Linear Algebra Flashcards
Define a field
A set F with two binary operations + and × is a field if both (F, +, 0) and
(F \ {0}, ×, 1) are abelian groups and the distribution law holds:
(a + b)c = ac + bc, for all a, b, c ∈ F.
Define the characteristic of F
The smallest integer p such that
1 + 1 + · · · + 1 (p times) = 0
is called the characteristic of F. If no such p exists, the characteristic of F is defined to be zero.
If such a p exists, it is necessarily prime.
Define a vector space V over a field F in terms of groups
A vector space V over a field F is an abelian group (V, +, 0) together with a scalar multiplication F × V → V such that for all a, b ∈ F, v, w ∈ V :
(1) a(v + w) = av + aw
(2) (a + b)v = av + bv
(3) (ab)v = a(bv)
(4) 1.v = v
Let V be a vector space over F
Define a set S ⊆ V being linearly independent
(1) A set S ⊆ V is linearly independent if whenever a1, · · · , an ∈ F, and
s1, · · · , sn ∈ S,
a1s1 + · · · + ansn = 0 ⇒ a1 = · · · = an = 0.
Let V be a vector space over F
Define what it means for a set S ⊆ V to be spanning
(2) A set S ⊆ V is spanning if for all v ∈ V there exists a1, · · · , an ∈ F and s1, · · · , sn ∈ S with
v = a1s1 + · · · + ansn
Let V be a vector space over F
Define what it means for a set S ⊆ V to be a basis of V
(3) A set B ⊆ V is a basis of V if B is spanning and linearly independent. The size of B is the
dimension of V
Define a linear map/transformation
Suppose V and W are vector spaces over F. A map T : V → W is a linear
transformation (or just linear map) if for all a ∈ F, v, v′ ∈ V ,
T(av + v’) = aT(v) + T(v’)
What is a bijective linear map called?
an isomorphism of vector spaces.
What is the assignment T → ᵦ’[T]ᵦ ?
Meant to be fancy B’ subscript
an isomorphism of vector spaces from Hom(V, W)
to the space of (m×n)-matrices over F. It takes composition of maps to multiplication of matrices.
In particular, if T : V → V and B and B′ are two different bases with ᵦ’[Id]ᵦ the change of basis matrix then:
ᵦ’[T]ᵦ’ = ???
all Bs are meant to be fancy Bs
ᵦ’[T]ᵦ’ = ᵦ’[Id]ᵦ ᵦ[T]ᵦ ᵦ[Id]ᵦ’ with ᵦ’[Id]ᵦ ᵦ[Id]ᵦ’ = I the identity matrix
Define a ring
A non-empty set R with two binary operations + and × is a ring if (R, +, 0) is an
abelian group, the multiplication × is associative and the distribution laws hold: for all a, b, c ∈ R,
(a + b)c = ac + bc and a(b + c) = ab + ac.
Define a commutative ring
The ring R is called commutative if for all a, b ∈ R we have ab = ba.
Define a ring homomorphism
A map φ : R → S between two rings is a ring homomorphism if for all
r, r′ ∈ R:
φ(r + r’) = φ(r) + φ(r’) and φ(rr’) = φ(r)φ(r’).
Define a ring isomorphism
A bijective ring homomorphism is called a ring isomorphism.
Define an ideal
A non-empty subset I of a ring R is an ideal if for all s, t ∈ I and r ∈ R we have s − t ∈ I and sr, rs ∈ I.
What is the first isomorphism theorem? (rings)
The kernel Ker(φ) := φ⁻¹(0) of a ring homomorphism φ : R → S is an ideal, its image Im(φ) is a subring of S, and φ induces an isomorphisms of rings R/Ker(φ) ∼= Im(φ)
Prove the first isomorphism theorem (rings)
Exercise
What is the “division algorithm” for polynomials?
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).
Prove the “division algorithm” for polynomials
If deg f(x) < deg g(x), put q(x) = 0, r(x) = f(x). Assume now that deg f(x) ≥ deg g(x)
and let
f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀
g(x) = bₖxᵏ + bₖ₋₁xᵏ⁻¹ + … + b₀
Then
deg( f(x) - aₙ/bₖ xⁿ⁻ᵏg(x) ) < n
By induction on deg f − deg g, there exist s(x), t(x) such that
f(x) - aₙ/bₖ xⁿ⁻ᵏg(x) = s(x)g(x) + t(x) and deg g(x) ? deg t(x)
Hence put q(x) = aₙ/bₖ xⁿ⁻ᵏ + s(x) and r(x) = t(x)
For all f(x) ∈ F[x] and a ∈ F,
f(a) = 0 ⇒ ???
For all f(x) ∈ F[x] and a ∈ F,
f(a) = 0 ⇒ (x − a)|f(x).
For all f(x) ∈ F[x] and a ∈ F,
f(a) = 0 ⇒ (x − a)|f(x).
