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CHAPTER 4. VECTOR SPACES
4.4
Linear Independence
We now turn to the problem of …nding an "optimal" spanning set for a given
vector space V that is a set with the smallest number of vectors that still spans
V . We begin with the following observation, which partly answers one of the
questions in the previous section.
Theorem 284 Let V denote a vector space and S = fu1 ; u2 ; :::; un g a spanning
set of V . Suppose further that one of the vectors of S is a linear combination
of the others. If we remove that vector from S, the resulting set still spans V .
Proof. Without loss of generality, we may assume that un is a linear combination of the other vectors of S. In other words, there exists scalars c1 ; c2 ; :::; cn 1
such that
un = c1 u1 + c2 u2 + ::: + cn 1 un 1
Given v 2 V;we need to prove that v 2 span (u1 ; u2 ; :::; un 1 ) in other words, we
need to prove that v can be expressed as a linear combination of u1 ; u2 ; :::; un 1 .
Since S = fu1 ; u2 ; :::; un g a spanning set of V , it follows that
v = d1 u1 + d2 u2 + ::: + dn un
for some scalars d1 ; d2 ; :::; dn . Since un = c1 u1 +c2 u2 +:::+cn
that
v
1 un 1 ,
it follows
= d1 u1 + d2 u2 + ::: + dn (c1 u1 + c2 u2 + ::: + cn 1 un 1 )
= (d1 + dn c1 ) u1 + (d2 + dn c2 ) u2 + ::: + (dn 1 + dn cn 1 ) un
Thus, v can be expressed as a linear combination of u1 ; u2 ; :::; un
1
1.
The set S in the theorem was not an "optimal" spanning set in the sense that
one of its vectors was redundant. The notion of linear independence addresses
this.
De…nition 285 Let V denote a vector space and S = fu1 ; u2 ; :::; un g be a
subset of V .
1. S is said to be linearly independent if
c1 u1 + c2 u2 + ::: + cn un = 0 () c1 = c2 = ::: = cn = 0
2. If c1 u1 + c2 u2 + ::: + cn un = 0 has a non-trivial solution, then S is said
to be linearly dependent.
To determine if a set S = fu1 ; u2 ; :::; un g is linearly independent, do the
following:
1. From the vector equation c1 u1 + c2 u2 + ::: + cn un = 0 form a system of
linear equations
4.4. LINEAR INDEPENDENCE
131
2. Using the techniques studied in the course, solve the system.
3. If the system only has the trivial solution, then it is linearly independent.
Otherwise it is linearly dependent.
You will recall that for the system to have only the trivial solution, the
coe¢ cient matrix must be invertible, that is have a determinant not equal to
zero.
Example 286 Let R3 be the underlying vector space. Is S = f(1; 1; 0) ; (0; 1; 0) ; (0; 0; 1)g
linearly dependent?
For this, we try to solve the system c1 (1; 1; 0) + c2 (0; 1; 0) + c3 (0; 0; 1) = 0. This
system is equivalent to
8
c1 = 0
<
c1 + c2 = 0
:
c3 = 0
The solution is
c1 = 0
c2 = 0
c3 = 0
Therefore, S is linearly independent.
Example 287 Let R3 be the underlying vector space. Is the set fv1 ; v2 ; v3 g
where v1 = (1; 1; 1), v2 = (1; 1; 0) and v3 = (0; 0; 1) linearly dependent?
For this, we try to solve the system c1 (1; 1; 1) + c2 (1; 1; 0) + c3 (0; 0; 1) = 0. This
system is equivalent to
8
< c1 + c2 + c3 = 0
c1 + c2 = 0
:
c3 = 0
The solutions are
c1
c3
=
=
c2
0
So the system is linearly dependent.
Example 288 Show that the polynomials p0 = 1, p1 = x, p2 = x2 , ::: pn = xn
form an independent set in Pn .
