Download - CH6.1-6.4 Real Vector Space
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Chapter 6
Real Vector Spaces
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Objective:
l Study many important properties of
vectors in
nR
l A carefully constructed generalization of
nR to many other important vector space.
l Application of vector space to the linearsystem and matrix.
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6.1. Vector Spaces
Definition of Vector Space
Definition:
A real vector space V is asetof elements
together with two operations, addition and
scalar multiplication , satisfying the
following properties:
Letu, v, andwbe vectors in V, and letcanddbe
scalars.
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Addition:
(a) Ifuandvare any elements ofV, then vu+
is inV(closed).
1.
uvvu +=+ .
2. wvuwvu ++=++ )()( .
3. V has a zero vector0such that for everyuin V, uuu =+=+ 00 .
4. For everyuinV, there is an element u- in
V, 0)( =-+ uu .
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Scalar multiplication:
(b) If uis any element ofVand cis any real
number, then cu is inV(closed).
5. ucdduc )()( = .
6. ducuudc +=+ )( .
7. cvcuvuc +=+
)(
.
8. uu=1 .
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Example :
nRV 1 together with standard vector addition and scalar
multiplication. Then, 1V
is a vector space since for any two vectors u and v in
n
and scalar c, both vu + and cu are innR . Therefore,
conditions (a) and (b) are satisfied. In addition, the
conditions (1) to (8) are satisfied (see the previoussubsection).
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Example :
2V the set consisting of all nm matrices together with
standard matrix addition and scalar multiplication. 2V is a
vector space since for any two nm
matrices u and vand scalar c, both vu+ and cu are nm matrices.
That is,both vu+ and cu are still in 2V .Therefore
conditions (a
) and (b
) are satisfied. In addition, theconditions (1) to (8) are satisfied (see section 2).
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Example:
3V the set consisting ofallpolynomials of degree 2 or less with
the form together with standard polynomial addition and scalar
multiplication. Is 3V a vector space?
We need to examine whether the conditions (a ), (b ), and the
conditions (1) to (8) are satisfied. Let
01
2
2 axaxau ++= ,
012
2 bxbxbv ++= ,
01
2
2 cxcxcw ++=
and let c and d be scalars. Then,
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Solution:
Addition:
(a):
30011
2
22
012
2012
2
)()()(
)()(
Vbaxbaxba
bxbxbaxaxavu
+++++=
+++++=+
since vu+ is a polynomial of degree 2 or less.
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Solution:
(1):
uvaxaxabxbxb
abxabxab
baxbaxba
bxbxbaxaxavu
+=+++++=
+++++=
+++++=
+++++=+
)()(
)()()(
)()()(
)()(
01
2
201
2
2
0011
2
22
00112
22
01
2
201
2
2
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Solution:
(2):
)(
)]()()[(
)()()(
)]()()[()()(
01
2
20011
2
22
000111
2
222
0011
2
2201
2
2
wvu
cxcxcbaxbaxba
cbaxcbaxcba
cbxcbxcbaxaxawvu
++=
++++++++=
++++++++=
++++++++=++
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Solution:
(3):
Let 00002 ++= xx . Then,
uaxaxa
uaxaxaaxaxau
=++=
+=+++++=+++++=+
01
2
2
01
2
201
2
2 0)0()0()0()0()0()0(0
(4):
Let ).()()( 012
2 axaxau -+-+-=- Then,
0000)]([)]([)]([ 200112
22 =++=-++-++-+=-+ xxaaxaaxaauu
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Solution:
Scalar multiplication:
(b):
301
2
2 )()()( Vcaxcaxcacu ++=
since cu is a polynomial of degree 2 or less.
(5):
cvcucbcaxcbcaxcbca
bacxbacxbacvuc
+=+++++=
+++++=+
)()()(
)]([)]([)]([)(
0011
2
22
0011
2
22
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Solution:
(6):
ducudaxdaxdacaxcaxca
adcxadcxadcudc
+=+++++=
+++++=+
)]()()[()]()()[(
])[(])[(])[()(
01
2
201
2
2
01
2
2
(7):
ucdacdxacdxacd
dacxdacxdacdaxdaxdacduc
)()()()(
)()()()]()()[()(
01
2
2
01
2
201
2
2
=++=
++=++=
(8):
uaxaxaaxaxau =++=++= 012
201
2
2 1111
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Note:
nP the set consisting ofallpolynomials of degree n
or less with the form together with standard
polynomial addition and scalar multiplication. Then,
nP is avector space. In addition,
the setconsisting ofallpolynomials with the form together
with standard polynomial addition and scalar
multiplication. Then, P is also avector space.
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Example:
4V the set consisting ofallreal-valuedcontinuous
unctionsdefined on the entire real line together with
standard addition and scalar multiplication. Is 4V a vector
space?
We need to examine whether the conditions (a), (b), andthe conditions(1) to (8) are satisfied. Let
)(xfu= , )(xgv= , )(xhw=
and letcanddbe scalars. Then,
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Solution:
Addition:
(a):
4)()( Vxgxfvu +=+
since )()( xgxf + is still a continuous function.
