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MATH 304
Linear Algebra
Lecture 10:
Linear independence.
Basis of a vector space.
Linear independence
Definition.LetVbe a vector space. Vectors
v
1,v2,...,vk?Vare calledlinearly dependentif
they satisfy a relation r
1v1+r2v2+···+rkvk=0,
where the coefficientsr1,...,rk?Rare not all equal to zero. Otherwise vectorsv1,v2,...,vkare calledlinearly independent. That is, if r
1v1+r2v2+···+rkvk=0=?r1=···=rk= 0.
An infinite setS?Vislinearly dependentif
there are some linearly dependent vectorsv1,...,vk?S.
OtherwiseSislinearly independent.
Examples of linear independence
Vectorse1= (1,0,0),e2= (0,1,0), and
e
3= (0,0,1) inR3.
xe1+ye2+ze3=0=?(x,y,z) =0 =?x=y=z= 0
MatricesE11=?1 00 0?
,E12=?0 10 0? E
21=?0 01 0?
, andE22=?0 00 1? aE
11+bE12+cE21+dE22=O=??a b
c d? =O =?a=b=c=d= 0
Examples of linear independence
Polynomials 1,x,x2,...,xn.
a
0+a1x+a2x2+···+anxn= 0 identically
The infinite set{1,x,x2,...,xn,...}.
Polynomialsp1(x) = 1,p2(x) =x-1, and
p
3(x) = (x-1)2.
a
1p1(x) +a2p2(x) +a3p3(x) =a1+a2(x-1) +a3(x-1)2=
= (a1-a2+a3) + (a2-2a3)x+a3x2.
Hencea1p1(x) +a2p2(x) +a3p3(x) = 0 identically
=?a1-a2+a3=a2-2a3=a3= 0 =?a1=a2=a3= 0
ProblemLetv1= (1,2,0),v2= (3,1,1), and
v
3= (4,-7,3). Determine whether vectors
v
2,v2,v3are linearly independent.
We have to check if there existr1,r2,r3?Rnot all
zero such thatr1v1+r2v2+r3v3=0. This vector equation is equivalent to a system???r
1+ 3r2+ 4r3= 0
2r1+r2-7r3= 0
0r1+r2+ 3r3= 0((
1 3 4 0 2 1-7 0 0 1 3 0))
The vectorsv1,v2,v3are linearly dependent if and
only if the matrixA= (v1,v2,v3) is singular.
We obtain that detA= 0.
TheoremThe following conditions are equivalent:
(i) vectorsv1,...,vkare linearly dependent; (ii) one of vectorsv1,...,vkis a linear combination of the otherk-1 vectors.
Proof:(i) =?(ii) Suppose that
r
1v1+r2v2+···+rkvk=0,
v i=-r1 riv1- ··· -ri-1rivi-1-ri+1rivi+1- ··· -rkrivk. (ii) =?(i) Suppose that v for some scalarssj. Then s
TheoremVectorsv1,v2,...,vm?Rnare linearly
dependent wheneverm>n(i.e., the number of coordinates is less than the number of vectors).
Proof:Letvj= (a1j,a2j,...,anj) forj= 1,2,...,m.
Then the vector equalityt1v1+t2v2+···+tmvm=0 is equivalent to the system ?a
11t1+a12t2+···+a1mtm= 0,
a
21t1+a22t2+···+a2mtm= 0,
a n1t1+an2t2+···+anmtm= 0. Note that vectorsv1,v2,...,vmare columns of the matrix (aij). The number of leading entries in the row echelon form is at mostn. Ifm>nthen there are free variables, therefore the zero solution is not unique.
Example.Consider vectorsv1= (1,-1,1),
v
2= (1,0,0),v3= (1,1,1), andv4= (1,2,4) inR3.
Two vectors are linearly dependent if and only if
they are parallel. Hencev1andv2are linearly independent.
Vectorsv1,v2,v3are linearly independent if and
only if the matrixA= (v1,v2,v3) is invertible. detA=??????1 1 1 -1 0 1
1 0 1??????
=-????-1 11 1???? = 2?= 0.
Thereforev1,v2,v3are linearly independent.
Four vectors inR3are always linearly dependent.
Thusv1,v2,v3,v4are linearly dependent.
Problem.Show that functionsex,e2x, ande3x
are linearly independent inC∞(R).
Suppose thataex+be2x+ce3x= 0 for allx?R, where
a,b,care constants. We have to show thata=b=c= 0.
Differentiate this identity twice:
ae x+be2x+ce3x= 0, ae x+ 2be2x+ 3ce3x= 0, ae x+ 4be2x+ 9ce3x= 0.
