Linear Algebra Flashcards

1
Q

Define a field

A

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.

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2
Q

Define the characteristic of F

A

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.

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3
Q

Define a vector space V over a field F in terms of groups

A

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

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4
Q

Let V be a vector space over F

Define a set S ⊆ V being linearly independent

A

(1) A set S ⊆ V is linearly independent if whenever a1, · · · , an ∈ F, and
s1, · · · , sn ∈ S,
a1s1 + · · · + ansn = 0 ⇒ a1 = · · · = an = 0.

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5
Q

Let V be a vector space over F

Define what it means for a set S ⊆ V to be spanning

A

(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

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6
Q

Let V be a vector space over F

Define what it means for a set S ⊆ V to be a basis of V

A

(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

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7
Q

Define a linear map/transformation

A

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’)

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8
Q

What is a bijective linear map called?

A

an isomorphism of vector spaces.

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9
Q

What is the assignment T → ᵦ’[T]ᵦ ?

Meant to be fancy B’ subscript

A

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.

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10
Q

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

A

ᵦ’[T]ᵦ’ = ᵦ’[Id]ᵦ ᵦ[T]ᵦ ᵦ[Id]ᵦ’ with ᵦ’[Id]ᵦ ᵦ[Id]ᵦ’ = I the identity matrix

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11
Q

Define a ring

A

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.

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12
Q

Define a commutative ring

A

The ring R is called commutative if for all a, b ∈ R we have ab = ba.

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13
Q

Define a ring homomorphism

A

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’).

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14
Q

Define a ring isomorphism

A

A bijective ring homomorphism is called a ring isomorphism.

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15
Q

Define an ideal

A

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.

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16
Q

What is the first isomorphism theorem? (rings)

A

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)

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17
Q

Prove the first isomorphism theorem (rings)

A

Exercise

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18
Q

What is the “division algorithm” for polynomials?

A
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).
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19
Q

Prove the “division algorithm” for polynomials

A

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)

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20
Q

For all f(x) ∈ F[x] and a ∈ F,

f(a) = 0 ⇒ ???

A

For all f(x) ∈ F[x] and a ∈ F,

f(a) = 0 ⇒ (x − a)|f(x).

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21
Q

For all f(x) ∈ F[x] and a ∈ F,
f(a) = 0 ⇒ (x − a)|f(x).
Prove it

A

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.

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22
Q

Assume f ≠ 0. If deg f ≤ n then f has [ ] roots

A

Assume f ≠ 0. If deg f ≤ n then f has at most n roots.

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23
Q

Assume f ≠ 0. If deg f ≤ n then f has at most n roots.

Prove it

A

Follows from
For all f(x) ∈ F[x] and a ∈ F,
f(a) = 0 ⇒ (x − a)|f(x).
and induction

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24
Q

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) =

A

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)

