About
Definition
Flashcards for M1P2 - Algebra I
Imperial College London
Subspace Sum & Intersection
Algebra I
Theorem
Algebra I
Definition
dim U + W = dimU + dimW − dim U ∩ W
Rank, Row-Rank, Column-Rank
Algebra I
Theorem
Algebra I
Definition
Row-Rank = Column-Rank
Linear Transformation
Algebra I
Proposition
Algebra I
Definition
dimW = dimV =⇒ ∃! T s.t. T (vi ) = wi
Kernels & Images
Algebra I
Proposition
Algebra I
Theorem
If {v1 , . . . , vn } is a basis for V ,
Im(T ) = Span{T v1 , . . . , T vn }
Algebra I
Rank Nullity Theorem
Algebra I
U + W = {u + w : u ∈ U, w ∈ W }
= Span{u1 , . . . , ur , w1 , . . . , ws }
These flashcards and the accompanying LATEX. For
brevity, often important detail is omitted from proofs
- you should make sure you can do them fully!
U ∩ W = {v : v ∈ U and v ∈ W }
P
P
Solve v = ar ui = bs ws to find U ∩ W .
Row-Rank is dim RSp(A). Column rank is dim
CSp(A). Rank is either since they’re equal.
RSp(A) = RSp(Aech )
bit.ly/imperialmaths
Proof. Let {v1 , . . . , vm } basis for U ∩ W , write U =
{v1 , .., vm , u1 , .., ur−m }, W = {v1 , .., vm , w1 , .., ws−m }.
Span B = {v1 , .., vm , u1 , w1 , .., ur−m , ws−m } = U +W .
X
X
X
X
γi w i = −
αi vi −
β i ui =
δi v i
{v1 , ..vm , w1 , ..ws−m } is a basis =⇒ γi = δi = 0
X
X
αi vi +
βi ui = 0
{v1 , ..vm , u1 , ..ur−m } is a basis =⇒ αi = βi = 0. So
B is a basis for U + W . Result follows.
”
T : V → W such that:
Proof. ri = (ai1 , ai2 , .., ain ), cj = (a1j , a2j , .., amj )T
RSp(A) has basis {v1 , . . . , vk }, ri = λ1i v1 + . . . λil vk .
If vi = (bi1 , . . . , bin ), then aij = λi1 b1j + · · · + λik bkj
a1j
1. T (v1 + v2 ) = T (v1 ) + T (v2 ) (preserves +)
cj =
2. T (λv) = λT (v) (preserves scalar ×)
Matrix Transformations are always linear.
amj
Ker(T ) = {v ∈ V : T (v) = 0}
λ11 b1j +···+λ1k bkj
!
=
!
..
.
λm1 b1j +···+λmk bkj
RSp(AT ) = CSp(A) ≤ k = RSp(A) = CSp(AT ).
Similarly RSp(A) ≤ CSp(A). So RSp(A) = CSp(A).
”
P
λi vi . Let T (v) = λi wi . Then ∀u, v ∈ V :
X
X
µi wi , T (v) =
λi wi
T (u) =
X
T (u + v) =
(µi + λi )wi , so T preserves + .
P
If π ∈ F . T (πv) = πλi vi = πT (v), so preserves ×.
If v =
Im(T ) = {T v : v ∈ V } (CSp(T ) if matrix)
..
.
P
For T : V → W : dim Im(T ) + dim Ker(T ) = dim(V )
Proof. If {u1 , . . . , us }, {w1 , . . . , wr } are bases for Ker
and Im, Let V = {v1 , . . . , vr } such that T (vi ) = wi .
Then I claim B = U ∪ V is basis for V ( =⇒ thm):
P
B
v ∈ V, Im 3 T v =
µi wi . Let
P Spans V : IfP
P v̄ =
µi vi ,PT v̄ = P
µi wi = T v. Ker 3 v − v̄ =
λi ui ,
so v =
µi vi + λP
i ui ∈ SpanB
P
Lin. Independent:
λi ui +
µi vi = 0. Apply T ,
P
µi wi = 0 =⇒ µi = 0. So λi = 0.
”
P
Proof. ∀w P
∈ Im(T ), ∃v ∈ V : T (v) = w. v =
λi vi .
w = Tv =
λi T vi ∈ Span{T v1 , . . . , T vn } ≥ Im(T ).
∀i, T vi ∈ Im(T ), so Span{T v1 , . . . , T vn } ≤ Im(T ).