Prove it
By division alg for polyn there exist q(x), r(x) such that
f(x) = q(x)(x − a) + r(x)
where r(x) is constant (as deg r(x) < 1). Evaluating at a gives
f(a) = 0 = q(a)(a − a) + r = r
and hence r = 0.
Assume f ≠ 0. If deg f ≤ n then f has [ ] roots
Assume f ≠ 0. If deg f ≤ n then f has at most n roots.
Assume f ≠ 0. If deg f ≤ n then f has at most n roots.
Prove it
Follows from
For all f(x) ∈ F[x] and a ∈ F,
f(a) = 0 ⇒ (x − a)|f(x).
and induction
Let a(x), b(x) ∈ F[x] be two polynomials. Let c(x) be a monic polynomial of highest degree dividing
both a(x) and b(x) and write c = gcd(a, b) (also wrote less commonly hcf(a, b)).
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) =
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)
Let a(x), b(x) ∈ F[x] be two polynomials. Let c(x) be a monic polynomial of highest degree dividing
both a(x) and b(x) and write c = gcd(a, b) (also wrote less commonly hcf(a, b)).
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)
Prove it
If c ≠ 1, divide a and b by c. We may thus assume deg(a) ≥ deg(b) and gcd(a, b) = 1, and
will proceed by induction on deg(a) + deg(b).
By the Division Algorithm there exist q, r ∈ F[x] such that
a = qb + r with deg(b) > deg(r).
Then deg(a) + deg(b) > deg(b) + deg(r) and gcd(b, r) = 1.
If r = 0 then b(x) = λ is constant since gcd(a, b) = 1. Hence
a(x) + b(x)(1/λ)(1 − a(x)) = 1.
Assume r ≠ 0.Then by the induction hypothesis, there exist s’, t′ ∈ F[x] such that
bs′ + rt′ = 1.
Hence,
bs′ + (a − qb)t′ = 1 and at′ + b(s’ − qt’) = 1
So, we may put t = t’ and s = s’− qt′
Let A ∈ Mn(F) and f(x) = aₖxᵏ + · · · + a₀ ∈ F[x]. Then
f(A) := [ ]
aₖAᵏ + · · · + a₀I ∈ Mn(F).
Let A ∈ Mn(F) and f(x) = aₖxᵏ + · · · + a₀ ∈ F[x]. Then
f(A) := aₖAᵏ + · · · + a₀I ∈ Mn(F).
Since AᵖAʳ = AʳAᵖ and λA = Aλ for p, q ≥ 0 and λ ∈ F, then for all f(x), g(x) ∈ F[x] we have that
f(A)g(A) =
Av = λv ⇒
f(A)g(A) = g(A)f(A); Av = λv ⇒ f(A)v = f(λ)v
For all A ∈ Mn(F), there exists a non-zero polynomial f(x) ∈ F[x] such that f(A) = ?
For all A ∈ Mn(F), there exists a non-zero polynomial f(x) ∈ F[x] such that f(A) = 0
For all A ∈ Mn(F), there exists a non-zero polynomial f(x) ∈ F[x] such that
f(A) = 0
Prove it
Note that the dimension dim Mn(F) = n × n is finite. Hence {I, A, A², · · · , Aᵏ} as a subset of Mn(F) is linearly dependent for k ≥ n². So there exist scalars ai ∈ F, not all zero, such that
aₖAᵏ + · · · + a₀I = 0,
and f(x) = aₖxᵏ + · · · + a₀ is an annihilating polynomialFor any (n × n)-matrix A,
the assignment f(x) 7→ f(A) defines a ring homomorphism
Eₐ: F[x] → ???
Capital subscript A
Eₐ: F[x] → Mₙ(F)
For any (n × n)-matrix A, the assignment f(x) 7→ f(A) defines a ring homomorphism Eₐ: F[x] → Mₙ(F)
“For all A ∈ Mn(F), there exists a non-zero polynomial f(x) ∈ F[x] such that f(A) = 0” tells us what about the kernel?
The kernel is non-zero
For any (n × n)-matrix A, the assignment f(x) 7→ f(A) defines a ring homomorphism Eₐ: F[x] → Mₙ(F)
As F[x] is commutative so is the [ ]
As F[x] is commutative so is the Eₐ, that is f(A)g(A) = g(A)f(A) for all polynomials f and g.
What is the minimal polynomial of A?
The minimal polynomial of A, denoted by mₐ(x), is the monic polynomial
p(x) of least degree such that p(A) = 0.
should be a capital subscript A
Thm: If f(A) = 0 then mₐ divides ... Furthermore mₐ is [ ] (hence showing that mₐ is well-defined)
If f(A) = 0 then mₐ|f Furthermore mₐ is unique hence showing that mₐ is well-defined)
If f(A) = 0 then mₐ|f Furthermore mₐ is unique hence showing that mₐ is well-defined)
Prove it
By the division algorithm, , there exist polynomials q, r with deg r < deg mₐ such that
f = qmA + r.