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25
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′
26
Let A ∈ Mn(F) and f(x) = aₖxᵏ + · · · + a₀ ∈ F[x]. Then | f(A) := [ ]
aₖAᵏ + · · · + a₀I ∈ Mn(F).
27
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 ```
28
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
29
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 polynomial ```
30
For 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)
31
``` 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
32
``` 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.
33
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
34
``` 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) ```
35
``` 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
36
Define the characteristic polynomial of A
The characteristic polynomial of A is defined as | χA(x) = det(A − xI).
37
χA(x) = (-1)ⁿxⁿ + .....????
χA(x) = (-1)ⁿxⁿ + (-1)ⁿ⁻¹tr(A)xⁿ⁻¹ + ... + det(A) | Proof see lin alg 2 (prelims)
38
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)
39
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.
40
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.
41
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.
42
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.
43
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.
44
What is the fundamental theorem of algebra
The field of complex numbers C is algebraically closed.
45
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)
46
Every field has an algebraic [ ]
Every field F has an algebraic closure F¯.
47
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 space ```
48
``` 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 space Prove ```
``` We 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. ```
49
``` Let E be a basis of U, and extend E to a basis B of V (we assume this is possible, which we certainly know to be the case at least for V finite dimensional). Define B¯ := {e + U | e ∈ B\E} ⊆ V /U. Fancy B bar What is the set B¯ a basis of? ```
V/U
50
``` Let E be a basis of U, and extend E to a basis B of V (we assume this is possible, which we certainly know to be the case at least for V finite dimensional). Define B¯ := {e + U | e ∈ B\E} ⊆ V /U. Fancy B bar The set B¯ is a basis of V/U Prove it ```
pg 12
51
Let U ⊂ V be vector 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
52
Let U ⊂ V be vector 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 Prove it
Exercise
53
If V is finite dimensional then | dim(V ) = dim(U) + [ ]
dim(V ) = dim(U) + dim(V /U).
54
Let T : V → W be a linear map of vector spaces over F. Then.... What is the first isomorphism theorem?
Then T¯ : V /Ker(T) → Im(T) v + Ker(T) → T(v) is an isomorphism of vector spaces.
55
``` Let T : V → W be a linear map of vector spaces over F. Then T¯ : V /Ker(T) → Im(T) v + Ker(T) → T(v) is an isomorphism of vector spaces. ``` Prove it
Proof. It follows from the first isomorphism theorem for groups that T¯ is an isomorphism of (abelian) groups. T¯ is also compatible with scalar multiplication. Thus T¯ is a linear isomorphism.
56
If T : V → W is a linear transformation and V is finite dimensional, then (Rank-nullity theorem)
dim(V ) = dim(Ker(T)) + dim(Im(T)).
57
If T : V → W is a linear transformation and V is finite dimensional, then dim(V ) = dim(Ker(T)) + dim(Im(T)). prove it
