theorems to prove

16 flashcardsby Layne
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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 ).
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}.
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
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.
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.
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).
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}.
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]).
forward direction: suppose it is consistent, that is there is some x such that Ax=b. Then
backward: If rank(a) = rank(A|b) then b is in the span of the columns of A, hence
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