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Sec 4.3 Linearlyindependent sets^ bases
Idea Identify subsets that span a^ vector space H
as (^) efficiently as (^) possible
I Def
Def let^ V^ be^ a^ vector^ space The set^ y
Vp in^ V^ is^ called
1 linearly independent
if
the equation c G 22T
CPV
GE (^) element in V
has only the^ trivial^ solution^4 0 22 0 Cp 0
(^2) linearly dependent (^) if
there are scalars di des dp not^ all zero^ such that
dive dare dpup 0
In this (^) case (^) the equation is^ called
a linear dependence relation among the elements V.V Vp
Recall IP polynomials in^ forms a^ vector space
Previously (^) for Sea^ 4.1^ Lecture^ we^ saw^ that^ we^ can^ write (^2) as
a linear combination of 1 I X and 1 2x x
x (^1 1 2) i x 7 1 2 1 2
So 1 1 2 1 x 1 2x^ x2^1
1
di de^ dis^
dk
Therefore the^ set^1 Itx^1 2
(^2) x is linearly dependent^ in^ IP
The set ORT of all continuous functions on t 2T
forms a^ vector^ space The set (^) sin t cost is
linearly independent in^ 0,2T
because (^) sint (^) and cos (^) t are (^) not multiples of^ one another in 0,2T i e (^) there (^) is (^) no (^) sialar c (^) such that cost sint (^) for all^ t^ in^ O^ 2T c (^) g (^) cos 12 1 and^ sin^0 jo Yftz and there^ is^ no^ scalar^ such^ that^1 CO Def Let^ H^ be (^) a (^) subspace of a^ vector^ space V (^) possibly H V
A set of vectors B in U is a basis
for H^ if
1 B is a
linearly independent^ set^ AND 2
H SpanB
i e
every element^ of the^ subspace^ H^ can^ be^ written^ as^ a^ linear^ combination^ of
the elements in the set^ B
Ex Let^ H^ IRP (^) and let A be (^) an non matrix
The Invertible Matrix Theorem^ sec^ 2.3^ pg 127 says
A (^) is (^) invertible (^) if and
only if
the columns^ of are linearly independent if and only if
the columns ofA span IR
So if A is invertible then the columns of A are
1 linearly independent and
2
span IR
and thus the columns
of A^ form a^ basis^ for IR
If A^ is^ non^ invertible^ the^ columns of A
are not linearly independent and also don'tspan IR
I (^) Bases (^) for Nul (^) A Col (^) A and Row A
Find a basis
for
Nul A
The method (^) for (^) finding the spanning set^ of Nul^ A^ in^ Sec^4 see also^ the^ method^ for solving a^ linear^ system in^ Sec^ 1. produces a^ linearly independent set^ and^ thus this (^) spanning set^ is^ a basis (^) for Nul (^) A Ex (^) Find a basis (^) for Nul A where A
f y
write the^ augmented matrix^ of the^ system A^ x̅
REEF
1 This (^) is the reduced echelon (^) form The (^) general solution^ is^
ovariables
3 2 4 0
Effi.it
itti
the non^ leading
variables are
the
free variables
is a (^) basis (^) for NulA
Find a^ basis^ for Col^ A step 1 Put^ A^ into^ row^ echelon^ form doesn't^ need^ to^ be^ reduced^ B Step 2 Check^ which^ columns^ of B^ have^ pivot positions Step^3 keep (^) only the^ columns^ of A^ which^ are^ pivot columns The result is^ a^ basis^ for Col^ A EI Find^ a^ basis^ for Col^ A^ where^
A
y as before 1 I 9 not reduced
pivot positions^
in (^) columns 1 and^3 A basis^ for Col A is Find a basis^ for Row^ A step 1 Put^ A^ into^ row^ echelon^ form doesn't^ need^ to^ be^ reduced^
B
step 2 The^ nonzero^ rows^ of B^ form a^ basis^ for Row^ A Ex (^) A basis for Row A (^) for the^ previous example is 1 2 13 00 3 6
Solutions to (^) Group Quiz pg 1 True or (^) false The set (^) of all (^) solutions to A is a (^) subspace of IR
A True since this set is
equal to^ the^ Nul^ A^ which^
is a (^) subspace of IR (^) by Them 2 Q True^ or^ false^ The^ set^ of^ all^ solutions^ to^ A^ is^ a^ subspace^ of^ IR
A False This set does not contain the zero vector in^ IR
since 8 8 E^
so this set is^ not a subspace
True or^ false Let H be the set (^) of all (^) vectors in 124 whose
coordinates a^ b^ c^ d^ satisfy the
equations a^ 2b^ 5C^ d
and c^ a^ b Then^ H^ is^ a^ subspace of 1124
A True^ Reasoning^ It^ is^ the^ set^ of all^ solutions^ of
3
which is a system of homogeneous linear
equations
By Thm^2 H^ is^ a^ subspace of^ 1124 1 True or (^) false Let (^) It be the set (^) of all vectors in (^124) whose
coordinates a b^ c^ d^ satisfy the
equations a^ 2b^ 5C^ d^1
and c a b Then^ H^ is^ a^ subspace of 1124
A False^ Reasoning^ It^ is^ the^ set^ of^ all^
solutions of
3 c did
which is a system of non homogeneous linear
equations
This is^ not a (^) subspace because the zero (^) vector is^ not in^ H
Solutions to^ Group Quiz pg
2 See also
Sec 4.2 Exercise 44
Define a^ linear^ map T (^) P2 1R in MML by (^) TCP
p s
(^9) What is the (^) kernel (^) of T Find two^ polynomials in (^) P2 that span the kernel^ of T 11
her T
p me (^12) p s^ o
Span
t 5 t^572 b what^ is^ the^ range of^
T
Find a^ vector^ in (^) IR which spans the range of^
T
sol
range
T
pÉ pnep
cer
Span (^) i
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