Use dim(V ) = dim(U) + dim(V /U), with U = ker(T). Then dim(V ) = dim(Ker(T)) + dim(V /Ker(T)). By the First Isomorphism Theorem also: dim(V /Ker(T)) = dim(Im(T)).
58
Let T : V → W be a linear map and let A ⊆ V, B ⊆ W be subspaces The formula T¯(v + A) := T(v) + B gives a well-defined linear map of quotients T¯ : V /A → W/B if and only if [ ]
T(A) ⊆ B.
59
Let T : V → W be a linear map and let A ⊆ V, B ⊆ W be subspaces The formula T¯(v + A) := T(v) + B gives a well-defined linear map of quotients T¯ : V /A → W/B if and only if T(A) ⊆ B. Prove it
Assume T(A) ⊆ B. Now T¯ will be linear if it is well-defined. Assume v + A = v′ + A. Then v = v′ + a for some a ∈ A. So T¯(v + A) = T(v) + B by definition = T(v′ + a) + B = T(v′) + T(a) + B as T is linear = T(v') + B as T(A) ⊆ B = T¯(v' + A). Hence T¯ is well-defined. Conversely, assume that T¯ is well-defined and let a ∈ A. Then B = 0𝓌/ᵦ= T¯(0ᵥ/ₐ) = T¯(A) = T¯(a + A) = T(a) + B. Thus T(a) ∈ B, and so T(A) ⊆ B.
60
Assume now that V and W are finite dimensional. Let B = {e₁, · · · , eₙ} be a basis for V with E = {e₁, · · · , eₖ} a basis for a subspace A ⊆ V (so k ≤ n). Let B′ = {e′₁, · · · , e′ₘ} be a basis for W with E′ = {e′₁, · · · , e′ℓ} a basis for a subspace B ⊆ W. The induced bases for V /A and W/B are given by B¯ = B¯' =
``` B¯ = eₖ₊₁ + A, · · · , eₙ + A and B¯′ = e'ₗ₊₁ + B, · · · , e′ₘ + B ```
61
Let T : V → W be a linear map such that T(A) ⊆ B. Then T induces a map T¯ on quotients by Lemma 3.7 and restricts to a linear map T|ₐ : A → B with T|ₐ(v) =[ ]
T(v) for v ∈ A.
62
What is the block matrix decomposition for ᵦ'[T]ᵦ?
Top left: ɛ'[T|ₐ]ɛ Top right: * Bottom left: 0 Bottom right: ᵦ¯'[T¯]ᵦ¯ where ᵦ¯'[T¯]ᵦ¯ = (aᵢⱼ) for l+1≤i≤m, k+1≤j≤n
63
Prove the block matrix decomposition for ᵦ'[T]ᵦ
For j ≤ k, T(eⱼ) ∈ B and hence aᵢⱼ = 0 for i > ℓ and aij is equal to the (i, j)-entry of ɛ′ [T|ₐ]ɛ for i ≤ ℓ. To identify the bottom right corner of the matrix, note that T¯(eⱼ + A) = T(eⱼ) + B = a₁ⱼe'₁ + ... + aₘⱼe'ₘ + B = aₗ₊₁,ⱼ(e'ₗ₊₁ + B) + ... + aₘⱼ(e'ₘ + B)
64
Let T : V → V be a linear transformation. | What does it mean for a subspace to be T-invariant?
A subspace U ⊆ V is called T-invariant if T(U) ⊆ U. | By the result of the previous section, such a T induces a map T¯ : V /U → V /U
65
``` Let T : V → V be a linear transformation. Let S : V → V be another linear map. If U is T- and S-invariant, then U is also invariant under the following maps: 1. the zero map 2. [ ] 3. aT, ∀a ∈ F 4. [ ] 5. [ ] ```
1. the zero map 2. the identity map 3. aT, ∀a ∈ F 4. S + T 5. S ◦ T
66
Let T : V → V be a linear transformation. Let S : V → V be another linear map. If U is T- and S-invariant, the U is invariant under any polynomial p(x) evaluated at [ ]. p(T) indices a map of quotients [ ]
evaluated at T p(T)¯ : V /U → V /U the whole p(T) is barred
67
Let T : V → V be a linear transformation and assume U ⊆ V is T-invariant. Then χT (x) = Note that this formula does not hold for the minimal polynomial
χT (x) = χT|U (x) × χT⁻(x) Note that this formula does not hold for the minimal polynomial
68
Let T : V → V be a linear transformation and assume U ⊆ V is T-invariant. Then χT (x) = χT|U (x) × χT⁻(x) Prove it
Middle pg 16 Extend a basis E for U to a basis B of V
69
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 ᵦ[T]ᵦ is [ ]
Upper triangular
70