(1):
uvxfxgxgxfvu +=+=+=+ )()()()(
(2):
[ ] [ ]wvu
xhxgxfxhxgxfwvu
++=
++=++=++
)(
)()()()()()()(
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Solution:
(3):
Let the zero vector 00 = . Then,
uxfxfu ==+=+ )(0)(0
(4):
Let )(xfu -=-
. Then,
[ ] 0)()()()()( =-=-+=-+ xfxfxfxfuu
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Solution:
(b):
4)( Vxcfcu =
since)(xcf
is still a continuous function.(5):
[ ] cvcuxcgxcfxgxfcvuc +=+=+=+ )()()()()(
(6):
ducuxdfxcfxfdcudc +=+=+=+ )()()()()(
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Solution:
(7):
[ ]
ucdxfcdxcdfxdfcduc )()()()()()( ====
(8):
[ ]
uxfxfu === )()(1)(1
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Note:
Let *
4V the set of all differentiable functions
defined on the entire real line and
**4V the set o
all integrable functions defined on the entire real
line. Both*
4V and**
4V are vector space under
standard addition and scalar multiplication.
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Example:
5V the set consisting of all integers with standard
addition and scalar multiplication. Is 5V a vector
space?
5V is not a real vector space since for 51 Vu = , and
7.0=c ,
57.017.0 Vcu == ,
condition(b)is not satisfied.
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Example:
6V the set consisting of all vectors in2R with
standard addition and nonstandard scalar
multiplication defined by
=
0
1
2
1 cx
x
xc
. Is 6V a vector space?
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Solution:
6V is not a real vector space since for2
2
1R
c
cu
= ,
0,, 221 cRcc ,
uc
cc
c
cu =
=
=
2
11
2
1
011
,
condition(8)is not satisfied.
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Example:7V the set consisting of only second degree
olynomials with standard addition and scalar
multiplication. Is 7V a vector space?
7V isnota real vector space since for2xu = and
2xv -=
7
22 0)( Vxxvu =-+=+ ,
condition(a
)is not satisfied.
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Example:
8V the set consisting of all real-valued continuous
unctions such that 3)1( =f . Suppose the operations are
standard addition and scalar multiplication. Is 8V a vectorspace?
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Solution:
8V isnota real vector space since for
8)( Vxfu = , 8)( Vxgv =
8)()(3633)1()1(
Vxgxfvugf
+=+=+=+
,
condition(a)is not satisfied.
Example 1 9 Page 273 276
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Example1 p273
n Consider the set Rn together with the
operations of vector addition and
scalar multiplication as defined inSection 4.2. Theorem 4.2 in Section
4.2 established the fact that Rn is a
vector space under the operations ofaddition and scalar multiplications of n-
vector.
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Example2 p273
n Consider the set V of all ordered triples
of real number of the form (x,yx,0)and
define the operations and by
(x,y,0)(x,y,0)=(x+x,y+y,0)
c(x,y,0)=(cx,cy,0)
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Example3 p273
n Consider the set V of all ordered triples ofreal numbers (x,y,z) and define theoperations and by
1.(x,y,z)(x,y,z) = (x+x,y+y,z+z).
2.c(x,y,z)=(cx,y,z).
3.c[(x,y,z)(x,y,z)]=(c(x+x), y+y,z+z).
4.c (x,y,z)d(x,y,z)= ((c+d)x,2y,2z).
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Example4 p274
n Consider the set M23 of all 23matrices.
n Similarly the ser of all mn matricesunder the usual operations of matrix
addition and scalar multiplication is a
vector space will be denoted by Mmn.
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Example5 p274
n Let F[a,b] be the set of all real-valuedfunctions that are defined on the interval[a,b]. If f and g are in V, we define fg and
cf by (fg)(t)f(t)+g(t).(cf)(t)cf(t).
n A polynomial (in t) is a function that is
expressible asp(t)antn+ an-1t
n-1++a1t+a0a0,a0,a1,are real numbers.
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Example6 p274
n The follow functions are polynomials
n p1(t)3t42t25t1.
n p2(t)2t1.n p3(t)4.
n f4(t)2 6 and f5(t)1/t22t1 are
not polynomials.t
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Example7 p275
n p1(t) : degree 4
n p2(t) : degree 1
n p3(t) : degree 0
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Example8 p275
n If p(t)antn+ an-1t
n-1++a1t+a0 .
n q(t)bntn+ bn-1t
n-1++b1t+b0 .
n p(t) q(t)
(an+bn)tn+(an-1+bn-1)tn-1++(a1+b1)t+a0+b0
n cp(t) cantn+ can-1t
n-1++ca1t+ca0 .
n (c+d)p(t)cp(t)dp(t).
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Example9 p276
n Let uvu-v , cucu ,
n If u3,v2 ,
uv-1. vu1
n If u4,c2,d3
(c+d)(2+3)420
cudv812-4
Theorem 6 1:
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Theorem 6.1:
(Important Result)
IfVis a vector space and letube any element of a
real vector spaceV. Then,
(a) .00 =u
(b) .0,,00 VRcc =
(c) .0or0.0 === uccu
(d) .)1( uu -=-
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6.2. Subspace
Definition of subspace:
Wis called a subspace of a real vector spaceV if
1. Wis a subset of the vector spaceV.
2. W is a vector space with respect to the
operations in V.Example 1(Page 279)
Example 2(Page 279)
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Example1 p279
n Every vector space has at least two
subspaces, itself and subspace{0}.
n The subspace {0} is called the zerosubspace.
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Example2 p279
n Let u(a1,b1,0), v (a2,b2,0)
uv(a1+a2,b1+b2,0)
cu(ca1,cb1,0)
Theorem 6.2: (Important
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Theorem 6.2: (Important
Result)
Wis a subspace of a real vector spaceV
1.Ifuandvare any vectors inW, then
Wvu +.
2.If c is any real number and u is any vector in W,
then Wcu .
