A0: Linear Algebra Flashcards
Theorem
First Isomorphism Theorem
The kernel Ker(∅) of a ring homomorphism ∅: R → S is an ideal, its image Im(∅) is a subring of S, and ∅ induces an isomorphism of rings
R/Ker(∅) ≅ Im(∅).
Rings & Polynomials
Proof
The kernel Ker(∅) of a ring homomorphism ∅: R → S is an ideal, its image Im(∅) is a subring of S, and ∅ induces an isomorphism of rings
R/Ker(∅) ≅ Im(∅).
4 points
First Isomorphism Theorem
- r̅ ↦ ∅(r)
- Check well-defined
- Homomorphism & surjectivity trivial
- Injective ⇔ Ker(∅) = {0}
Theorem
Division Algorithm for Polynomial
Let f(x), g(x) ∈ 𝔽[x] be two polynomials with g(x) ≠ 0.
Then there exists q(x), r(x) ∈ 𝔽[x] such that
f(x) = q(x) g(x) + r(x) and deg r(x) < deg g(x).
Proof
Let f(x), g(x) ∈ 𝔽[x] be two polynomials with g(x) ≠ 0.
Then there exists q(x), r(x) ∈ 𝔽[x] such that
f(x) = q(x) g(x) + r(x) and deg r(x) < deg g(x).
3 points
Divsion Algorithm for Polynomials
- Induct on deg f(x) - deg g(x)
- If deg f(x) < deg g(x), let q(x) = 0 and r(x) = f(x)
- Otherwise, there exist s(x) and t(x) such that f(x) - (aₙ / bₖ) xⁿ⁻ᵏ g(x) = s(x) g(x) + t(x) and deg g(x) > deg t(x)
Proof
Let a, b ∈ 𝔽[x] be non-zero polynomials and let gcd(a, b) = c.
Then there exist s, t ∈ 𝔽[x] such that:
a(x) s(x) + b(x) t(x) = c(x).
5 points
- WLOG assume deg a(x) ≥ deg b(x) and gcd(a, b) = 1
- Induct on deg a(x) + deg b(x)
- a = q b + r by the division algorithm
- If r = 0 then b is contant
- Otherwise, there exist s’, t’ ∈ 𝔽[x] such that b s’ + r t’ = 1
Proof
For all A ∈ Mₙ(𝔽), there exists a non-zero polynomial f(x) ∈ 𝔽[x] such that f(A) = 0.
1 point
{I, A, A², …, A^(n²)} is linearly dependent
Proof
The set of cosets
V/U = {v + U | v ∈ V}
with the operations
(v + U) + (w + U) := v + w + U
a (v + U) := a v + U
for v, w ∈ V and a ∈ 𝔽 is a vector space, calledd the quotient space.
2 point
- Show operations are well-defined
- Operations in V satisfy vector space axioms so these ones must
Proof
Let 𝓔 be a basis for U, and extend 𝓔 to a basis B of V.
The set
B̅ := {e + U | e ∈ B \ 𝓔} ⊆ V / U
is a basis for V / U.
3 points
Proposition
- Use B is spanning to show B̅ is spanning
- a₁ e₁ + … + aᵣ eᵣ = b₁ e₁’ + … + bₛ eₛ’ for e₁, …, eᵣ ∈ B \ 𝓔, e₁’, …, eᵣ’ ∈ B, a₁, …, aᵣ, b₁, …, bᵣ ∈ 𝔽
- Use linear independence of B
Proof
Let U ⊂ V be ector spaces, with 𝓔 a basis for U, and 𝓕 ⊂ V a set of vectors such that
{v + U: v ∈ 𝓕}
is a basis for the quotient V / U.
Then the union
𝓔 ∪ 𝓕
is a basis for V.
2 points
Proposition
- For spanning, v + U = a₁ f₁ + … + aₖ fₖ + U for v ∈ V, f₁, …, fₖ ∈ 𝓕, a₁, …, aₖ ∈ 𝔽
- For linear independence, a₁ f₁ + … + aₖ fₖ + U = U
Theorem
First Isomorphism Theorem
Quotient Spaces
Let T: V → W be a linear map of vector spaces over 𝔽.
Then
T̅: V / Ker(T) → Im(T)
T̅: v + Ker(T) ↦ T(v)
is an isomorphism of vector spaces.
Proof
Let T: V → W be a linear map of vector spaces over 𝔽.
Then
T̅: V / Ker(T) → Im(T)
T̅: v + Ker(T) ↦ T(v)
is an isomorphism of vector spaces.
4 points
First Isomorphism Theorem
- Show well-defined
- Show linear
- Show surjective
- Show injective by Ker(T̅) = {Ker(T)}
Theorem
Rank-Nullity Theorem
If T: V → W is a linear transformation and V is finite-dimensional, then
dim(V) = dim(Ker(T)) + dim(Im(T)).
Proof
If T: V → W is a linear transformation and V is finite-dimensional, then
dim(V) = dim(Ker(T)) + dim(Im(T)).
2 points
Rank-Nullity Theorem
- dim(V) = dim(Ker(T)) + dim(V / Ker(T))
- First isomorphism theorem
Proof
Let T: V → W be a linear map and A ⊆ V, B ⊆ W be subspaces.
The formula T̅(v + A) := T(v) + B gives a well-define linear map of quotients T̅: V / A → W / B if and only if T(A) ⊆ B.
2 points
Lemma
- T̅ is linear iff it is well-defined
- B = T̅(a + A) for a ∈ A
- T(eⱼ) ∈ B for j ≤ k
- T̅(eⱼ + A) = a₁ⱼ e₁’ + … + aₘⱼ eₘ’ + B for k + 1≤ j ≤ n
Proof
Let T: V → V be a linear transormation on a finite dimensional space and assume U ⊆ V is T-invariant.
Then
χ_T(x) = χ_(T|_U)(x) × χ_T̅(x).
2 points
Proposition
- Extend a basis 𝓔 for U to a basis 𝓑 of V
- _𝓑[T]_𝓑 is an upper-triangular block matrix
Proof
Let V be a finite-dimensional vector space, and let T: V → V be a linear map such that its characteristic polynomial is a product of linear factors.
Then, there exists a basis 𝓑 of V such that _𝓑[T]_𝓑 is upper-triangular.
3 points
Theorem
- Induct on dimension of V
- Let U = ⟨v₁⟩ where T v₁ = λ v₁
- Induce T̅: V / U → V / U and apply inductive hypothesis
Proof
Let A be an upper triangular (n × n)-matrix with diagonal entries λ₁, …, λₙ.
Then
∏ⁿᵢ₌₁(A – λᵢ I) = 0.
2 points
Proposition
- (A – λₙ I) v ∈ ⟨e₁, …, eₙ₋₁⟩ for v ∈ 𝔽ⁿ
- Im(A – λₙ I) ⊆ ⟨e₁, …, eₙ₋₁⟩
Theorem
Cayley-Hamilton
If T: V → V is a linear transformation and V is a finite-dimensional vector space, then χ_T(T) = 0.
Hence, in particular, m_T(x) | χ_T(x).
Proof
If T: V → V is a linear transformation and V is a finite-dimensional vector space, then χ_T(T) = 0.
Hence, in particular, m_T(x) | χ_T(x).
3 points
Cayley-Hamilton
- χ_T(x) = ∏(x – λᵢ) for λᵢ ∈ F̅
- A is a matrix for T and upper-triangularisable over F̅
- χ_B(B) = 0 for B = S⁻¹ A S