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 ᵦ[T]ᵦ is Upper triangular Prove it
By induction on the dimension of V | End pg 16
71
If A is an n×n matrix with a characteristic polynomial that is a product of linear factors, then there exists an (n × n)-matrix P such that P⁻¹AP is [ ]
upper triangular.
72
Let A be an upper triangular (n × n)-matrix with diagonal entries λ₁, . . . , λₙ. Then ⁿ∏ᵢ₌₁ ( A - λᵢI) = ??
ⁿ∏ᵢ₌₁ ( A - λᵢI) = 0
73
Let A be an upper triangular (n × n)-matrix with diagonal entries λ₁, . . . , λₙ. Then ⁿ∏ᵢ₌₁ ( A - λᵢI) = 0 Prove it
``` Let e₁, . . . , eₙbe the standard basis vectors for F n. Then (A − λnI)v ∈ for all v ∈ Fⁿ and more generally (A − λᵢI)w ∈ for all w ∈ . Hence, since Im(A − λₙI) ⊆ Im(A − λₙ₋₁I)(A − λnI) ⊆ and so on, we have that ⁿ∏ᵢ₌₁ ( A - λᵢI) = 0 as required. ```
74
What is the Cayley-Hamilton theorem?
If T : V → V is a linear transformation and V is a finite dimensional vector space, then χT (T) = 0. Hence, in particular, mT (x) | χT (x).
75
Prove the Cayley-Hamilton theorem
Bottom pg 19
76
V is a vector space | What does it mean for V to be the direct sum of subspaces W1, ..., Wr
V = W1 ⊕ ... ⊕ Wr if every vector v ∈ V can be written uniquely as a sum v = w1 + · · · + wr with wi ∈ W
77
If V is the direct sum of the subspaces W1, ... Wr. Describe the basis of V in terms of the bases of Wi/
For each i, let Bi be a basis for Wi. Then B = ∪ᵢBᵢ is a basis for V.
78
If V is the direct sum of the subspaces W1, ... Wr. Assume from now on that V is finite dimensional. If T : V → V is a linear map such that each Wi is T-invariant, then the matrix of T with respect to the basis B is block diagonal .... (What does ᵦ[T]ᵦ look like?) What is the relationship between Xₜ(x) in terms of the characteristic polynomial of the subspaces?
pg 21 top
79
``` Assume f(x) = a(x)b(x) with gcd(a, b) = 1 and f(T) = 0. Then V = Ker(a(T)) ⊕ [ ] is a T-invariant direct sum decomposition ```
V = Ker(a(T)) ⊕ Ker(b(T))
80
Assume f(x) = a(x)b(x) with gcd(a, b) = 1 and f(T) = 0. Then V = Ker(a(T)) ⊕ Ker(b(T)) is a T-invariant direct sum decomposition. Furthermore, if f = mT is the minimal polynomial of T and a and b are monic, then mₜ|ₖₑᵣ₍ₐ₍ₜ₎₎ (x) = mₜ|ₖₑᵣ₍ᵦ₍ₜ₎₎ (x) =
mₜ|ₖₑᵣ₍ₐ₍ₜ₎₎ (x) = a(x) | mₜ|ₖₑᵣ₍ᵦ₍ₜ₎₎ (x) = b(x)
81
Assume f(x) = a(x)b(x) with gcd(a, b) = 1 and f(T) = 0. Then V = Ker(a(T)) ⊕ Ker(b(T)) is a T-invariant direct sum decomposition. Furthermore, if f = mT is the minimal polynomial of T and a and b are monic, then mₜ|ₖₑᵣ₍ₐ₍ₜ₎₎ (x) = a(x) mₜ|ₖₑᵣ₍ᵦ₍ₜ₎₎ (x) = b(x) Prove it
pg 21
82
What is the Primary Decomposition Theorem?
Let mT be the minimal polynomial and write it in the form mT (x) = fᵃ¹₁(x)· · · fᵃʳᵣ(x) where the fᵢ are distinct monic irreducible polynomials. Put Wi := Ker(fᵃᶦᵢ(T)). Then 1) ) V = W1 ⊕ · · · ⊕ Wr 2) Wi is T-invariant 3) mₜ|𝓌ᵢ = fᵃᶦᵢ
83
Prove the primary decomposition theorem
pg 22
84
There exists unique distinct irreducible monic polynomials f1, ... fr ∈ F[x] and positive integers ni ≥ ai > 0 (1 ≤ i ≤ r) such that mT(x) = and XT =
``` mT(x) = fᵃ¹₁(x)· · · fᵃʳᵣ(x) XT = ± fⁿ¹₁ ... fⁿʳᵣ ```
85
There exists unique distinct irreducible monic polynomials f1, ... fr ∈ F[x] and positive integers ni ≥ ai > 0 (1 ≤ i ≤ r) such that mT(x) = fᵃ¹₁(x)· · · fᵃʳᵣ(x) and XT = ± fⁿ¹₁ ... fⁿʳᵣ Prove it
Mid pg 22
86
T is triangularisable (over a given field) ⇐⇒ χT [ ] ⇐⇒ [ ] ⇐⇒ mT [ ]