Consider the solution of
c0 p0 + c1 p1 + ::: + cn pn = 0
that is
c0 + c1 x + c2 x2 + ::: + cn xn = 0 for all x in ( 1; 1)
Since a polynomial of degree n has at most n distinct roots, we must have c0 =
c1 = c2 = ::: = cn = 0 otherwise c0 + c1 x + c2 x2 + ::: + cn xn = 0 would be a
polynomial of degree n with more than n distinct roots.
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CHAPTER 4. VECTOR SPACES
Theorem 289 Let V denote a vector space and S = fu1 ; u2 ; :::; un g be a subset
of V , assume that n 2. Then S is linearly dependent if and only if one of the
vectors of S can be expressed as a linear combination of the others.
Proof. We prove both directions.
(=)) Here, we assume that S is linearly dependent. Then, there exists
scalars c1 ; c2 ; :::; cn , not all zero, such that c1 u1 + c2 u2 + ::: + cn un = 0.
Without loss of generality, we may assume that c1 6= 0. Then, we have
c1 u1 + c2 u2 + ::: + cn un
c1 u1
u1
= 0
=
c2 u2
c2
u2
=
c1
:::
cn un
cn
:::
un
c1
So, we see that u1 is a linear combination of the other vectors.
((=) Here, we assume that one of the vectors of S is a linear combination
of the others. Without loss of generality, we may assume that u1 is a linear
combination of the other vectors. Then, we have
u1 = d2 u2 + ::: + dn un
Therefore,
u1 + d2 u2 + ::: + dn un = 0
and at least one of the coe¢ cients (the coe¢ cient of u1 ) is not zero. It
follows that S is linearly dependent.
Corollary 290 Two vectors are linearly dependent if and only if one is a scalar
multiple of the other.
Proof. See exercises below.
We can use the theorem to redo an example we did above.
Example 291 Let R3 be the underlying vector space. Is the set fv1 ; v2 ; v3 g
where v1 = (1; 1; 1), v2 = (1; 1; 0) and v3 = (0; 0; 1) linearly dependent?
We note that v1 = v2 +v3 , this means one of the vectors is a linear combination
of the other two, hence the system is linearly dependent.
Remark 292 It should be obvious to the reader that if one of u1 ; u2 ; :::; un
is the zero vector, that the set S = fu1 ; u2 ; :::; un g will be linearly dependent.
Suppose that ui = 0 then we can always have c1 u1 + c2 u2 + ::: + cn un = 0 with
ck = 0 except for k = i.
Remark 293 The term linearly dependent suggests that the vectors depend on
one another, meaning that one can be written in terms of others. This is exactly
what the theorem told us.
4.4. LINEAR INDEPENDENCE
133
One might ask, is the fact that a set of vector is linearly independent or not
only a matter of what vectors it contains, or does it depend on other factors
too? The next theorem tell us that in Rn the size of the set is important.
Theorem 294 Let S = fu1 ; u2 ; :::; ur g be a set of vectors in Rn . If r > n then
S must be linearly dependent.
Proof. If one writes the system corresponding to c1 u1 + c2 u2 + ::: + cr ur , one
will have a homogeneous system of n equations in r unknowns. We know that
such a system has in…nitely many solutions.
Thus, in Rn , a set which is linearly independent cannot contain more than
n vectors. However, it does not mean that every set containing n vectors or
less will be linearly independent. For example, in R2 , any set of more than two
vectors is linearly dependent. In R3 , any set with more than three vectors is
linearly dependent.
Example 295 The set f(1; 0) ; (2; 4) ; (5; 1)g is linearly dependent because it is
a set of three vectors in R2 .
4.4.1
Problems
1. Do # 1, 2, 4, 7, 9, 10, 12, 15, 16, 18 on pages 248, 249.
2. Prove corollary 290
3. Let V be a vector space and S a subset of V . Answer the questions below.
Either prove your statement, or give a counter-example.
(a) If S is linearly independent, and span (S) = V , can we remove a
vector from S and have S remain a spanning set for V ? If yes, which
vector can be removed?
(b) If S is linearly dependent, and span (S) = V , can we remove a vector
from S and have S remain a spanning set for V ? If yes, which vector
can be removed?