Number Theory
Denition: Leta, bbe integers, a6= 0. We say thata dividesb, (orb is a multiple ofa), if there is an integerkfor whichb=ka.
Notation: a|b.
Denition: The integer|p|>1 is a prime, ifp=ab, then either p=aor p=b(i.e. phas no proper divisors, only 1 and itself).
Proposition: For an integerp,pis a prime if and only if it has the following property: ifpdivides ab, then eitherpdividesaorpdividesb.
Theorem (Fundamental Theorem of Algebra): Every integer|n|>1 can be written as a product of primes in a unique way (up to the order and the signs of the primes).
Corollary: Every integer n >1 has a unique canonical form,n=pe11·pe22· · · · ·perr, where p1 < p2<
· · ·< prare primes, and ei>0 for alli= 1,2, . . . , r.
Proposition: Ifn=pe11·pe22· · · · ·perr andm=pf11·pf22· · · · ·pfrr withei, fi≥0, then 1. m|nif and only iffi≤ei for alli= 1,2, . . . , r,
2. g.c.d.(m, n) =pmin{e1 1,f1}·pmin{e2 2,f2}· · · · ·pmin{er r,fr}, 3. l.c.m.(m, n) =pmax{e1 1,f1}·pmax{e2 2,f2}· · · · ·pmax{er r,fr}. Denition: mandnare relatively prime, if g.c.d.(m, n) = 1.
Denition: For an integern >1,d(n)is the number of divisors of n. Proposition: 1. d(n)≥2; and d(n) = 2if and only if nis a prime.
2. Ifn=pe11·pe22· · · · ·perr thend(n) = (e1+ 1)(e2+ 1). . .(er+ 1).
3. Thed(n)function is multiplicative, i.e. if g.c.d.(m, n) = 1, thend(mn) =d(m)·d(n).
Denition: For an integer n >1, ϕ(n), the value of the Euler's ϕ function, is the number of positive integers less thannwhich are relatively prime ton.
Proposition: 1. ϕ(n)≤n−1; andϕ(n) =n−1 if and only ifnis a prime.
2. Ifn=pe11·pe22· · · · ·perr thenϕ(n) = (pe11−pe11−1)·(pe22−pe22−1)· · · · ·(perr−perr−1) =
=n·(1−p1
1)·(1−p1
2)· · · ·(1−p1
r).
3. Theϕ(n)function is multiplicative, i.e. if g.c.d.(m, n) = 1, thenϕ(mn) =ϕ(m)·ϕ(n).
Theorems about primes
Theorem (Euclid): There are innitely many primes.
Theorem: There are arbitrarily large gaps between consecutive primes.
Theorem (prime number theorem): ifπ(n)is the number of primes less thann, thenπ(n)∼n/lnn, i.e. π(n)/(n/lnn)→1, asn→ ∞.
Theorem (Dirichlet): If g.c.d.(a, b) = 1, then there are innitely many primes of the form ak+b, wherekis an integer.
Congruences
Denition: Form >1, anda, b∈Z we say thatais congruent tobmodulo m, ifm dividesa−b. Notation: a≡b (mod m). mis the modulus of the congruence.
Proposition: The congruence mod m is compatible with the usual operations on integers (addition, subtraction, multiplication, exponentiation), i.e. ifa≡b (modm)andc≡d (modm), then
a+c≡b+d (mod m), a−c≡b−d (mod m), ac≡bd (modm)
andak≡bk (mod m)for everyk∈N.
Proposition (cancellation in congruences): ac≡bc (mod m)if and only ifa≡b(mod g.c.d.(c,m)m ). Denition: A congruenceax≡b (mod m)with unkownxis called a linear congruence.
Theorem: 1. If g.c.d.(a, m)6 |b, then the linear congruenceax≡b (modm)has no solutions.
2. If g.c.d.(a, m)|b, then the linear congruenceax≡b (modm)has g.c.d.(a, m)solutions modm. Denition: x≡a1 (mod m1), x≡a2 (mod m2)is a simultaneous congruence system.
Theorem: The simultaneous congruence system x≡a1 (modm1),x≡a2 (mod m2) has a solution if and only if g.c.d.(m1, m2)|a1−a2, and in this case it has a unique solution mod l.c.m.(m1, m2). Denition: A reduced residue system modm is a set ofϕ(m)pairwise non-congruent integers mod m, each relatively prime tom, i.e. a set of integersa1, a2, . . . , ak, s.t.
1. g.c.d.(ai, m) = 1for alli= 1,2, . . . , k, 2. ai6≡aj (modm), ifi6=j, i, j= 1,2, . . . , k, 3. k=ϕ(m).
Lemma: If g.c.d.(a, m) = 1anda1, a2, . . . , aϕ(m)is a reduced residue system modm, thena1a, a2a, . . . , aϕ(ma is also a reduced residue system modm.