It follows thatA(x)v=0, where
A(x) =((
exe2xe3x e x2e2x3e3x e x4e2x9e3x)) ,v=(( a b c))
A(x) =((exe2xe3x
e x2e2x3e3x e x4e2x9e3x)) ,v=((a b c)) detA(x) =ex??????1e2xe3x
1 2e2x3e3x
1 4e2x9e3x??????
=exe2x??????1 1e3x
1 2 3e3x
1 4 9e3x??????
=exe2xe3x??????1 1 11 2 31 4 9?????? =e6x??????1 1 11 2 31 4 9?????? =e6x??????1 1 10 1 21 4 9?????? =e6x??????1 1 10 1 20 3 8?????? =e6x????1 23 8???? = 2e6x?= 0.
Since the matrixA(x) is invertible, we obtain
A(x)v=0=?v=0=?a=b=c= 0
Wronskian
Letf1,f2,...,fnbe smooth functions on an interval
[a,b]. TheWronskianW[f1,f2,...,fn] is a function on [a,b] defined by
1(x)f2(x)···fn(x)
f f (n-1)
1(x)f(n-1)
TheoremIfW[f1,f2,...,fn](x0)?= 0 for some
x
0?[a,b] then the functionsf1,f2,...,fnare
linearly independent inC[a,b].
Theorem 1Letλ1,λ2,...,λkbe distinct real
numbers. Then the functionseλ1x,eλ2x,...,eλkx are linearly independent.
Theorem 2The set of functions
{xmeλx|λ?R,m= 0,1,2,...} is linearly independent.
Spanning set
LetSbe a subset of a vector spaceV.
Definition.Thespanof the setSis the smallest
subspaceW?Vthat containsS. IfSis not empty thenW=Span(S) consists of all linear combinations r1v1+r2v2+···+rkvksuch that v
1,...,vk?Sandr1,...,rk?R.
We say that the setSspansthe subspaceWor
thatSis aspanning setforW.
Remark.IfS1is a spanning set for a vector space
VandS1?S2?V, thenS2is also a spanning set
forV. Basis
Definition.LetVbe a vector space. A linearly
independent spanning set forVis called abasis.
Suppose that a setS?Vis a basis forV.
"Spanning set" means that any vectorv?Vcan be represented as a linear combination v=r1v1+r2v2+···+rkvk, wherev1,...,vkare distinct vectors fromSand r
1,...,rk?R. "Linearly independent" implies that the above
representation is unique: =?(r1-r?1)v1+ (r2-r?2)v2+···+ (rk-r?k)vk=0 =?r1-r?1=r2-r?2=...=rk-r?k= 0
Examples.Standard basis forRn:
e
1= (1,0,0,...,0,0),e2= (0,1,0,...,0,0),...,
e n= (0,0,0,...,0,1). Indeed, (x1,x2,...,xn) =x1e1+x2e2+···+xnen.
Matrices?1 00 0?
,?0 10 0? ,?0 01 0? ,?0 00 1? form a basis forM2,2(R).?a b c d? =a?1 00 0? +b?0 10 0? +c?0 01 0? +d?0 00 1?
Polynomials 1,x,x2,...,xn-1form a basis for
P n={a0+a1x+···+an-1xn-1:ai?R}.
The infinite set{1,x,x2,...,xn,...}is a basis
forP, the space of all polynomials.
Bases forRn
Letv1,v2,...,vkbe vectors inRn.
Theorem 1Ifk v 1,v2,...,vkdo not spanRn.
Theorem 2Ifk>nthen the vectors
v 1,v2,...,vkare linearly dependent.
Theorem 3Ifk=nthen the following conditions
are equivalent: (i){v1,v2,...,vn}is a basis forRn; (ii){v1,v2,...,vn}is a spanning set forRn; (iii){v1,v2,...,vn}is a linearly independent set. Example.Consider vectorsv1= (1,-1,1),
v 2= (1,0,0),v3= (1,1,1), andv4= (1,2,4) inR3.
Vectorsv1andv2are linearly independent (as they
are not parallel), but they do not spanR3. Vectorsv1,v2,v3are linearly independent since??????1 1 1 -1 0 1 1 0 1??????
=-????-1 1 1 1????
=-(-2) = 2?= 0. Therefore{v1,v2,v3}is a basis forR3.
Vectorsv1,v2,v3,v4spanR3(becausev1,v2,v3
already spanR3), but they are linearly dependent.quotesdbs_dbs17.pdfusesText_23
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