Finding Vectors That Span a Solution Set
Linear Independence and Span
Span
We have seen in the last discussion that the span of vectors v1 , v2 , ... , v n is the set of linear combinations
c1 v1 + c2 v2 + ... + cn v n
and that this is a vector space.
We now take this idea further. If V is a vector space and S = {v1 , v2 , ... , v n ) is a subset of V, then is Span(S) equal to V?
Definition
Let V be a vector space and let S = {v1, v2, ... , vn) be a subset of V. We say that S spans V if every vector v in V can be written as a linear combination of vectors in S .
v = c1v1 + c2v2 + ... + cnvn
Example
Show that the set
S = {(0,1,1), (1,0,1), (1,1,0)}
spans R3 and write the vector (2,4,8) as a linear combination of vectors in S.
Solution
A vector in R3 has the form
v = (x, y, z)
Hence we need to show that every such v can be written as
(x,y,z) = c1(0, 1, 1) + c2(1, 0, 1) + c3(1, 1, 0)
= (c2 + c3, c1 + c3, c1 + c2)
This corresponds to the system of equations
c2 + c3 = x
c1 + c3 = y
c1 + c2 = z
which can be written in matrix form
We can write this as
Ac = b
Notice that
det(A) = 2
Hence A is nonsingular and
c = A-1 b
So that a nontrivial solution exists. To write (2,4,8) as a linear combination of vectors in S, we find that
so that
We have
(2,4,8) = 5(0,1,1) + 3(1,0,1) + (-1)(1,1,0)
Example
Show that if
v1 = t + 2 and v2 = t2 + 1
and S = {v1 , v2 }
then
S does not span P2
Solution
A general element of P2 is of the form
v = at2 + bt + c
We set
v = c1 v1 + c2 v2
or
at2 + bt + c = c1(t + 2) + c2(t2 + 1) = c2t2 + c1t + c1 + c2
Equating coefficients gives
a = c2
b = c1
c = c1 + c2
Notice that if
a = 1 b = 1 c = 1
there is no solution to this. Hence S does not span V.
Example
Let
Find a spanning set for the null space of A.
Solution
We want the set of all vectors x with
Ax = 0
We find that the rref of A is
The parametric equations are
x1 = 7s + 6t
x2 = -4s - 5t
x3 = s
x4 = t
We can get the span in the following way. We first let
s = 1 and t = 0
to get
v1 = (7,-4,1,0)
and let
s = 0 and t = 1
to get
v2 = (6,-5,0,1)
If we let S = {v1 ,v2 } then S spans the null space of A.
Linear Independence
We now know how to find out if a collection of vectors span a vector space. It should be clear that if S = {v1 , v2 , ... , v n ) then Span(S) is spanned by S. The question that we next ask is are there any redundancies. That is, is there a smaller subset of S that also span Span(S). If so, then one of the vectors can be written as a linear combination of the others.
vi = c1 v1 + c2 v2 + ... + ci -1 vi -1 + ci+1 vi+1 + ... + cn v n
If this is the case then we call S a linearly dependent set. Otherwise, we say that S is linearly independent. There is another way of checking that a set of vectors are linearly dependent.
Theorem
Let S = {v1, v2, ... , vn) be a set of vectors, then S is linearly dependent if and only if 0 is a nontrivial linear combination of vectors in S . That is, there are constants c1, ..., cn with at least one of the constants nonzero with
c1 v1 + c2 v2 + ... + cn v n = 0
Proof
Suppose that S is linearly dependent, then
vi = c1 v1 + c2 v2 + ... + ci -1 vi -1 + ci+1 vi+1 + ... + cn v n
Subtracting vi from both sides, we get
c1 v1 + c2 v2 + ... + ci -1 vi -1 + vi + ci+1 vi+1 + ... + cn v n = 0
In the above equation ci = 1 which is nonzero, so that 0 is a nontrivial linear combination of vectors in S.
Now let
c1 v1 + c2 v2 + ... + ci -1 vi -1 + ci vi + ci+1 vi+1 + ... + cn v n = 0
with ci nonzero. Divide both sides of the equation by c i and let aj = -cj / ci to get
-a1 v1 - a2 v2 - ... - ai -1 vi -1 + vi- ai+1 vi+1 - ... - an v n = 0
finally move all the terms to the other right side of the equation to get
vi = a1 v1 + a2 v2 + ... + ai -1 vi -1 + ai+1 vi+1 + ... + an v n
Example
Show that the set of vectors
S = {(1, 1, 3, 4), (0, 2, 3, 1), (4, 0, 0, 2)}
are linearly independent.
Solution
We write
c1(1, 1, 3, 4) + c2(0, 2, 3, 1) + c3(4, 0, 0, 2) = 0
We get four equations
c1 + 4c3 = 0
c1 + 2c2 = 0
3c1 + 3c2 = 0
4c1 + c2 + 2c3 = 0
The matrix corresponding to this homogeneous system is
and
Hence
c1 = c2 = c3 = 0
and we can conclude that the vectors are linearly independent.
Example
Let
S = {cos2 t, sin2 t, 4)
then S is a linearly dependent set of vectors since
4 = 4cos2t + 4sin2t
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Finding Vectors That Span a Solution Set
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