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Example:
1W the subset of3R consisting of all vectors of the
form,
Raa
,
0
0
,
together with standard addition and scalar multiplication.
Is 1W a subspace of3
R ?
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Solution:
We need to check if the conditions (1) and (2) are satisfied.Let
Rc
a
v
a
u
=
= ,00,
00
21
.
Then,
(1):
1
2121
0
0
0
0
0
0 W
aaaa
vu
+
=
+
=+.
.
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Solution:
(2):
1
1
0
0 W
ca
cu
=
.
1W is a subspace of 3R .
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Example:
Let the real vector space V be the set consisting of allnn matrices together with the standard addition and
scalar multiplication. Let
2W the subset of V consisting of all nn diagonalmatrices.
Is 2W a subspace of V?
Let
2
22
11
00
00
00
W
a
a
a
u
nn
=
L
MOMM
L
L
,2
22
11
00
00
00
W
b
b
b
v
nn
=
L
MOMM
L
L
, and
Rc .
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Solution:
(1):
2
2222
1111
00
00
00
W
ba
ba
ba
vu
nnnn
+
+
+
=+
LMOMM
L
L
since vu + is still a diagonal matrix.
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Solution:
(2):
2
22
11
00
00
00
W
ca
ca
ca
cu
nn
=
LMOMM
L
L
since cu is still a diagonal matrix.
2W is a subspace of V.
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Example:
nP the set consisting ofallpolynomials of degree n or
less with the form together with standard polynomial
addition and scalar multiplication. nP
is a vectorspace.
P the set consisting of allpolynomials withthe form together with standard polynomial addition and
scalar multiplication. Then, is also a vector space.
Then, nP is a subspace of P.
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Example:
4V the set consisting of all real-valued continuous
unctions defined on the entire real line together with
standard addition and scalar multiplication. Let
*4Vthe
set of all differentiable functions defined on the entire real
line together with standard
addition and scalar multiplication. Then, 4V
is a realvector space. Also.
*
4V is a subspace of 4V .
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Example:
3W the subset of 3R consisting of all vectors of the form,
Rba
b
a
a
,,2
,
together with standard addition and scalar multiplication. Is
3W a subspace of3R ?
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Solution:
(1):
( ) 3
21
2
21
21
21
2
2
2
1
21
2
2
2
2
1
2
1
1
W
bb
aa
aa
bb
aa
aa
b
a
a
b
a
a
vu
+
+
+
+
+
+
=
+
=+
.
Therefore, 3Wvu + .
3W isnota subspace of3R .
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Example:
3V the set consisting ofallpolynomials of degree 2 or less
with the form together with standard polynomial addition
and scalar multiplication. 3V is a vector space. Let
4W the subset of 3V consisting of all polynomials of the
form
2,2 =++++ cbacbxax .
Is 4W a subspace of 3V ?
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Solution:
Let
401
2
2 Waxaxau ++=
and
4012
2 Wbxbxbv ++= .
Then, 2012 =++ aaa and 2012 =++ bbb . Thus,
40011
2
22
01
2
201
2
2
)()()(
)()(
Wbaxbaxba
bxbxbaxaxavu
+++++=
+++++=+
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Solution:
since
( ) ( ) ( ) ( ) 422)( 012012001122 =+=+++++=+++++ bbbaaabababa .
4W isnota subspace of 3V .
Example 3--10(Page 280--282)
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Example3 p280
=
11
11
0
0
dc
bau
=
22
22
0
0
dc
bav
++
++=+
2121
2121
0
0
ddcc
bbaavu
=
11
11
0
0
kdkc
kbkaku
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Example5 p282
n Let u(a1,b1,1), v (a2,b2,1)
uv(a1+a2,b1+b2,2)
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n We let Pn denote the vector space
consisting of all polynomials of defree
n and the zero polynomial. And let Pdenote the vector space of akkpolynomials. It is easy to verify that P2is a subspace of P3 and in general
that Pn is a subspaces of Pn+1. Pn is a
subspace of P.
Example6 p282
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Example7 p282
n Let V be the set all polynomials of
degree exactly2.V is a subset P2 but
it is not a subspace of P2 , since thesum of polynomials 2t2+3t+1 and
-2t2+t+2 , a polynomial of degree 1, is
not in V.
E l 8 282
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Example8 p282
n Let C[a,b] denote the set of all real-
values continuous functions that are
defined on the interval [a,b]. If f and gare in C[a,b] then f+g is in C[a,b].
E l 9 282
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Example9 p282
n Consider the homogeneous system
Ax=0, A is an mn matrix.A solution consists of avector x in Rn. Let W be the subset of Rn consistingof all solutions to the homogeneous system. Since
A0=0, we conclude that W is not empty. To checkthat W us a subspace of Rn, we verify properties ()and () of Them 6.2. Thus let x and y be solutions.
Ax=0 and Ay=0
A(x+y)=Ax+Ay=0A(cx)=c(Ax)=0
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Example10 p282
n Let v1 and v2 be fixed vectors in a
vector space V and let W be the set of
all linear combinations of v1 and v2.w1=a1v1+a2v2
w2=b1v1+b2v2
w1+w2= (a1+b1)v1+ (a2+b2)v2
cw1=(ca1)v1+ (ca2)v2
D fi iti f li bi ti
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Definition of linear combination:
Let kvvv ,,, 21 K be vectors in a real vector space V. A
vector v inV is called a linear combination o
kvvv ,,, 21 K
if
kkvcvcvcv +++= L2211 ,
where kccc ,,, 21 K are real numbers.
