theorems to prove

16 Karteikartenvon Layne
Diese Karteikarten lernenKostenlos · Spaced Repetition · iPhone, Android, Mac & Web

Alle Karten

1
Theorem 1.4. Any intersection of subspaces of a vector space V is a subspace of V.
Take two arbitrary subsets, intersect them, show non-empty, closed under addition and scalar multiplication
2
Theorem 1.7. Let S be a linearly independent subset of a vector space V , and let v be a vector in
V that is not in S. Then S ∪{v} is linearly dependent if and only if v ∈ span(S).
forward: v must be a linear combination of vectors in S, thus v in span
backward: v in span means v is linear combination of vectors in s, thus linearly dependent
3
Theorem 1.10 (Replacement Theorem). Let V be a vector space that is generated by a set G
containing exactly n vectors, and let L be a linearly independent subset of V containing exactly m
vectors. Then m ≤ n and there exists a subset H of G containing exactly n−m vectors such that
L∪H generates V.
induction on m
base case: m=0 so L is empty. take H=G and done
induction step: assume true for m, prove for m+1. Subsets of L are linearly independent, construct vm+1
4
Theorem 2.1. Let V and W be vector spaces and T : V → W be linear. Then N(T) and R(T) are subspaces of V and W, respectively
show for each 0 in there, closed under addition, closed under scalar multiplication.
5
Theorem 2.2. Let V and W be vector spaces, and let T : V → W be linear. If β = {v1,v2,...,vn}
is a basis for V , then R(T) = span(T(β)) = span{T(v1),T(v2),...,T(vn)}.
Clearly T(vi) in R(T). R(T) a subspace, so it contains the span({T(Vi)}. then do the other direction taking a w in R(T) and showing it is in the span of T(B)
6
Theorem 2.3 (Dimension Theorem). Let V and W be vector spaces, and let T : V → W be linear. If V is finite-dimensional, then nullity(T) + rank(T) = dim(V ).
define dim(V)=n, dim(N(T))=k and a basis for N(T). this can be extended to a basis B for V. T(B) is a basis for R(T), thus rank(T)=n-k
7
Theorem 2.4. Let V and W be vector spaces, and let T : V → W be linear. Then T is one-to-one
if and only if N(T) ={0}.
forwards: suppose t is one-to-one, there can only be one thing in the null space. as T linear, it must be 0.
backwards: suppose N(T)={0} take w1,w2 in R(T) with w1=w2. let v1=T(w1) and v2=T(w1) then T(v1-v2)=T(v1)-T(v2)=w1-w2=0, thus v1-v2=0 so v1=v2
8
Theorem 2.5. Let V and W be finite-dimensional vector spaces of equal dimension, and let T :
V →W be linear. Then the following are equivalent.
(a) T is one-to-one.
(b) T is onto.
dimension theorem, T is one to one iff N(T) = {0} iff N(T)=0 iff rank(T)=dim(V) iff rank(T)=dim(W) iff dim(R(T))=dim(W). As R(T) is a subset of W, dim(R(T))=dim(W) iff R(T) is onto.
9
Theorem 2.6 (Bread and butter theorem). Let V and W be vector spaces over F, and suppose
that {v1,v2,...,vn} is a basis for V . For w1,w2,...,wn in W, there exists exactly one linear
transformation T : V → W such that
T(vi) = wi for i = 1,2,...,n.
define T from V to W as T(x)= sum from i=1 to n of aiwi where ai are the cooefficients in the linear combination of basis vectors in V to get x.
Show it is linear, T(vi)=wi, and it is unique.
10
Theorem 2.14. Let V and W be finite-dimensional vector spaces having ordered bases β and γ,
respectively, and let T : V → W be linear. Then, for each u ∈ V, we have
[T(u)]γ = [T]γ
β[u]β.
Fix u in V and define f: F to V by f(a)=au and g:F to W by g(a)=aT(u) for a in F.
Then coordinate map equalities
11
Theorem 2.19. Let V and W be finite-dimensional vector spaces (over the same field). Then V
is isomorphic to W if and only if dim(V) = dim(W).
forward direction: define an isomorphism. it is one-to-one and onto, so dimV=nullity+rank=0+dimW
backward direction: dim(V)=dim(W), define bases and the unique linear transformation between them T. Then R(T)=span(T(b))=span(a)=w (onto). equivalently T is one-to-one
12
Theorem 2.21. For any finite-dimensional vector space V with ordered basis β, φβ is an isomor
phism.
φβ : V->F^n by φβ (v)=[v]b
13
Theorem 3.5. The rank of any matrix equals the maximum number of its linearly independent
columns; that is, the rank of a matrix is the dimension of the subspace generated by its columns.
rank(A)=rank(LA)=dim(R(LA))=dim(span({a1,a2,...,an})
14
Theorem 3.7. Let T : V → W and U : W → Z be linear transformations on finite-dimensional
vector spaces V , W, and Z, and let A and B be matrices such that the product AB is defined. Then
(a) rank(UT) ≤ rank(U).
(b) rank(UT) ≤ rank(T).
(a) R(T) subset of W, so R(UT) = UT(V)=U(T(V))=U(R(T)) which is a subset of U(W)=R(U)
(b) define ordered bases of V,W,Z coordinate matrices for each transformation.
15
Theorem 3.9. Let K be the solution set of a consistent system of linear equations Ax = b, and let
KH be the solution set of the corresponding homogeneous system Ax = 0. Then for any solution s
to Ax =b
K ={s}+KH ={s+k:k∈KH}.
Let s be a solution. If w in K then Aw=b. Hence A(w-s)=Aw-As=b-b=0, so w-s is in KH
converse direction w=s+k for some k in KH, Aw=A(s+k)=As+Ak=b+0 so w in K
16
Theorem 3.11. Let Ax = b be a system of linear equations. Then the system is consistent if and
only if
rank(A) = rank([A | b]).
Ax=b is consistent iff b in span of columns of a iff span(columns of a)=span(columns of a,b) iff dim(span(columns))=dim(span(columns,b)) iff rank(A)=rank(A|b).
Alle 16 Karten kostenlos lernen

Mehr Karteikarten: Mathe

Sommes et produits12 KartenMathe · FranzösischLaraProbabilités10 KartenMathe · FranzösischCarolineIdeal gas equations9 KartenMathe · EnglischAbdulEconomics Midterm146 KartenMathe · EnglischDaraMultiplication table 80 KartenMathe · FinnischAnnacj research final49 KartenMathe · Englischstride124Effect Size38 KartenMathe · EnglischAnaïsNormal Distribution28 KartenMathe · EnglischAnaïsTrig Identities 23 KartenMathe · SpanischJustAGuyChapter 7 Econ Behav19 KartenMathe · EnglischshyannekendraECON3029 DD - Group Preferences16 KartenMathe · EnglischFathiaLecture 1 DSP Drei15 KartenMathe · EnglischDiandrei

Alle Decks: Mathe

Lerne diese Decks in der App

Kostenlos auf iPhone, Android, Mac und im Web. Füge jedes Deck mit einem Tipp zu deinem Konto hinzu.

Laden im App StoreJetzt bei Google Play
Alle 16 Karten kostenlos lernen