Theorem (Euler-Fermat): If g.c.d.(a, m) = 1thenaϕ(m≡1 (modm). Theorem (Fermat): 1. Ifp6 |athenap−1≡1 (modp).
2. ap≡a (mod p)for every integera.
Euclidean algorithm for determining the g.c.d. ofmandn,m > n: Letm=k1·n+r1, 0< r1< n,
n=k2·r1+r2, 0< r2< r1, r1=k3·r2+r3, 0< r3< r2, ...
rl−1=kl+1·rl+rl+1, 0< rl< rl+1, rl=kl+2·rl+1+ 0,
then g.c.d.(m, n) =rl+1.
Geometry of 3-space
Points and vectors in 3D have 3 coordinates.
Equation of a plane
A plane is determined by its normal vectorn= (A, B, C)and one of its pointsP0(x0, y0, z0). IfP(x, y, z)is a point on the plane, thenn·−−→
P0P = 0.
The equation of the plane (with coordinates): A(x−x0)+B(y−y0)+C(z−z0) = 0, orAx+By+Cz=D. System of equations of a line
A line is determined by its direction vectorv= (a, b, c)and one of its points P0(x0, y0, z0). IfP(x, y, z)is a point on the line, then −−→
P0P =t·vfor some t∈R.
The system of equations of the line (with coordinates): x−x0=ta, y−y0=tb, z−z0=tc, t∈R, or
x−x0
a = y−yb 0 = z−zc 0, if none of a, b, cis 0. Ifa= 0, thenx=x0, y−yb 0 = z−zc0, and ifa=b= 0, then x=x0, y=y0, z∈R.
The vector space R
nDenition 1: Rn is the set of all column vectors withnreal numbers (ncoordinates).
The operations onRnare the (coordinatewise) addition of vectors and the (coordinatewise) multiplication by a scalar(=real number).
Proposition: Ifu, v, w∈Rn, λ, µ∈R, then (1)u+v=v+u(addition is commutative),
(2)(u+v) +w=u+ (v+w)(addition is associative), (3)λ(u+v) =λu+λv,
(4)(λ+µ)u=λu+µu, (5)(λµ)u=λ(µu).
Denition 2: W ⊆Rn, W 6=∅is a subspace ofRn, if it satises (1) ifu, v∈W thenu+v∈W (i.e. W is closed under addition), and
(2) ifu∈W, λ∈R, thenλu∈W (i.e. W is closed under multiplication by a scalar).
Denition 3: Letu1, . . . , uk ∈Rn. Then for someλ1, . . . , λk∈R, the vectorv=λ1u1+. . . , λkuk is a linear combination ofu1, . . . , uk.
(In other words,v can be expressed usingu1, . . . , uk.)
Denition 4: Letu1, . . . , uk ∈Rn. ThenW, which is the set of all the linear combinations ofu1, . . . , uk is the subspace spanned (or generated) byu1, . . . , uk.
We say thatu1, . . . , uk is a generating system for the subspaceW. Notation: W =hu1, . . . , uki
Proposition: W above is really a subspace of Rn (according to Denition 2).
Denition 5: Letu1, . . . , uk ∈Rn. u1, . . . , uk are linearly independent, if none of u1, . . . , uk is a linear combination of the remaining vectors.
u1, . . . , uk are linearly dependent, if they are not linearly independent, i.e. at least one of the vectors u1, . . . , uk is a linear combination of the remaining vectors.
Denition 6: u1, . . . , uk are linearly independent, if the zero vector can be expressed using them only in the trivial way (i.e. when all the coecients are 0).
Proposition: Denitions 5 and 6 are equivalent.
Lemma: If the vectorsu1, . . . , uk are linearly independent, butu1, . . . , uk, uk+1 are linearly dependent, thenuk+1∈ hu1, . . . , uki.
Lemma: (Exchange theorem) Let W ⊆Rn be a subspace, u1, . . . , uk ∈W linearly independent, and v1, . . . , vm∈W a generating system ofW. Then for each1≤i≤k there exists a1≤j≤m such that u1, . . . , ui−1, vj, ui+1, . . . , uk∈W are also linearly independent.
Theorem: (I-G inequality) Let W ⊆ Rn be a subspace, u1, . . . , uk ∈ W linearly independent, and v1, . . . , vm∈W a generating system ofW. Thenk≤m.
Denition 7: LetW ⊆Rn be a subspace. B={b1, . . . , bk}is a basis inW, if it is linearly independent and a generating system inW.
Theorem: LetW ⊆Rn be a subspace. Ifb1, . . . , bk andc1, . . . , cm are both bases inW, thenk=m. Denition 8: LetW ⊆Rn be a subspace. Ifb1, . . . , bk is a basis inW, then the dimension ofW isk. Notation: dim(W) =k.
Remark: The dimension ofW is well-dened because of the previous theorem.
Proposition: {u1 = (1,0, . . . ,0)T, u2 = (0,1, . . . ,0)T, . . . , un = (0,0, . . . ,1)T} is a basis in Rn, the standard basis. Thereforedim(Rn) =n.
