Basis and Dimension Round 2

Definition: A set B = {u₁, …, u_m} is a basis for a subspace S if

• B spans S,
• B is linearly independent

Theorem: Let B = {u₁, …, u_m} be a basis for a subspace S. Then every s ∈ S can be written as a linear combination of u₁, …, u_m in a unique way.

Example: We’ll use the same subspace as last time. Let S ⊆ ℝ⁴ be the subspace spanned by u₁ = ( − 1, 2, 3, 1), u₂ = ( − 6, 7, 5, 2), u₃ = (4,  − 3, 1, 0). From last class, we determined that a basis for S is given by {v₁, v₂} where v₁ = ( − 1, 2, 3, 1), v₂ = (0, 5, 13, 4). It is clear that u₃ ∈ S. How do we express u₃ as a linear combination of v₁, v₂? This amounts to solving [v₁, v₂|u₃].

v1 = vector([-1,2,3,1])
v2 = vector([0,5,13,4])
u3 = vector([4,-3,1,0])
A = matrix([v1,v2]).transpose()
A, u3

(
[-1  0]               
[ 2  5]               
[ 3 13]               
[ 1  4], (4, -3, 1, 0)
)

A \ u3

(-4, 1)

Most sets of n vectors in ℝⁿ are a basis.

Example: Take S as before with basis B = {v₁, v₂}. How can we extend B to be a basis for ℝⁿ?

• Eyeball it
• Add 2 random vectors

In this case, we see that {v₁, v₂, e₁, e₂} will form a basis for ℝ⁴. But so will {v₁, v₂, ( − 213, π, 4, 2), (4, π², 3, 4)}.