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Example:
Let
=
-=
-=
=
12
80,
31
02,
21
31,
01
20321 vvvv
be vectors in the vector space consisting of all 22
matrices. Then,
321 231
02)1(
21
312
01
201
12
80vvvv -+=
--+
-+
=
= .
That is, v is a linear combination of 321 ,, vvv .
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Example:
For linear system
b
x
x
x
Ax =
=
-
=
1
1
1
123
012
101
3
2
1
.
-=1
3
2
x is a solution for the above linear system
Thus,
Example:
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Example:
bAcolAcolAcol
A
=
=+-=
-+
-
=
-
-=
-
1
11
)()(3)(2
1
0
1
1
2
1
0
3
3
2
1
2
1
3
2
123
012
101
1
3
2
121
That is, b is a linear combination of the column vectorsofA,
)(),(),( 321 AcolAcolAcol
Note:
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Note:
For a linear system 11 = mnnm bxA , the linear
system has solution or solutions
b is a linear
combination of the column vectors ofA,)(,),(),( 21 AcolAcolAcol nL .
For example, if
=
nc
c
c
c M
2
1
is a solution of
11 = mnnm bxA ,
Note:
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Note:
then
bAcolcAcolcAcolc nn =+++ )()()( 2211 L .
On the other hand, the linear system has no solution
b isnota linear combination of the columnvectors ofA
Example:
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Example:
Is the vector
=
5
5
4
va linear combination of the vectors
=
-
=
=
2
3
3
,
4
1
1
,
3
2
1
321 vvv
.
Solution:
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Solution:
We need to find the constants 321 ,, ccc such that
332211321
2
3
3
4
1
1
3
2
1
5
5
4
vcvcvccccv ++=
+
-
+
=
=
.
we need to solve for the linear system
=
-
=
5
5
4
243
312
311
3
2
1
3
2
1
c
c
c
c
c
c
A.
Solution:
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Solution:
The solutions are
Rttctctc =-=+-= ,,1,32 321 .
Thus,
( ) ( ) Rttvvtvtv +-++-= ,132 321v is a linear combination of 321 ,, vvv withinfinite number
of expressions.
Example:
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Example:
Is the vector
-
-=
6
4
3
va linear combination of the vectors
=
-
-
-
=
=
5
4
1
,
2
1
1
,
3
2
1
321 vvv
.
Solution:
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Solution:
We need to find the constants 321 ,, ccc such that
332211321
5
4
1
2
1
1
3
2
1
6
4
3
vcvcvccccv ++=
+
-
-
-
+
=
-
-=
.
Solution:
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Solution:
we need to solve for the linear system
-
-=
-
-
-
=
6
4
3
523
412
111
3
2
1
3
2
1
c
c
c
c
c
c
A.
The linear system hasnosolution.
v isnota linear combination of 321 ,, vvv
Note:
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Note:
Let nvvv ,,, 21 K and v be vectors in m and let Am*nbe
the matrix with column vectors njvAcol jj ,,2,1,)( K== .
Thus,
vAx= has solution or solutions v is a linea
combinationof nvvv ,,, 21 K .
vAx= has no solution v is not a linearcombinationof nvvv ,,, 21 K .
Definition of spanning set:
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Definition of spanning set:
Let { }
kvvvS ,,, 21 K= be a set of vectors in a real vector
space V. Then, the span of S, denoted by )( Sspan , is
the set consisting of all the vectors that are linear
combinations of kvvv ,,, 21 K . That is,
{ }RcccvcvcvcSspan kkk +++= ,,,|)( 212211 KL .
If VSspan =)( , it is said that V is spanned by S or S
spans V.
Example:
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Example:
Let
{ }321321 ,,and,
1
0
0
,
0
1
0
,
0
0
1
eeeSeee =
=
=
=.
Then,
3
321
3
2
1
332211 ,,|)( RRccc
c
c
c
ecececSspan =
=++=,
Example: Example 1(Page 292)
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Example: Example 1(Page 292)
Let
{ }321321 ,,and,01
1
,20
1
,12
1
vvvSvvv =
=
=
= .
Does3)( RSspan = ?
Solution:
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Solution:
3)( RSspan = For any vector3R
c
b
a
v
=
, there
exist real numbers 321 ,, ccc such that
332211321
0
1
1
2
0
1
1
2
1
vcvcvcccc
c
b
a
v ++=
+
+
=
=.
we need to solve for the linear system
Solution:
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Solution:
=
c
b
a
c
c
c
3
2
1
021
102
111
.
The solution is
3
24,
3,
3
22321
cbac
cbac
cbac
--=
+-=
++-=
.
3213
2433
22 vcbavcbavcbav
--+
+-+
++-= .
That is, every vector in 3R can be a linear combination of
321 ,, vvv
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Example 10(Page 282)Example 11(Page 283)
Example 12(Page 284)
Theorem 6.3: (Important result)
Let { }kvvvS ,,, 21 K= be a set of vectors in a real vector
space V. Then, )( Sspan is a subspace of V.
Example 13(Page 285)
Example11 p283
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p p
n v1=(1,2,1), v2=(1,0,2), v3=(1,1,0), v=(2,1,5)
n If c1v1+c2v2 +c3v3 =v ,find c1 c2 c3 ?
n c1
(1,2,1) + c2
(1,0,2) + c3
(1,1,0) =(2,1,5)
c1+c2 +c3=2
2c1+0c2 +c3=1
c1+2c2 +0c3=5
So c1=1,c2=2,c3=-1
Example12 p284
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p p
=
100
000,
010
000,
000
010,
000
001S
=
+
+
+
dc
badcba
0
0
100
000
010
000
000
010
000
001
Where a,b,c ,and d are real number
Section 6.3
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Linear Independence
Definition:
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The vectors 1, 2, , k in a vector space are said tospan if
every vector in is a linear combination of 1, 2,, k. Moreover,
ifS={1, 2, , k}, then we also say that the setSspans , or that
{1, 2, , k}spans , or that isspannedbyS, or in the
language of Section 6.2, spanS=.
