qualify for basis
DESCRIPTION
quality for basisTRANSCRIPT
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Basis and Dimensions of Vector Spaces
A set of vectors S = {v1, v2, , vn} are said to form a basis if the following conditions are met. 1. The set S spans all of their vector space. 2. The set S is linearly independent. To show that a set of vectors for a basis, just show that they span all of their dimension OR show that they are linearly independent. Both concepts imply each other because they both require the use of the determinant. Example: Determine if the vectors (2,2,2), (0,0,3), and (0,1,1) form a basis for R3. Spanning: We must be able to write every vector as a linear combination of the vectors given to us. Let v = (a, b, c) be an arbitrary vector in R3. Thus, we have the following system of equations:
( )( )
( )
++=
+=
=
+++=
++=
++=
321
31
1
321311
332111
321
32
2
2
)32,2,2(,,
),,0()3,0,0()2,2,2(,,
)1,1,0()3,0,0()2,2,2(,,
cccc
ccb
ca
cccccccba
cccccccba
ccccba
Using Gaussian Elilmination, we have the following:
ab
bc
a
ab
ac
a
ab
ac
a
ac
ab
a
ac
ab
a
c
b
a
RRR
RRR
RR
RR
3/3/
2/
100
010
001
3/3/
2/
100
3/110
001
2/
100
130
0012/
130
100
001
130
100
002
132
102
002
33
122
3
1
231
2
1
13122
So we have the equations c1 = a/2, c2 = c/3 b/3, and c3 = b a. For any value of a, b, c, the system always remain consistent. Therefore, the vectors span R3.
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Linear Independence: We must be able to write the zero vector in R3 as a linear combination of the vectors given to us and the ALL constants must be equal to zero. Thus, we have the following equation:
( )( )
( )
++=
+=
=
+++=
++=
++=
321
31
1
321311
332111
321
320
20
20
)32,2,2(0,0,0
),,0()3,0,0()2,2,2(0,0,0
)1,1,0()3,0,0()2,2,2(0,0,0
ccc
cc
c
cccccc
cccccc
ccc
Using Gaussian Elilmination, we have the following:
0
0
0
100
010
001
0
0
0
100
3/110
001
0
0
0
100
130
001
0
0
0
130
100
001
0
0
0
130
100
002
0
0
0
132
102
002
33
122
3
1
231
2
1
13122
RRR
RRR
RR
RR
So we have the equations c1 = 0, c2= 0, and c3 = 0. Clearly, we can see that c1 = c2 = c3 = 0. Therefore, the vectors are linearly independent. We can also use the determinant to show that these vectors form a basis in R3. The coefficient matrix associated with the system is given by. Computing the determinant using cofactor expansion yields:
06)2)(3(12
02)3()1(
132
102
00223 === +
Therefore, the vectors form a basis for R3
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Next, we talk about the dimensions of different types of vectors. The number of vectors in any set given determines if it will be a basis for its vector space.
Vector Object Dimension of Vector Space
Vectors (x1, x2, , xn)
n
Matrices
mnmm
n
n
aaa
aaa
aaa
K
MOMM
K
K
21
22221
11211
mn
Polynomials a0 + a1x + a2x
2 + + a1xn
n + 1