Theorem: B ={b1, . . . , bk} is a basis in W if and only if each vector in W is a linear combination of b1, . . . , bk in a unique way.
Denition: The coordinate vector ofv∈W in a given basisB={b1, . . . , bk}in W is(λ1, λ2, . . . , λk)T, ifv=λ1b1+. . . , λkbk.
Notation: [v]B= (λ1, λ2, . . . , λk)T.
Theorem: Let W ⊆ Rn be a subspace. If u1, . . . , uk are linearly independent vectors in W, then u1, . . . , uk can be extended to a basis inW (with nitely many vectors, maybe 0).
Corollary 1: Every subspace ofRn has a basis (and a dimension).
Corollary 2: LetW ⊆Rnbe a subspace of dimensionk. Ifu1, . . . , ukareklinearly independent vectors inW then they are a basis inW.
Corollary 3: IfV ⊂W, V 6=W are subspaces inRn, thendim(V)<dim(W).
Proposition: Let W ⊆ Rn be a subspace. If u1, . . . , uk is a generating system inW, then there is a subset ofu1, . . . , uk which is a basis inW.
Corollary: Let W ⊆ Rn be a subspace of dimension k. If u1, . . . , uk is a generating system in W consisting ofkvectors then they are a basis inW.
Linear mappings
Denition: A mapping f : Rn → Rk is a linear mapping, if there is a k×n matrix A for which f(x) =A·xfor every x∈Rn.
Ifn=k, thenf is a linear transformation.
Ais the matrix off, notation: A= [f].
Theorem: f :Rn→Rkis a linear mapping if and only if it preserves the addition and the multiplication by a scalar, i.e.
1.) f(u+v) =f(u) +f(v)for allu, v∈Rn, and 2.) f(λu) =λf(u)for allλ∈R, u∈Rn.
In this case theith column of the matrix off isf(ui)fori= 1,2, . . . , n, whereu1, . . . , un is the standard basis inRn.
Denition: Iff :Rn →Rk is a linear mapping, then the kernel off is the set of vectors in Rn whose image is the zero vector inRk.
The image off is the set of vectors inRk which are images underf of some vector inRn. Notation: Kerf andImf.
Proposition: Kerf is a subspace in Rn andImf is a subspace in Rk.
Theorem: (Dimension theorem) Iff :Rn→Rk is a linear mapping, thendim Kerf+ dim Imf =n. Denition: Iff :Rn →Rk andg:Rk →Rm are linear mappings, then the product (or composition) of them isg◦f :Rn→Rm, for which(g◦f)(x) =g(f(x))for everyx∈Rn.
Theorem: Iff :Rn→Rk andg:Rk →Rmare linear mappings, theng◦f :Rn→Rmis also a linear mapping, and its matrix is[g◦f] = [g]·[f].
Denition: The inverse of a mappingf :A→B isg:B →A, iff(x) =y ⇐⇒g(y) =x.
Theorem: A linear transformationf :Rn→Rn is invertible if and only if det[f]6= 0. In this casef−1 is also a linear transformation and[f−1] = [f]−1.
Remark: f is invertible⇐⇒ Kerf ={0} ⇐⇒ Imf =Rn.
Theorem: Letf :Rn→Rn be a linear transformation,B={b1, . . . , bn}a basis inRn, andBthen×n matrix whose columns areb1, . . . , bn. Then the mappingg: [x]B→[f(x)]Bis also a linear transformation.
Denition: In this case we say that the matrix off in the basisB is[g]. Notation: [g] = [f]B.
Theorem: With the above notations, 1.) [f]B=B−1·[f]·B,
2.) [f(x)]B = [f]B·[x]B,
3.) theith column of[f]B is[f(bi)]B fori= 1,2, . . . , n.
Denition: LetA be ann×n matrix. If A·x=λ·xholds for a nonzero vector x∈Rn andλ∈R thenxis an eigenvector ofA andλis an eigenvalue ofA.
Theorem: λ∈Ris an eigenvalue of the square matrixA if and only ifdet(A−λI) = 0, where Iis the identity matrix. In this case the eigenvectors belonging toλare the nontrivial solutions of the system of equations(A−λI)x= 0.
Denition: The characteristic polynomial of the square matrixAisdet(A−λI), where λis a variable.
Proposition: (Diagonalisation of the matrix of a linear transformation) Letf : Rn →Rn be a linear transformation, andB ={b1, . . . , bn} a basis in Rn. Then[f]B is a diagonal matrix if and only if each vectorbi inB is an eigenvector of [f].
Determinants
Denition: IfAis ann×n(square) matrix with entriesai,j, i, j= 1,2, . . . , n, then
det(A) = X
all permutationsπ
(−1)I(π)·a1,π(1)·a2,π(2)· · · · ·an,π(n), whereI(π)is the number of inversions of the permutationπ(1), π(2), . . . , π(n).