The procedure to check if the vectors 1, 2, , k spanthe vector
space is as follows.
Definition:
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Step 1. Choose an arbitrary vector in .
Step 2. Determine if is a linear combination of the given
vectors. If it is, then the given vectors span . If it is not,
they do not span .
Example 2--6(Page 292--294)
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Question:
Let { }kvvvS ,,, 21 K= and WSspan =)( . Is it possible to
find a smaller (or even smallest) set, for example,
{ }121 ,,, -* = kvvvS K ,such that
)()( *== SspanWSspan ?To answer this question, we need to introduce the concept o
linear independence and linear dependence.
Linear Independence
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Definition of linear dependence and linear
independence:
The vectors
kvvv ,,, 21K
in a vector space V arecalled linearly dependent if there exist constants,
kccc ,,, 21 K , not all 0, such that
02211 =+++ kkvcvcvc L .
Definition of linear dependence
and linear independence:
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kvvv ,,, 21 K are linearly independent if
00 212211 =====+++ kkk cccvcvcvc LL
The procedure to determine if kvvv ,,, 21 K
are
linearly dependent or linearly independent:
Definition of linear dependence
and linear independence:
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1.Form equation 02211 =+++ kkvcvcvc L , which lead to a
homogeneous system.
2.If the homogeneous system has only the trivial solutionthen the given vectors are linearly independent if it
has a nontrivial solution, then the vectors are linearly
dependent.
Example:
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{ }321321 ,,and,
1
0
0
,
0
1
0
,
0
0
1
eeeSeee =
=
=
=. Are
21 , ee and 3e linearly independent?
Solution:
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0
100
010
001
1
0
0
0
1
0
0
0
1
3
2
1
321332211 =
=
+
+
=++
c
c
c
cccececec
=
0
0
0
3
2
1
c
c
c
.
Therefore, 21 , ee
and 3e
are linearly
independent.
Example:
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.
10
6
8
,
1
1
2
,
3
2
1
321
=
-
=
= vvv. Are 21 , vv and 3v
linearly independent?
Solution:
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0
1013
612
821
10
6
8
1
1
2
3
2
1
3
2
1
321332211 =
-=
+
-+
=++
c
c
c
cccvcvcvc
Rtt
c
c
c
-
-=
,
1
2
4
3
2
1
.
Therefore, 21 , vv
and 3v
are linearly dependent.
Example:
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Determine whether the following set of vectors in
the vector space consisting of all 22 matrices is
linearly independent or linearly dependent.
{ }
==
02
01,
12
03,
10
12,, 321 vvvS
.
Solution:
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=
+
+
=++
00
00
02
01
12
03
10
12321332211 cccvcvcvc
.
Thus,
0
022
0
032
21
32
1
321
=+=+
=
=++
cc
cc
c
ccc
=
+
+
0
0
0
0
0
2
0
1
1
2
0
3
1
0
1
2
321 ccc
.
Solution:
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The homogeneous system is
=
0
0
0
0
011
220
001
132
3
2
1
c
c
c
.
The associated homogeneous system has only the trivial
solution
=
0
0
0
3
2
1
c
c
c
.
Therefore, 21 , vv and 3v are linearly independent.
Example:
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Determine whether the following set of vectors in
the vector space consisting of all polynomials o
degree n is linearly independent or linearlydependent.
{ } 223,2,2,, 222321 +++++== xxxxxxvvvS .
Solution:
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( ( ( 022322 232221332211 =+++++++=++ xxcxxcxxcvcvcvc .Thus,
022
02
032
31
321
321
=++
=++
=++
cc
ccc
ccc
=
+
+
0
0
0
2
2
3
0
1
2
2
1
1
321 ccc.
Solution:
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The associated homogeneous system is
=
0
0
0
202
211
321
3
2
1
c
c
c
.
The homogeneous system has infinite number of solutions,
.,
1
1
1
3
2
1
Rtt
c
c
c
-
=
Therefore, 21 , vv and 3v
are linearly dependent since
Rttvtvtv =-+ ,0321 .
Example 7--12(Page 295--296)
Example7 p295
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-
-
1
1
02
0
0
11
and
=
-
+
-
0
0
00
1
1
02
0
0
11
21 cc
c1=c2=0. the vectors arelinear independent.
Example8 p295
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n Are the vectors v1=(1,0,1,2),
v2=(0,1,1,2) , v3=(1,1,1,3) independent?
c1+c
3=0
c2+c3=0
c1+ c2+c3=0
2 c1+2 c2+3c3=0
The onlysolutions
c1=c2=c3 =0. the
vectors are
linear
independent.
Example9 p295
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n Are the vectors v1=(1,2,-1), v2=(1,-2,1) ,
v3=(-3,2,-1), v4=(2,0,0), independent?
c1+ c2+-3c3+2c4=0
2c1+-2c2+2c3=0
-c1+ c2-c3=0
The only solutionsc1=1,c2=2, c3 =1.
c4 =0 or
c1=1,c2=1, c3 =0.
c4 =-1 so thevectors are linear
dependent.