”
Definition
Definition
Vector with respect to basis, B
Matrix with respect to basis, B
Algebra I
Proposition
Algebra I
Definition
Eigenvectors & Eigenvectors
[T v]B = [T ]B [v]B
Algebra I
Proposition
Algebra I
Proposition
1. Eigenvalues of T are same as [T ]B
2. Eigenvectors v of T are vectors v s.t. [v]B
eigenvector of [T ]B
[T ]B is diagonal iff every basis vector in B is
an eigenvector for T
Algebra I
Definition
Algebra I
Definition
Diagonalisable
Change of Basis Matrix, P
Algebra I
Proposition
Algebra I
Proposition
1. P is invertible, with P −1 change of basis
from C to B
1. P = [T ]B where T : T (vi ) = wi
2. ∀v ∈ V, P [v]C = [v]B
2. [T ]C = P −1 [T ]B P
Algebra I
Algebra I
T vi =
X
aij vi for basis B = {v1 , . . . , vn }
v=
Then
a11 a12 ...
..
.
[T ]B =
!
a1n
..
.
X
λi vi for basis B = {v1 , . . . , vn }
Then
[V ]B = (λ1 , . . . , λn )T
an1 an2 ... ann
So the ith column is [T vi ]B
P
Proof.P[T ]B = (aijP
), B P
= {v1 , . . . ,P
vn },
λi vi .
Plet v =
Tv =
λi T vi =
λi aji vj = ( λi aji )vj , so:
v is an eigenvector if v 6= 0 and T v = λv. Then λ is
the v 0 s eigenvalue.
P
[T v]B =
P
λi a1i
..
.
λi ani
a11 ... a1n
!
=
..
.
..
.
an1 ... ann
!
λ1
..
.
!
= [T ]B [v]B
λn
”
Proof.
Proof. For standard basis e1 , . . . , en , A is diagonal iff
ei ’s are eigenvectors of A. So [T ]B is diagonal iff all the
ei are eigenvectors. But ei = 0v1 +· · ·+1vi +· · ·+0vn =
[vi ]B . So ei eigenvector iff vi is eigenvector for T . ”
T v = λv ⇐⇒ [T v]B = [λv]B
⇐⇒ [T ]B [v]B = λ[v]B
⇐⇒ [V ]B is e-vector for [T ]B with e-value λ
”
If B = {v1 , . . . , vn }, C = {w1 , . . . , wn }, wj =
!
λ11 ... λ1n
..
..
PB to C =
= (λij )
.
.
P
λij vi
T diagonalisable iff ∃ basis of V such that every element of V is an eigenvector of T .
λn1 ... λnn
So jth column of P is [wj ]B
Proof. Let Q be change of basis from C to B. So
Q[v]B = [v]C .
P Qei = P Q[vi ]B = P [vi ]C = [vi ]B = ei =⇒ Q = P
−1
Also P −1 [T ]B P ei = P −1 [T ]B P [wi ]C − P −1 [T ]B [wi ]B
= P −1 [T wi ]B = [T wi ]C = [T ]C [wi ]C = [T ]C ei .
Thus P −1 [T ]B P = [T ]C
”
Proof. We know that [T ]B [vi ]B = [T vi ]B = [wi ]B .
Since P [vi ]B = P ei = [wi ]B , it follows P = [T ]B
P
P
If v ∈ V , P
v=
ai wi .P
[v]C = (a1 , . . P
. , an )T =
ai ei .
P [v]C =
ai P ei =
ai [wi ]B = [ ai wi ]B = [v]B
”
Definition
Proposition
1. G has exactly one identity.
2. Every element has exactly one inverse.
Group
3. Left Cancel.: x ∗ y = x ∗ z =⇒ y = z
4. Right Cancel.: x ∗ z = y ∗ z =⇒ x = y
Algebra I
Definition
Algebra I
Definition
Abelian Group
Permutation
Algebra I
Definition
Algebra I
Definition
Symmetric Group
Subgroup
Algebra I
Proposition
Algebra I
Proposition
Subgroup Criteria: H ≤ G iff
1. e ∈ H, where e is identity of G
2. If a, b ∈ H, then ab ∈ H
3. If a ∈ H, then a
−1
1. (g n )−1 = g −n
2. g m g n = g m+n
3. (g m )n = g mn
∈ H.
Algebra I
Algebra I
Definition
Definition
Cyclic Subgroup
Cyclic Group
Algebra I
Algebra I
Proof. If e, f identity, e ∗ f = f = e.
If y, z inverses for x, z = (y ∗ x) ∗ z = y ∗ (x ∗ z) = y.
