Show that αx = 0 implies α = 0 or x = 0 for α in F and x in V.
Solution:
Assume αx = 0. If α = 0, the condition is satisfied.
If α ≠ 0, then its multiplicative inverse α⁻¹ exists in the field F. Multiplying both sides by α⁻¹:α⁻¹(αx) = α⁻¹(0)
(α⁻¹α)x = 0
1x = 0 ⇒ x = 0.
Thus, either α = 0 or x = 0.
Justify whether W = {(x, y, 2) | x, y in R} is a subspace of R³.
Solution:
For W to be a subspace, it must contain the zero vector (0, 0, 0).
In this set, the third component is fixed at 2. Since (0, 0, 0) does not belong to W (because 0 ≠ 2), W is not a subspace of R³.Define linearly dependent/independent sets. Show {(1,0,0), (0,1,0), (0,0,1), (1,1,1)} is linearly dependent.
Definitions:
Justification:
Notice that (1,1,1) = 1(1,0,0) + 1(0,1,0) + 1(0,0,1).
Since one vector can be expressed as a linear combination of the others, the set is linearly dependent.Prove dim(W₁ + W₂) = dim W₁ + dim W₂ - dim(W₁ ∩ W₂).
Proof Strategy:
Define nullity and rank of a linear transformation.
Rank: The dimension of the range (image) of a linear transformation T, denoted as ρ(T).
Nullity: The dimension of the kernel (null space) of T, denoted as ν(T).
State and prove rank-nullity law.
Statement: If V is a finite-dimensional vector space and T: V → W is linear, then rank(T) + nullity(T) = dim(V).
Proof: Let {v₁, ..., vₖ} be a basis for Ker(T), so nullity(T) = k.
Extend this to a basis {v₁, ..., vₖ, vₖ₊₁, ..., vₙ} for V, where n = dim(V). It can be shown that {T(vₖ₊₁), ..., T(vₙ)} is a basis for Range(T). Thus, rank(T) = n - k. Therefore, rank(T) + nullity(T) = (n-k) + k = n.State and prove Cayley-Hamilton theorem.
Statement: Every square matrix satisfies its own characteristic equation.
Proof Outline: Let p(λ) = det(A - λI) be the characteristic polynomial.
Using the property that B * adj(B) = det(B) * I, where B = (A - λI), and equating coefficients of the powers of λ, we show that p(A) = O (zero matrix).Define an inner product space. Show ||x|| ≥ 0 and ||x|| = 0 iff x = 0.
An inner product space is a vector space V over F with a function (u, v) satisfying:
Norm property: Since ||x|| = √(x, x) and (x, x) ≥ 0 by definition, ||x|| ≥ 0. Also, ||x|| = 0 implies √(x, x) = 0 ⇒ (x, x) = 0, which by the last axiom means x = 0.