Example10 p296
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n The vector e1 and e2 in R2 , defined in EX4, are
linearly independent since
c1(1,0)+c2(0,1)=(0,0)
only if c1=c2=0
n We form the matrix A, whose columns are the
given n vectors. Then the given vectors arelinearly independent if and only if det(A)0.
=10
01A
Example11 p296
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n Consider the vectors
n p1(t)=t2+t+2, p2(t)=2t
2+t, p3(t)=3t2+2t+2
n c1+2 c2+3c3=0n c1+ c2+2c3=0
n 2 c1+2c3=0
c1=1,c2=1, c3=-1.
Linear dependent
Example12 p296
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n If v1, v2, v3, vk are k vectors in andy
vector space and vi is the zero vector.
n Then S={v1
, v2
, v3
, vk
} is linearly
dependent.
Note:
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In the examples with,
10
68
,
1
12
,
3
21
321
=
-=
= vvv
or with
{ } 223,2,2,, 222321 +++++== xxxxxxvvvS, 21
, vv
an
3v are linearly dependent. Observe that 3v in both
examples are linear combinations of 21 , vv ,
Note:
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233 24
1
12
2
3
21
4
10
68
vvv -=
--
=
=
and
21
222
3 22223 vvxxxxxxv +=++++=++= .
As a matter of fact, we have the following generalresult.
Theorem 6.4: (Importantresult)
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The nonzero vectors kvvv ,,, 21 K in a vector space V
are linearly dependent if and only if one of the
vectors 2, jvj , is a linear combination of the
preceding vectors 121 ,,, -jvvv K .
Note:
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Every set of vectors containing the zero vector is
linearly dependent. That is, kvvv ,,, 21 K are k vectors
in any vector space and iv is the zero vector, then
kvvv ,,, 21 K are linearly dependent.
Example 13 on page 298
Example 14 on page 299
Example13 p298
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n If v1, v2, v3, v4 are as in Ex9,then we
find that v1+ v2+ 0v3- v4=0
n So v1
, v2
, v3
, v4
are linear dependent.
We then hace v4= v1+ v2
Example14 p299
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2144321 vSinceS.spanlet w
0
1
1
2
,
0
1
1
0
,
0
1
0
1
,
0
0
1
1
vvvvvv +==
=
=
=
=
We conclude thar W=span S1 where
S1={ v1,v2,v3 }
6.4. Basis and Dimension
vvv
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The vectors
kvvv ,,,
21K
in a vector space V are said to
form abasisof V if
(a) kvvv ,,, 21 K span V (i.e., Vvvvspan k =),,,( 21 K ).
(b)
kvvv ,,, 21 K are linearly independent.
Example 1(Page 303)
Natural basis or standard basis
Example1 p303
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n The vector e1=(1,0) and e2=(0,1) from
a basis for R2.Each of these set of
vectors is called the natural basis or
standard basis.
Example:
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{ }321321 ,,and,
1
00
,
0
10
,
0
01
eeeSeee =
=
=
=
. Are 21 , ee
and 3e a basis in3
R ?
Solution:
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21 , ee and 3e form a basis in 3R since
(a)3
321 ),,()( ReeespanSspan == (see the example in the
previous section).
(b) 21 , ee and 3e are linearly independent (also see the
example in the previous section).
Example:
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=
=
=
43,
10,
01
321 vvv . Are 21 , vv and 3v a basis in2?
[solution:]
21 , vv and 3v arenota basis of 2 since 21 , vv and 3v are
linearly dependent,
043 321 =-+ vvv .
ote that2
321 ),,( Rvvvspan = .
Example:
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.
10
68
,
1
12
,
3
21
321
=
-=
= vvv
. Are 21 , vv and 3v a
basis in3
R ?
Solution:
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21 , vv and 3v are not a basis in3R since 21 , vv and 3v
are linearly dependent,
233 24
1
1
2
2
3
2
1
4
10
6
8
vvv -=
-
-
=
=.
Example 2(Page 303)
Example 3(Page 304)
Example 4(Page 304)
Example2 p303
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n S={ v1,v2,v3,v4} where
n v1=(1,0,1,0), v2=(0,1,-1,2)
n v3=(0,2,2,1) ,v
4=(1,0,0,1)
n Show that S is a basis for R4
part1
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n c1+c4=0
n c2+2c3 =0
n c1-c
2+2c
3=0
n 2 c2+c3 +c4=0
The only solutions
c1=c2=c3=c4=0,
Showing that S islinear indepeddent
part2
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n Let v =(a,b,c,d) be any vector in R4.
n We now seek constants k1,k2,k3,k4
such that k1v
1+k
2v
2+ k
3v
3+ k
4v
4=v
n Substituting for v1 v2 v3 v4 and v we find
a solution for k1,k2,k3,k4 to the resulting
linear system for any a,b,c,d . Hence S
spans R4 and is a vasis for R4.
Example3 p304 part 1
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n To show S={t2+1,t-1,2t+2} is a basisfor the vectorspace P2.
n at2+bt+c=a1t2+(a2+
a3)t+(a1-a2+2a3)
n a1=a
n a2+2a3=bn a1-a2+2a3=c
4,
2, 321
abca
cbaaaa
-+=
-+==
13622
++ ttgiven
a1=2 , a2=-5/2 ,a3=17/4
a1(t2+1)+a2(t-1)+a3(2t+2)=0
Example3 p304 part 2
2 ( 2 ) ( 2 ) 0
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Then a1t2+(a2+2a3)t+(a1-a2+2a3)=0
a1=0
a2+2a3=0
a1
-a2
+2a3
=0
The only solutions a1=a2=a3=0,
S is linear indepeddent
Example4 p304
Fi d b i f th b V f
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n Find a basis for the subspace V ofP2 ,consisting of all vector of the format2+bt+c ,where c=a-b.
n at2+bt+a-b
n a(t2+1)+b(t-1)+cn So the t2+1and t-1 span V.
n a(t2+1)+b(t-1)=0
n Since this equation id to hold all values of t,
we must have a1=a2=0.