Set G with binary operation ∗ (a function mapping
S × S → S) with
1. Associativity:(a ∗ b) ∗ c = a ∗ (b ∗ c)
If x ∗ y = x ∗ z, let w = x−1 , then
w ∗ (x ∗ y) = w ∗ (x ∗ z) =⇒ (w ∗ x) ∗ y = (w ∗ x) ∗ z
2. Identity: ∃e ∈ G s.t. x ∗ e = e ∗ x = x
3. Inverses: ∀x ∈ G, ∃y ∈ G s.t. x ∗ y = e
=⇒ y = z, similar for right cancellation
”
A bijective function f : X → X
A group G such that ab = ba (a and b commute)
A subset of G which is itself a group under ∗,
denoted H ≤ G.
Sym(X) is the set of all permutations of X.
Proof. Obvious.
G cyclic if ∃ a generator, g ∈ G s.t. hgi = G,
”
Proof. blah blah blah
hgi = {g n : n ∈ Z} ≤ G
”
Definition
Proposition
|hgi| = ord(g)
(ord = ∞ iff hgi is infinite)
Order of g
Algebra I
Algebra I
Proposition
Definition
Every permutation in Sn can be written as a
product of disjoint cycles.
Cycle Shape
Algebra I
Proposition
Algebra I
Proposition
If a, b commute, then
ord(g) = d then for all k ∈ Z,
g k = e ⇐⇒ d | k
1. a−1 b = ba−1
2. ai bj = bj ai
3. (ab)k = ak bk
Algebra I
Proposition
Algebra I
Definition
If f is a permutation with cycle shape
(r1 , . . . , rk ), then ord(f ) = lcm(r1 , . . . , rk )
Right Coset & Index
Algebra I
Theorem
Algebra I
Definition
Lagrange’s Theorem
Residue Class of a modulo m
Algebra I
Algebra I
Proof. If ord(g) = ∞. Suppose wlog n > m,
let k = n − m.
g n = g m+k = g m g k .
If
n
m
k
g = g
=⇒ g = e, contradicting ord = ∞,
so hgi is infinite.
If ord(g) = k, write n = pk + q, 0 ≤ q ≤ k − 1.
g n = g pk+q = (g k )p g q = g q ∈ {g 0 , . . . , g k−1 } = hgi.
If g i = g j , 0 ≤ i < j ≤ k − 1, let l = j − i. Then
g i = g i+l =⇒ g l = e, contradicting ord(g) = k
”
Sequence of cycle lengths in descending order written
in disjoint cycle notation, including 1-cycles.
”
Proof. Easy.
ord(g) = min k ∈ Z+ s.t. g k = e (if 6 ∃k, ord = ∞)
Proof. blah blah.
”
Proof. By Euclid’s Lemma, k = xd + y, 0 ≤ y < d.
g k = g xd+y = (g d )x g y = g y = e. y = 0 ⇐⇒ d | k. ”
Right coset is Hx = {hx : h ∈ H}
Proof. Easy.
Index is
|G|
|H| ,
”
number of right cosets of H in G.
[a]m = {s ∈ Z : s ≡ a (mod m)}
∀m ∈ N, [0]m ≤ (Z, +). Other classes are cosets.
For H ≤ G, |H| divides |G|
Proof. Consider S = {Hx : x ∈ G}, setPof cosets.
x ∈ G is in one member of S. Thus |G| = X∈S |X|.
Every coset has size |H|. So |G| = |S||H|.
”
Proposition
Proposition
(Z∗m , ×) is a group iff m is prime.
∗
= Zm \{[0]m })
(Zm
(Zm , +) is a group.
Algebra I
Algebra I
Theorem
Proposition
If 2m − 1 is prime, then m is prime.
Fermat’s Little Theorem
Algebra I
Proposition
Algebra I
Definition
Let n = 2p − 1. If q is a prime divisor of n,
then q ≡ 1 (mod p)
Symmetry & Dihedral Group
Algebra I
Definition
Algebra I
Definition
Ring
Polynomial Ring
Algebra I
Definition
Algebra I
Theorem
(R× , ×) is the unit group of R
Unit
Algebra I
Algebra I
”
Proof. Obvious.
Proof. Suppose m is not prime, so m = ab. Now
m
ab
a b
a
a(b−1)
2 −1 = 2 −1 = (2 ) −1 = (2 −1)(2
Proof. Obvious.