Evaluating both sides at A gives r(A) = 0. By the minimality property of mₐ,
r = 0 and mₐ divides f.
To show uniqueness, let m be another monic polynomial of minimal degree and m(A) = 0. Then by the above mₐ|m. Also m and mₐ must have the same degree, and so
m = amₐ for some a ∈ F. Since both polynomials are monic it follows that a = 1 and m = mₐ
Define the characteristic polynomial of A
The characteristic polynomial of A is defined as
χA(x) = det(A − xI).
χA(x) = (-1)ⁿxⁿ + …..????
χA(x) = (-1)ⁿxⁿ + (-1)ⁿ⁻¹tr(A)xⁿ⁻¹ + … + det(A)
Proof see lin alg 2 (prelims)
λ is an eigenvalue of A
⇔ ?? (χA(x))
⇔ ???(mₐ(x))
λ is an eigenvalue of A
⇔ λ is a root of χA(x)
⇔ λ is a root of mₐ(x)
λ is an eigenvalue of A
⇔ λ is a root of χA(x)
⇔ λ is a root of mₐ(x)
Prove it
χA(λ) = 0 ⇔ det(A − λI) = 0 ⇔ A − λI is singular ⇔ ∃ v ≠ 0 : (A − λI)v = 0 ⇔ ∃ v ≠ 0 : Av = λv ⇒ mₐ(λ)v = mₐ(A)v = 0 ⇒ mₐ(λ) = 0 (as v ≠ 0)
Conversely, assume λ is a root of mₐ. Then mₐ(x) = g(x)(x − λ) for some polynomial g. By minimality of mₐ, we have g(A) ≠ 0. Hence there exists w ∈ Fⁿ such that g(A)w ≠ 0. Put
v = g(A)w then
(A − λI)v = mₐ(A)w = 0,
and v is a λ-eigenvector for A.
Let C, P, A be (n × n)-matrices such that C = P⁻¹AP. Then m𝒸(x) = mₐ(x) for:
f(C) = f(P⁻¹AP) = P⁻¹f(A)P
for all polynomials f. Thus
0 = m𝒸(C) = [ ] and so m𝒸(A) = 0, and mₐ|m𝒸. Likewise m𝒸|mₐ and therefore mₐ = [ ] as both are monic.
Let C, P, A be (n × n)-matrices such that C = P⁻¹AP. Then m𝒸(x) = mₐ(x) for:
f(C) = f(P⁻¹AP) = P⁻¹f(A)P
for all polynomials f. Thus
0 = m𝒸(C) = [P⁻¹m𝒸(A)P] and so m𝒸(A) = 0, and mₐ|m𝒸. Likewise m𝒸|mₐ and therefore mₐ = [m𝒸] as both are monic.
Let V be a finite dimensional vector space and T : V → V a linear transformation. Define the minimal polynomial
Define the minimal polynomial of T as
mₜ(x) = mₐ(x)
where A = ᵦ[T]ᵦ with respect to some basis B of V . As mₐ(x) = mP⁻¹AP (x) the definition of
mₜ(x) is independent of the choice of basis.
For a linear transformation T : V → V define its characteristic polynomial
define its characteristic polynomial
as χT (x) = χA(x)
where A = ᵦ[T]ᵦ with respect to some basis B of V . As χA(x) = χP ⁻¹AP (x) the definition of χT (x) is independent of the choice of basis.
What does it mean for a field to be algebraically closed?
A field F is algebraically closed if every non-constant polynomial in F[x] has a root in F.
What is the fundamental theorem of algebra
The field of complex numbers C is algebraically closed.
What is an algebraic closure of F?
An algebraically closed field F¯ containing F with the property that there does not
exist a smaller algebraically closed field L with
F¯ ⊇ L ⊇ F
is called an algebraic closure of F
F¯ is F bar (all fancy Fs)
Every field has an algebraic [ ]
Every field F has an algebraic closure F¯.
Let V be a vector space over a field F and let U be a subspace.
What is the quotient space?
The set of cosets
V /U = {v + U | v ∈ V }
with the operations
(v + U) + (w + U) := v + w + U
a(v + U) := av + U
for v, w ∈ V and a ∈ F is a vector space, called the quotient spaceThe set of cosets
V /U = {v + U | v ∈ V }
with the operations
(v + U) + (w + U) := v + w + U
a(v + U) := av + U
for v, w ∈ V and a ∈ F is a vector space, called the quotient space
ProveWe need to check that the operations are well-defined. Assume v + U = v' + U and w + U = w' + U. Then v = v' + u, w = w'+ ˜u for u, u˜ ∈ U. Hence: (v + U) + (w + U) = v + w + U = v' + u + w'+ ˜u + U as u + ˜u ∈ U = v'+ w' + U = (v' + U) + (w' + U). Similarly, a(v + U) = av + U = av′ + au + U as au ∈ U = av′ + U = a(v' + U) That these operations satisfy the vector space axioms follows immediately from the fact that the operations in V satisfy them.