⇐⇒ χT factors as a product of linear polynomials (over that field) ⇐⇒ each fi is linear ⇐⇒ mT factors as a product of linear polynomials
87
T is diagonalisable ⇐⇒ mT [....]
mT factors as a product of distinct linear polynomials
88
T is diagonalisable ⇐⇒ mT factors as a product of distinct linear polynomials. Prove it
Bottom pg 22
89
Let V be finite dimensional and T : V → V be a linear transformation What does nilpotent mean?
If Tⁿ = 0 for some n > 0 | then T is called nilpotent.
90
If T is nilpotent, then its minimal polynomial has the form mT (x) = [ ]
If T is nilpotent, then its minimal polynomial has the form mT (x) = xᵐ for some m
91
If T is nilpotent, then its minimal polynomial has the form mT (x) = xᵐ for some m and there exists a basis B of V such that: ᵦ[T]ᵦ =
pg 24 top
92
If T is nilpotent, then its minimal polynomial has the form mT (x) = xᵐ for some m and there exists a basis B of V such that: ᵦ[T]ᵦ = (matrix, mostly 0s, one diag of 1s and 0s) pg 24 Prove it
Pg 24-26: | Very long
93
Let V be finite dimensional and T : V → V be a linear transformation. Assume mT (x) = (x−λ)ᵐ for some m. Then, there exists a basis B of V such that ᵦ[T]ᵦ is block diagonal with blocks of the form:
Jᵢ(λ) := λIᵢ + Jᵢ = [matrix, 0 bottom right, lambdas down the diagonal, 1s down the next diag, 0 top right] and 1≤i≤m
94
Let V be finite dimensional and T : V → V be a linear transformation. Assume mT (x) = (x−λ)ᵐ for some m. Then, there exists a basis B of V such that ᵦ[T]ᵦ is block diagonal with blocks of the form: Jᵢ(λ) := λIᵢ + Jᵢ = [matrix, 0 bottom right, lambdas down the diagonal, 1s down the next diag, 0 top right] and 1≤i≤m Prove it
pg 27 middle
95
Let V be finite dimensional and let T : V → V be a linear map with minimal polynomial mT (x) = (x − λ1)ᵐ¹· · ·(x − λr)ᵐʳ Then there exists a basis B of V such that ᵦ[T]ᵦ is a [ ] and each diagonal block is of the form [ ]
block diagonal | Ji(λj ) for some 1 ≤ i ≤ mj and 1 ≤ j ≤ r.
96
Let V be a vector space over F. What is a dual?
Its dual V' is the vector space of linear maps | from V to F, i.e V' = Hom(V, F).
97
Let V be a vector space over F. What is a linear functional?
Its dual V' is the vector space of linear maps | from V to F, i.e V' = Hom(V, F). Its elements are called linear functionals
98
``` Let V be finite dimensional and let B = {e1, . . . , en} be a basis for V . Define the dual e'i of ei (relative to B) by e'i(ej) = δij. What is the dual basis? Describe the assignment ei → e'i dim V = ? ```
Then B′ := {e'1, . . . , e′n} is a basis for V', the dual basis. In particular, the assignment ei → e'i defines an isomorphism of vector spaces. In particular, dim V = dim V'
99
Let V be finite dimensional and let B = {e1, . . . , en} be a basis for V . Define the dual e'i of ei (relative to B) by e'i(ej) = δij. Then B′ := {e'1, . . . , e′n} is a basis for V', the dual basis. In particular, the assignment ei → e'i defines an isomorphism of vector spaces. In particular, dim V = dim V' Prove it
Bottom pg 29
100
Let V be a finite dimensional vector space. Then, V → (V')' =: V'' defined by v → Eᵥ is a natural linear [ ] How is Eᵥ defined? What does natural mean?
isomorphism Ev(f) := f(v) for f ∈ V' “Natural” here means independent of a choice of basis
101
Let V be a finite dimensional vector space. Then, V → (V')' =: V'' defined by v → Eᵥ is a natural linear isomorphism. Ev(f) := f(v) for f ∈ V' Prove it