Example:
L
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Let
{ }321321 ,,and,
0
1
1
,
2
0
1
,
1
2
1
vvvSvvv =
=
=
=.
Are S a basis in3R ?
Solution:
( )
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(a)
3)( RSspan = For any vector3R
c
b
a
v
=, there
exist real numbers 321 ,, ccc such that
332211321
0
1
1
2
0
1
1
2
1
vcvcvcccc
c
b
a
v ++=
+
+
=
=
.
Solution:
d t l f th li t
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we need to solve for the linear system
=
c
b
a
c
c
c
3
2
1
021
102
111
.
The solution is
3
24
,3,3
22321
cba
c
cba
c
cba
c
--
=
+-
=
++-
= .
Solution:
Th
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Thus,
321 3
24
33
22
v
cba
v
cba
v
cba
v
--+
+-+
++-=
.
That is, every vector in3R can be a linear
combination of321 ,, vvv
and
3)( RSspan =.
Solution:
(b)
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(b)
Since
0
0
0
0
2
2 321
21
31
321
332211 ===
=
+
+
++
=++ ccc
cc
cc
ccc
vcvcvc,
321 ,, vvv are linearly independent.
By (a) and (b), 321 ,, vvv are a basis of3R .
Theorem 6.5:(Important result)
If { }
kvvvS 21 i b i f t V th
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If
{ }kvvvS ,,, 21 K= is a basis for a vector space V, thenevery vector in V can be written in an unique(one
and only one) way as a linear combination of the
vectors in S.
Example:
{ }
001
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{ }321321 ,,and,
1
0,
0
1,
0
0 eeeSeee =
=
=
=. S is a basis o
3R . Then, for any vector
=
c
b
a
v
,
321
1
0
0
0
1
0
0
0
1
cebeaecba
c
b
a
v ++=
+
+
=
=
is uniquely determined.
Theorem 6.6:(Importantresult)
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Let { }kvvvS ,,, 21 K= be a set of nonzero vectors in a
vector space V and let { }kvvvspanW ,,, 21 K= . Then,
some subset of S is a basis of W.
Example:
Let
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Let
{ }
==
2
3
1
,
1
0
0
,
0
3
2
,
0
1
0
,
0
0
1
,,,, 23121 aeaeeS
and ( )3RSspan = .Please find subsets of S which
form a basis of 3R .
Solution:
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We first check if 1e and 2e are linearly independent.
Since they are linearly independent, we continue to check i
1e
, 2e
and 1a
are linearly independent.Since032 121 =-+ aee ,
Solution:
we delete 1a from S and form a new set 1S , { }23211 ,,, aeeeS = .
Th i h k if d li l
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Then, we continue to check if 1e, 2e and 3e are linearly
independent.They are linearly independent. Thus, we finally
check if 1e, 2e 3e and 2a are linearly independent.Since
023 2321 =-++ aeee ,we delete
1a from S and form a new set 2S , { }3212 ,, eeeS = .
Therefore,
{ }3212 ,, eeeS =
is the subset of S which form a basis of form a basis of3R .
How to find a basis (subset of S) of W?
There are two methods:
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There are two methods:ethod 1:
The procedure based on the proof of the above
important result.
ethod 2:
Step 1: Form equation
02211 =+++ kk vcvcvc L .
How to find a basis (subset of S) of W?
Step 2: Construct the augmented matrix associated with the
eq ation in step 1 and transform this a gmented matri
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equation in step 1 and transform this augmented matrix
to the reduced row echelon form.
Step 3: The vectors corresponding to the columnscontaining the leading 1s form a basis. For example, i
6=k and the reduced row echelon matrix is
How to find a basis (subset of S) of W?
01
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0000000
0000000
01000
010001
MMMMMMM ,
then the 1st, the 3nd, and the 4th columns contain
a leading 1 and thus
431 ,, vvv are a basis of { }621 ,,, vvvspanW K= .
Example:
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Let
{ }
==
2
3
1
,
1
0
0
,
0
3
2
,
0
1
0
,
0
0
1
,,,, 23121 aeaeeS
and
( )
3
RSspan =
.Please find subsets of S whichform a basis of 3R .
Solution:
ethod1:
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We first check if 1e and 2e are linearly independent.Since they are linearly independent, we continue to check i
1e, 2e and 1a are linearly independent.Since
032 121 =-+ aee ,
Solution:
we delete 1a from S and form a new set 1S , { }
23211 ,,, aeeeS = .
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we delete a from S and form a new set ,
{ }.Then, we continue to check if 1e, 2e and 3e are linearly
independent.They are linearly independent. Thus, we
finally check if 1e, 2e 3e and 2a are linearly
independent.Since
023 2321 =-++ aeee ,
Solution:
we delete 1a from 1S and form a new set2S ,
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we delete from and form a new set ,
{ }3212 ,, eeeS = . Therefore,
{ }3212 ,, eeeS =
is the subset of S which form a basis of form a
basis of3R .
