Vector Spaces, Subspaces & Linear Independence
“Vector axioms, subspace closure criteria, spanning sets, linear independence, and basis dimensions.”
1.Vector Spaces, Closure Axioms & Subspace Criteria
§ 1.1.A Vector Space V over field F (typically R) is a set equipped with vector addition and scalar multiplication satisfying 8 core axioms: commutativity, associativity, additive identity 0, additive inverse, multiplicative identity 1, and distributivity.
The Two-Step Subspace Criterion:
A subset W of V is a subspace if and only if:
1. Zero vector: 0 in W.
2. Closure under addition: For all u, v in W, u + v in W.
3. Closure under scalar multiplication: For all u in W and c in R, c * u in W.
Combined in one step: For all u, v in W and alpha, beta in R, alpha u + beta v in W.
Linear Independence & Spanning Sets:
Vectors {v1, ..., vk} are linearly independent if the only solution to Sum_{i=1}^k c_i v_i = 0 is c1 = c2 = ... = ck = 0.
If at least one non-zero scalar exists, the vectors are linearly dependent, meaning at least one vector can be written as a linear combination of the others.
Basis & Dimension:
A set B is a basis for V if it is linearly independent and spans V. The number of vectors in B is the dimension dim(V).
2. Worked Practice Problem: Subspace Verification & Basis Extraction
Ensure you solve for the number of free variables; dim(W) = n - number of independent linear constraints.
Problem: Let W be the subset of R^4 defined by: W = { (x, y, z, w) in R^4 : x + 2y - z = 0 and y - w = 0 }. Verify that W is a subspace of R^4, find its dimension, and construct a basis.
Step 1: Check subspace criteria:
Homogeneous linear equations always contain the origin: (0, 0, 0, 0) satisfies 0 + 0 - 0 = 0 and 0 - 0 = 0.
Linear combinations of solutions to homogeneous linear equations are also solutions. Thus, W is a valid subspace.
Step 2: Express dependent variables in terms of free variables:
Constraint 1: x + 2y - z = 0 => x = z - 2y
Constraint 2: y - w = 0 => y = w
Let the free variables be z = s and w = t (where s, t in R).
Then: y = t, and x = s - 2t.
Step 3: Parameterize any vector v in W:
v = [ x, y, z, w ]^T = [ s - 2t, t, s, t ]^T
v = s [ 1, 0, 1, 0 ]^T + t [ -2, 1, 0, 1 ]^T
Step 4: Check linear independence of basis candidates:
Vector 1: b1 = [ 1, 0, 1, 0 ]^T
Vector 2: b2 = [ -2, 1, 0, 1 ]^T
Neither vector is a scalar multiple of the other (component 4 is 0 in b1 but 1 in b2).
Conclusion: Basis for W is { [1, 0, 1, 0]^T, [-2, 1, 0, 1]^T }, and dim(W) = 2.
Primary Academic References & Exam Sources
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