”
If p 6 | n, np−1 ≡ 1 (mod p)
a
+· · ·+2 +1) Proof. p 6 | n =⇒ [n] ∈ Z∗ . So ord[n] divides |Z∗ | =
p
p
p
p
p − 1 by Lagrange’s Theorem. Thus [n]pp−1 = [1]p ”
a
m
So 2 − 1 divides 2 − 1, so is not prime.
”
Symmetry group H = {A ∈ GLn (R) :
A preserves S}(Preserves iff AS = {Av : v ∈ S} = S)
D2n is the symmetry group of Pn , regular n-gon.
Proof. q | n =⇒ q | 2p − 1. So [2]pq = [1]q . So
ord([2]q ) | p, so it must be 1 or p. 2 6≡ 1 (mod q) so
[2]q 6= [1]q , so ord([2]q ) = p. So p divides |Z∗q | = q − 1
by Lagrange’s Theorem.
”
A set R with + and × such that
1. (R, +) is an abelian group
F [x] is set of polynomials in x with coefficients from
F , a field with usual + and ×.
2. × is associative
3. × distributes over +
Degree of f (x) is highest power of x in f (x)
4. ∃ identity for × (ring with unity)
5. × is commutative. (commutative ring)
Proof. Easy from the group axioms.
”
u, a multiplicatively invertible element of R,
i.e. ∃w ∈ R such that uw = wu = 1.
R× is the set of units in R.
Definition
Definition
Field
Zero Divisor & Integral Domain
Algebra I
Proposition
Algebra I
Defintion
If a | b and b | a then ∃u ∈ R such that
b = au
Irreducible
Algebra I
Defintion
Algebra I
Defintion
√
Ring Z[ d]
Norm Map
Algebra I
Proposition
Algebra I
Definition
r is a unit iff N (r) = ±1
Highest Common Factor
Algebra I
Proposition
Algebra I
Lemma
If c = hcf(a, b), then d= hcf(a, b) iff d = cu
Algebra I
∃q(x), r(x) ∈ F [x] s.t. f (x) = q(x)g(x) + r(x)
with 0 ≤ deg r < deg g
Algebra I
r is a zero divisor if ∃s 6= 0 ∈ R such that rs = 0.
A unit is not a zero-divisor.
F , a commutative ring such that F × = F ∗ = F \{0}
(Every non-zero element is a unit)
An integral domain is a ring with no zero-divisors
(except for 0). F ⊂ ID.
If r is a non-zero, non-unit in R, an ID, r is irreducible
if 6 ∃s, t ∈ R non-units such that r = st. Otherwise it
is reducible.
√
√
N : Z[ d] → Z by N (x + y d) = x2 − dy 2
Proof. Trivial.
”
√
√
Z[ d] = {x + y d : x, y ∈ Z}, usual + and ×.
c = hcf(a, b) iff
√
√
Proof. N : Z[ d] → Z by N (x + y d) = x2 − dy 2
1. c | a and c | b (so a common factor)
”
2. If d | a and d | b then d | c
Proof. Suppose that c adn d are hcfs, by definition
d = cu.
Proof. Do stuff.
”
Let d = cu. Then e | a and e | b =⇒ e | c. So
e | cu = d
”
Proposition
Definition
a = bq + r Then d= hcf(a, b) iff d = hcf(b, r)
Euclidean Function & Domain
Algebra I
Proposition
Algebra I
Lemma
√
f (a) = |N (a)| is Euclidean function for Z[ d]
if d ∈ {−2, −1, 2, 3}
Bezout’s Lemma
Algebra I
Definition
Algebra I
Definition
Coprime
Unique Factorisation Domain
Algebra I
Definition
Algebra I
Proposition
In a UFD, p is irreducible =⇒ p is prime
(non-unit such that p | ab =⇒ p | a or p | b)
Prime
Algebra I
Algebra I
Euclidean Function, f :
1. f (ab) ≥ f (a) for a, b ∈ R∗
2. For all a, b ∈ R, b 6= 0, ∃q, r s.t. a = qb + r with
0 ≤ f (r) < f (b).
Proof. Blah.
”
Proof. VERY LONG!
”
A Euclidean Domain is an ID with such a function.
∃s, t ∈ R s.t. as + bt = d
Proof. Like M1F.
”
An ID such that
1. a = p1 . . . ps
If 1 is a hcf for a and b in an ID.
2. Uniqueness of factorisation
Euclidean Domain ⊂ UFD (proof in M2P2)
Proof. Final proof.
”
p is prime iff p | ab =⇒ p | a or p | b