Solution:
ethod 2:
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Step 1:
The equation is
0
3
2
1
1
0
0
0
3
2
0
1
0
0
0
1
54321 =
+
+
+
+
ccccc
.
Solution:
Step 2:
The augmented matrix and its reduced row echelon matrix is
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The augmented matrix and its reduced row echelon matrix is
031000
020310
010201
.
The 1st, the 2nd and 4th columns contain the leading 1s.
Thus,
{ }321 ,, eee forms a basis.Example 5(Page 309)
Example5 p309
n S={ v1,v2,v3,v4,v5}, v1= (1,2,-2,1) ,
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S { v1,v2,v3,v4,v5}, v1 (1,2, 2,1) ,v2=(3-,0,-4,3), v3=(2,1,1,-1), v4=(-3,3,-9,6), v5=(9,3,7,-6).
n Find a subset of S that is a basis forW=span S.
n Step1:
n c1
(1,2,-2,1) + c2
(3-,0,-4,3)+ c3
(2,1,1,-1) + c4 (-3,3,-9,6) + c5(9,3,7,-6)=0
Step2:
331
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--
000000
000000
025
23
2110
02
3
2
3
2
101
Step3:
n The leading 1s appear in columns 1
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g ppand 2, so {v1,v2}is a basis for W= span
S.
Theorem 6.7: (Important result)
Let { }
nvvvS ,,, 21K=
be a basis for a vector space V and{ }
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{ } plet { }rwwwT ,,, 21 K= is a linear independent set o
vectors in V.Then, nr .
.
Corollary 6.1:
Let { }
nvvvS ,,, 21K=
and { }
mwwwT ,,, 21K=
be two
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{ },,,
{ }bases for a vector space V. Then, mn = .
Note:
For a vector space V, there are infinite bases. But
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pthe number of vectors in two different bases are
the same.
Example:
For the vector space
3
R ,111
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{ }321321 ,,,
0
1
1
,
2
0
1
,
1
2
1
vvvSvvv =
=
=
=is a basis
for3R (9094).
Example:
previous example). Also,
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{ }321321 ,,,
1
0
0
,
0
1
0
,
0
0
1
eeeTeee =
=
=
=is basis for
3R .
There are 3 vectors in both S and T.
Definition of dimension:
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The dimension of a vector space V is the number
of vectors in a basis for V. We often write dim(V
for the dimension of V. dim({0}) = 0.
Example:
{ }
001
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{ }321321 ,,,
1
0,
0
1,
0
0 eeeTeee =
=
=
=is basis for 3R .
The dimension of
3
R is 3.
Example 6--8(Page 310)
Example6 p310
n The dimension of R2 is 2
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n The dimension of R3 is 3
n The dimension of Rn is n.
Example7 p310
n The dimension of P2 is 3
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n The dimension of P3 is 4
n The dimension of Pn is n+1
Example8 p310
n The subspace w of R4 considered inE 5 h di i 2
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Ex5 has dimension 2.
Theorem 6.8: (Important result)
If S is a linearly independent set of vectors in a
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IfS is a linearly independent set of vectors in a
finite-dimensional vector space V, then there is a
basisTforV, which containsS.
Example 9(Page311)
Example9 p311
n Find the basis for R4 that contains thevector v =(1 0 1 0) and v =(-1 1 -1 0)
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vector v1=(1,0,1,0) and v2=(-1,1,-1,0).
n Let {e1, e2, e3, e4, } be the natural basis
for R
4
wheren e1=(1,0,0,0), e2=(0,1,0,0)
n e3=(0,0,1,0) e4=(0,0,0,0)
n
c1v1+ c2v2+ c3e1+ c4e2+ c5e3+ c6e4=0
Example9 p311
00010100011001
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-
01000000010100
0001010
Since leading 1s appear in columns 1,2,3 and 6 we
conclude that {v1,v2,e1.e4} is a basis for R4 containing
v1and v2
Theorem 6.9: (Important result)
Let V be an n-dimensional vector space, and let
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{ }nvvvS ,,, 21 K= be a set of n vectors in V.
(a) If S is linearly independent, then S is a basis for
V.
(b) If S spans V, then S is a basis for V.
Example:
Is{ }321321 ,,,1
1
,0
1
,2
1
vvvSvvv =
=
=
=a basis
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Is 021
a basis
for
3
R ?
Note:
n
.
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.
Solution:
Since
3
R is a 3-dimensional vector space, not likein the previous example we only need to examine
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in the previous example, we only need to examine
whether S is linearly independent or S spans3R .
We dont need to examine S being both linearly
independent and spans V.
Example:
421
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Is
{ }321321 ,,,
1
2,
0
1,
1
3 vvvSvvv =
=
=
-
=a basis for
3R ?
Solution:
Since3R is a 3-dimensional vector space, we only need to
examine whether S is linearly independent or S spans 3.Because
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0
0
0
0
23
42
321
31
321
321
332211 ===
=
+-
++
++
=++ ccc
cc
ccc
ccc
vcvcvc
,
321 ,, vvv are linearly independent. Therefore, 321 ,, vvv are a
basis of
3
R .Example 10(Page 312)
Example10 p312
n In Ex5. W= span S is a subspace of R4,sodim W4. Since S contains five vectors we
l d b C ll 6 1 th t S i t
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conclude by Corollary 6.1 that S is not abasis for W. In Ex2, since dim R4 =4, and
the set S contains four vectors, it is possiblefor S to be a basis for R4.If S is linearlyindependent or spans R4,it is a basisotherwise it is not a basis. Thus we needonly check one of the conditions in Thm 9.6not both.