pg 30
102
When V has dimension n, the kernel of a non-zero linear functional f : V → F is of dimension n − 1. The preimage f⁻¹({c}) for a constant c ∈ F is a called [ ] (not necessarily containing zero) of dimension n − 1.
hyperplane
103
When V has dimension n, the kernel of a non-zero linear functional f : V → F is of dimension n − 1. The preimage f⁻¹({c}) for a constant c ∈ F is a called hyperplane (not necessarily containing zero) of dimension n − 1. When V = Fⁿ (column vectors) every hyperplane is defined by an equation:
a1b1 + · · · + anbn = c | for a fixed scalar c and fixed b = (b1, . . . , bn) ∈ (Fⁿ)ᵗ(row vectors)
104
Let U ⊆ V be a subspace of V | What is an annihilator of U?
Define the annihilator of U to be: | U⁰ = {f ∈ V': f(u) = 0 for all u ∈ U}.
105
Annihilators: | f ∈ V' iff [ ]
f |ᵤ = 0
106
Annihilators: | [ ] is a subspace of V
U⁰
107
Prove that U⁰ is a subspace of V'
Top pg 31
108
Let V be finite dimensional and U ⊆ V be a subspace. Then dim(U⁰) = [ ]
dim(V ) − dim(U)
109
Let V be finite dimensional and U ⊆ V be a subspace. Then dim(U⁰) = dim(V ) − dim(U) Prove it
mid pg 31
110
Let U, W be subspaces of V . Then 1) U ⊆ W ⇒ [ ] 2) (U + W)⁰= [ ] 3) U⁰ + W⁰ ⊆ ([ ])⁰ and equal if [ ]
1) U ⊆ W ⇒ W⁰ ⊆ U⁰ 2) (U + W)⁰= U⁰ ∩ W⁰ 3) U⁰ + W⁰ ⊆ (U ∩ W)⁰ and equal if dim(V ) is finite.
111
Let U, W be subspaces of V . Then 1) U ⊆ W ⇒ W⁰ ⊆ U⁰ 2) (U + W)⁰= U⁰ ∩ W⁰ 3) U⁰ + W⁰ ⊆ (U ∩ W)⁰ and equal if dim(V ) is finite. Prove it
pg 32
112
Let U be a subspace of a finite dimensional vector space V . Under the natural map V → V''(:= (V')') given by v → Eᵥ, how is U mapped?
It is mapped isomorphically to | U⁰⁰ (:= (U⁰)⁰)
113
Let U be a subspace of a finite dimensional vector space V . Under the natural map V → V''(:= (V')') given by v → Eᵥ, U is mapped isomorphically to U⁰⁰ (:= (U⁰)⁰) Prove it
Bottom pg 32
114
Let U ⊆ V be a subspace. Then there exists a natural isomorphism U⁰ ≃ (V/U)' given by f → f¯ where f¯(v + U) :=
f¯(v + U) := f(v) for v ∈ V
115
Let U ⊆ V be a subspace. Then there exists a natural isomorphism U⁰ ≃ (V/U)' given by f → f¯ where f¯(v + U) := f(v) for v ∈ V Prove it
pg 33
116
What is a dual map?
Let T : V → W be a linear map of vector spaces. Define the dual map by T': W′ → V', f → f ◦ T Note that f ◦ T : V → W → F is linear, and hence f ◦ T ∈ V'
117
The dual map T' is a [ ] map
linear
118
Prove that T' is a linear map
``` Let f, g ∈ W′, λ ∈ F. We need to show T'(f + λg) = T'(f) + λT′(g) (an identity of functionals on V ). So let v ∈ V . Then, T'(f + λg)(v) = ((f + λg) ◦ T)(v) = (f + λg)(T v) = f(T v) + λg(T v) = T'(f)(v) + λT′(g)(v) = (T'(f) + λT′(g))(v), as required. ```
119
Let V and W be two finite dimensional vector spaces. The assignment T → T' defines a natural isomorphism from [ ] to [ ]
hom(V, W) | hom(W', V')
120
Let V and W be two finite dimensional vector spaces. The assignment T → T' defines a natural isomorphism from hom(V, W) to hom(W', V') Prove it
pg 34
121
Let V and W be finite dimensional, and let B𝓌 and Bᵥ be bases for W and V . Then, for any linear map T : V → W (ᵦ𝓌[T}ᵦᵥ)ᵗ =
(ᵦ𝓌[T}ᵦᵥ)ᵗ = ᵦ'ᵥ[T]ᵦ'𝓌 | where B'𝓌 and B'ᵥ are the dual bases
122
Let V and W be finite dimensional, and let B𝓌 and Bᵥ be bases for W and V . Then, for any linear map T : V → W (ᵦ𝓌[T}ᵦᵥ)ᵗ = (ᵦ𝓌[T}ᵦᵥ)ᵗ = ᵦ'ᵥ[T]ᵦ'𝓌 where B'𝓌 and B'ᵥ are the dual bases Prove it
end pg 34