Filetype PDF | Posted on 27 Jan 2023 | 2 years ago
Authentication
digital resources
134x Filetype PDF File size 0.07 MB Source: www.wright.edu
1 Section 1.7. Linear Independence
De
nition 1 Asetofpvectors~u ;~u ;:::;~u in Rm is called linearly independent if the vector
1 2 p
equation
~
x1~u1 + x2~u2 + ::: + xp~up = 0
has only the trivial solution x = 0; i = 1;2;:::;p: Otherwise, the set is called linearly depen-
i
dent; the coe¢ cients x1;:::;xn are called a linear relation.
Example 2 Anyonesingle vector ~u is always independent. two vectors ~u ;~u are dependent
1 2
i¤ ~u1 = ~u2.
The notation of linear dependence (or independence) is closely related to homogeneous
systems. In fact, let A = [~a ;~a ;:::;~a ] be a m n matrix with column vectors ~a ; then we
1 2 n i
have the following relation:
~
Claim 3 A~x = 0 has NO non-trivial solution i¤ its column vectors ~a ;~a ;:::;~a are linearly
1 2 n
independent.
Example 4 (1) Determine whether the set of the following three vectors is dependent:
2 1 3 2 2 3 2 4 3
~u =425; ~u =4 1 5; ~u =4 5 5:
1 2 3
3 0 6
(2) Find a linear relation.
~
Solution. (1) We need to determine if A~x = 0 has a non-trivial solution, where A =
[~u ; ~u ; ~u ]
1 2 3 2 3
1 2 4
A=42 1 5 5:
3 0 6
h ~i
To this end, we perform row operation on the augmented matrix A;0 :For simplicity, we
only work on A :
2 1 2 4 3 h ~i 2 1 2 4 0 3
A!40 3 3 5; since A;0 !4 0 3 3 0 5: (1)
0 0 0 0 0 0 0
~
Apparently, x3 is a free variable. Thus, A~x = 0 has non-trivial solution, and the set
f~u ;~u ;~u g is linearly dependent.
1 2 3
Solution. (2) Finding a linear relation ~u ;~u ;~u , i.e.,
1 2 3
~
x1~u1 + x2~u2 + x3~u3 = 0
1
means to
nd a non-trivial solution of
~
A~x = 0:
~
From (1), we
nd that A~x = 0 reduces to
x1 +2x2 +4x3 = 0
3x2 x3 =0:
Since we only need one non-zero solution, we take x3 = 3 (we could have choose any number
here. The only reason I chose 3 is to avoid fraction.) Then x2 = 1; x1 = (2x2 +4x3) =
10; and ~
10~u1 ~u2 +3~u3 = 0:
From the above equation, we see that, for instance,
~u2 = 10~u1 + 3~u3;
i.e., when f~u ;~u ;~u g is linearly dependent, ~u is a linear combination of f~u ;~u g: In general,
1 2 3 2 1 3
Theorem 5 (Characterization of linear dependence) Vectors ~u ;~u ;:::;~u are linearly de-
1 2 p
pendent i¤ at least one is a linear combination of the rest.
Proof. Suppose that one is a linear combination of the rest. Without loss of generality, we
say that ~u is a linear combination of ~u ;:::;~u ; i.e., there exist ;:::; such that
1 2 p 2 p
~u =~u +:::+ ~u :
1 2 2 p p
It follows that
~u + ~u +:::+ ~u = 0:
1 2 2 p p
This implies that ~u ;~u ;:::;~u are linearly dependent. On the other hand, suppose that
1 2 p
~u ;~u ; :::; ~u are linearly dependent. Then, we have
1 2 p
~u + ~u +:::+ ~u = 0;
1 1 2 2 p p
where at least one of p constants f ; ;:::; g is not zero. Without loss of generality, we
1 2 p
say 6= 0: Then we can solve for ~u as
1 1
~u = 2 ~u +:::+ p ~u :
1 2 p
1 1
Hence, ~u is a linear combination of ~u ;:::;~u ; the rest.
1 2 p
Notethatwhen~u ;~u ;:::;~u are linearly dependent, it does NOT mean ANY one member
1 2 p
is a linear combination of the rest. For instance,
1 ; 1 ; 1 are linearly dependent.
2 2 0
2
But
1 is not a linear combination of 1 ; 1 :
0 2 2
In summary, to determine whether ~u ;~u ;:::;~u in Rm are linearly independent, we need
1 2 p
~
to see whether the system A~x = 0 has only trivial solution, where A is the m p column
matrix [~u ;~u ;:::;~u ]:
1 2 p
De
nition 6 We call the number of pivots of a mp matrix A the RANK of A; and denote
it by r (A):
Obviously, the rank cannot exceed the number of rows or columns, i.e., r(A) m;
r(A) p:
The following theorem follows directly from the last conclusion in Section 1.5.
Theorem 7 A set of vectors ~u ;~u ;:::;~u in Rm is linearly dependent i¤ p > r(A); where
1 2 p
Ais the mp column matrix [~u ;~u ;:::;~u ]: In particular, it is linearly dependent if p > m;
1 2 p
i.e., the number of vectors is more than the dimension.
Example 8 Determine whether the following set is linearly dependent. If it is linearly
dependent
nd a set of linearly independent vectors ~v1;~v2;::: such that Spanf~v1;~v2;::g =
Spanf~u ;~u ;::g:
1 2
2 1 3 2 4 3 2 2 3 2 1 3
~u =425; ~u =4 5 5; ~u =4 1 5; ~u =4 1 5:
1 2 3 4
3 6 0 1
Solution. The answer is yes, it is linearly dependent because the number of vectors is
great than the dimension.
~
Wenext describe all vectors b 2 Spanf~u ;~u ;::g using parametric vector representation.
1 2
~ ~
AswedidinExample??,weneedtodescribebsuchthat[~u ;~u ;~u ;~u ] ~x = b has a solution.
h i1 2 3 4
~
To this end, we perform row operation on ~u ;~u ;~u ;~u ;b to arrive at a Echelon form:
1 2 3 4
2 3 2 3
1 4 2 1 b 1 4 2 1 b
1 R 2R !R 1
4 5 2 1 2 4 5
2 5 1 1 b 0 3 3 1 b 2b
2 R 3R !R 2 1
3 6 0 1 b 3 1 3 0 6 6 2 b 3b
3 ! 3 1
2 3
1 4 2 1 b
1
4 5
R 2R !R 0 3 3 1 b 2b :
3 2 3 2 1
! 0 0 0 0 b 2b +b
1 2 3
The system is consistent i¤ b 2b +b = 0; or
1 2 3
b =2b b :
1 2 3
3
Therefore, 2 3 2 3 2 3 2 3
b 2b b 2 1
1 2 3
~ 4 5 4 5 4 5 4 5
b = b = b =b 1 +b 0 ;
2 2 2 3
b b 0 1
3 3
and the set of 2 3 2 3
2 1
4 5 4 5
~v1 = 1 ; ~v2 = 0
0 1
is linearly independent, and Spanf~v ;~v g = Spanf~u ;~u ;~u ;~u g:
Homework 1.7: 1 2 1 2 3 4
#5,11,21,23,27,33,37
4
no reviews yet
Please Login to review.
