vector.

Theorem Let A = [a₁ … n]^t be an n × k matrix and B be a k × m matrix. Then the rows of AB are a₁B, …, a_nB.

Example Let’s consider the square matrices A = [2,  − 1; 1, 3] and B = [4,  − 2;  − 1, 1]. We can think of AB in two ways. Either A is acting on B by combining the rows or B is acting on A by combining the columns.

We should think about what the standard basis does. So what is e_i^tB?

We should think of [2,  − 1] as 2e₁^t − e₂^t.

3.3 Inverses

The compositon of f and g is the function (f ∘ g) where (f ∘ g)(x) = f(g(x)).

Definition: Let A, B be sets and f : A → B be a function. Then the (two-sided) inverse of f is a function g : B → A such that g ∘ f is the identity on B (which means (g ∘ f)(x) = x) and f ∘ g is the identity on A. We often denote g here by f^− 1.

The goal is to invert linear transforms. Let T : ℝ^m → ℝⁿ be linear transform. When can T be invert? It has to be one-to-one at least. Suppose T(x₁) = T(x₂). Then by applying T^− 1, we have x₁ = x₂. This implies n ≥ m. By using the same argument, T^− 1 : ℝⁿ → ℝ^m must be one-to-one as well so m ≥ n. Therefore, n = m.

Theorem: Suppose T : ℝ^m → ℝⁿ is a linear transform such that T(x) = Ax. Then T is invertible if and only if m = n and the columns of A are linearly independent (or spanning).

Let’s T : ℝ² → ℝ² be the linear transform given by T(x) = [1, 2; 3, 4]x. So we know T(1, 0) = (1, 2) and T(0, 1) = (3, 4). This implies T^− 1(1, 2) = (1, 0) and T^− 1(3, 4) = (0, 1). To determine T^− 1, we just need to know what T^− 1(1, 0) and T^− 1(0, 1) are. Row reduction!

The determinant of a 2,2 matrix is blah. Here’s the formula for the inverse of a 2,2 matrix.

Theorem: Let A and B be invertible matrices and C and D be matrices. Then

• A^− 1 is also invertible.
• AB is invertible. The inverse is given by (AB)^− 1 = B^− 1A^− 1.
• If AC = AD then C = D
• If CA = DA then C = D

Theorem: Let A be a n × n matrix. Let S be the columns of A. Let T(x) = Ax. Then the following are equivalent:

• S spans ℝⁿ
• S is linearly indepedent
• Ax = b has a unique solution for all b ∈ ℝⁿ given by x = A^− 1b.
• T is onto
• T is one-to-one
• T is invertible
• A is invertible

Subspaces

Here’s the defintion of subspaces. Here’s an example of what is a subspace. Here’s an example of what isn’t a subspaces.

The solutions to homoegenous equations are subspaces. This is why we care.

Introduce kernel and range.