Real vector spaces

Question 1

a real vector space

Answer 1

a non-empty set V with operations
- addition
- scalar multiplication

Question 2

properties of a real vector space

Answer 2

• element 0 such that 0 + V = V + 0 = V
• v + w = w + v
• for each v in V theres a -v such that v + -v = -v + v = 0
• v+ (w + u) = (v+w) + u
• 0v = 0
• 1v = v
• (λµ)v = λ(µv)
• (λ + µ)v = λv + µv
  µ(v+w) = µv + µw

Question 3

vector subspace

Answer 3

is a non empty subset U of V which is closed under addition and scalar multiplication (as used in V)
equivalently a vector subspace of V is a non empty subset U which is a vector space using same operations

Question 4

basis

Answer 4

of V is a set of elements in V which are
- LI
- spanning

Question 5

span and vector subspace

Answer 5

u1..uk are vectors in V
Span(u1..uk) is a vector subspace of V

Question 6

span if one of elements can be written as a linear combination

Answer 6

u1..uk are vectors in V
if one can be written as a linear combination of the others say u1
then Span(u1…uk) = Span(u2..uk)

Question 7

span of a non LI set removing a vector

Answer 7

{ u1..uk} is not LI then for some i
ui is an element of Span {u1…ui-1, ui+1..uk}
so span(u1..uk) = Span(u1…ui-1, ui+1..uk) - same set without ui

hence if we have a finite spanning set thats not LI we can reduce no vectors and still get a span - continuing to do this till its LI then we get a basis

Question 8

basis and finite spanning set

Answer 8

if V has a finite spanning set it has a basis

Question 9

basis, max LI set, min span

Answer 9

u1..uk are vectors in V the following are equivalent:
- {u1..uk} are a basis
- {u1…uk} are a maximal LI set - adding any other vector makes it not LI
- any v in V can be written uniquely as µ1 u1 + …+ µk uk
- {u1…uk} is a minimal spanning set - removing any uis and it wont span

Question 10

LI and span adding a vector

Answer 10

if {u1…uk} are LI and u isnt in span
then {u,u1.. uk} are LI

Question 11

what is a basis - all equivalent statements

Answer 11

• {u1…uk} are a basis for V
• the uis are a max LI set
• every u in V has a unique expression ∑ λu
• the uis are a minimal spanning set

Question 12

finite dimensional

Answer 12

a vector space is finite dimensional if it has a finite bases

Question 13

steinitz exchange theorem (SET)

Answer 13

let S = {v1…vk} be a spanning set for V
let {u1…uk} be a LI set in V
then there are l distinct vectors {vi1, vi2…vil} such that for each j, 0<= j <= l
the set S \ {vi1…vij} U {u1..uj} still spans V
- we must have k>= l
- any spanning set in V is at least as large as any LI set

Question 14

corollary of SET

Answer 14

if B1 = {a1..am} and B2 = {b1..bn} are both bases for V then m=n

Question 15

dimensions

Answer 15

if V is a finite dimensional vector space call ist dimensions the size of any basis

Question 16

dimensions of vector subspaces

Answer 16

V is a finite dimensional vector space and U is a vector subspace of V
- U is also finite dimensional
- Dim U <= Dim V
- Dim U = Dim V iff U = V

Question 17

dimensions and spanning or LI

Answer 17

V is a vector space {v1…vl} are l vectors in V
let k = dimV
- if k<l the {v1..vl} are not LI
- if k>l then {v1…vl} dont span
- if k = l then the following are equivalent:
-> the vectors are a basis
-> are LI
-> span V

Question 18

R^n and the number of vectors in a subset of it and LI / span / basis

Answer 18

if there are k vectors {u1..uk} c R^n
- k>n - not LI
- k<n - dont span
- k = n are a basis <=> matrix (u1..uk) = A is non singular <=> detA ≠ 0

Question 19

column space

Answer 19

the span of vectors given by the columns of a matrix

Question 20

column rank

Answer 20

column rank of A is the dimensions of the column space

Question 21

row space

Answer 21

the span of vectors given by the rows of a matrix

Question 22

row rank

Answer 22

row rank of A is the dimensions of the row space

Question 23

null space

Answer 23

null space of a matrix A is the solution set of Ax = 0

Question 24

nullity

Answer 24

dimensions of the null space of A

Question 25

matrix and LI

Answer 25

let v1..vk be LI vectors in R^n let P be an invertible matrix Pv1...Pvk are also LI

Question 26

A and PA ranks

Answer 26

- A and PA have same row space - A and PA may have diff column spaces but have same dimension (same column rank) - A and PA have same null space

Question 27

row rank and column rank

Answer 27

row rank = column rank common number called the rank

Question 28

rank nullity theorem

Answer 28

if A is a matrix M nxk rank + nullity = k (no. collumns)

Question 30

rank and nullity in terms of GJ

Answer 30

rank = no. leading 1s nullity = no columns without leading 1s

Question 31

sum of vector subspaces

Answer 31

U+W = { v in V | v = u + w for u in U and w in W}

Question 32

intersection of vector subspaces

Answer 32

UnW = { v in V| v in U and v in W}

Question 33

dim(U+W) if U and W vector subspaces

Answer 33

dim(U + W) = dim(U) + dim(W) - dim(UnW)

Question 34

direct sum

Answer 34

V = U ⊕ W if V = U + w and UnW = {0} V is direct sum of U and W

Question 35

equivalent to direct sum

Answer 35

if U and W vector subspaces of V the following are equivalent V = U ⊕ W if {u1...us} and {w1..wt} are basis then {u1..us,w1..wt} is a basis of V V = U+W and dimV = dimU + dimW every v in V can be written uniquely as v = u + w

Question 36

coordinate map

Answer 36

let B = {v1..vn} be an ordered basis for V any v in V has unique expression v = x1v1+ ..+ xnvn given B to specify v we just need the xis which we can write as a column vector (coordinates) the function from V-> R^n that goes from v to this column vector of xis is the coordinate map

Question 37

problem of 2 basis

Answer 37

we have standard basis S and another basis B suppose v is in S basis with xis given the xis how do we find the yis to write v in B? first write bis in S coordinates as the columns of an nxn matrix then do the inverse of that matrix onto the xis in S to find the it in B coords