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5.7 Vectors; Linear Dependence and Linear Independence
In Section 5.4 vectors were defined either as 1 × n matrices, which we called row vectors, or as m × 1
matrices, which we called column vectors. In this section we want to think of vectors as simply
being ordered n-tuples of real numbers. Most of the time will write vectors as a row vectors, but
occasionally it will be convenient to write them as columns. Our main interest is in the concept of
linear dependence/linear independence of sets of vectors. At the end of the section we will extend
this concept to linear dependence/independence of sets of functions.
We denote the set of ordered n-tuples of real numbers by the symbol R
n
. That is,
R
n = {(a 1 , a 2 , a 3 ,... , an) | a 1 , a 2 , a 3 ,... , an are real numbers}
In particular, R 2 = {(a, b) | a, b real numbers}, which we can identify with the set of all points in
the plane, and R
3 = {(a, b, c) | a, b, c real numbers}, which we can identify with the set of points
in space. We will use lower case boldface letters to denote vectors. The entries of a vector are called
the components of the vector.
The operations of addition and multiplication by a number (scalar) that we defined for matrices
in Section 5.4 hold automatically for vectors since a vector is a matrix. Addition is defined only for
vectors with the same number of components. For any two vectors u = (a 1 , a 2 , a 3 ,... , an) and
v = (b 1 , b 2 , b 3 ,... , bn) in R n , we have
u + v = (a 1 , a 2 , a 3 ,.. ., an) + (b 1 , b 2 , b 3 ,... , bn) = (a 1 + b 1 , a 2 + b 2 , a 3 + b 3 ,... , an + bn)
and for any real number λ,
λ v = λ (a 1 , a 2 , a 3 ,... , an) = (λa 1 , λa 2 , λa 3 ,... , λan).
Clearly, the sum of two vectors in R
n is another vector in R
n and a scalar multiple of a vector in
R
n is a vector in R n
. A sum of the form
c 1 v 1 + c 2 v 2 + · · · + ckvk
where v 1 , v 2 ,... , vk are vectors in R
n and c 1 , c 2 ,... , ck are real numbers is called a linear
combination of v 1 , v 2 ,... , vk.
The set R n together with the operations of addition and multiplication by a number is called
a vector space of dimension n. The term “dimension n” will become clear as we go on.
The zero vector in R
n , which we’ll denote by 0 is the vector
For any vector v ∈ R
n , we have v + 0 = 0 + v = v; the zero vector is the additive identity in R
n .
Linear Dependence and Linear Independence
In Chapter 3 we said that two functions f and g are linearly dependent if one of the functions
is a scalar multiple of the other; f and g are linearly independent if neither is a scalar multiple
of the other. Similarly, let u and v be vectors in R
n
. Then u and v are linearly dependent
if one of the vectors is a scalar multiple of the other (e.g., u = λv for some number λ); they are
linearly independent if neither is a scalar multiple of the other.
Suppose that u, v ∈ R n are linearly dependent with u = λv. Then
u = λ v implies u − λ v = 0.
This leads to an equivalent definition of linear dependence: u and v are linearly dependent if there
exist two numbers c 1 and c 2 , not both zero, such that
c 1 u + c 2 v = 0.
Note that if c 1 u + c 2 v = 0 and c 1 6 = 0, then u = (c 2 /c 1 )v = λv.)
This is the idea that we’ll use to define linear dependence/independence in general.
DEFINITION 1. The set of vectors {v 1 , v 2 ,... , vk } in R n is linearly dependent if there exist
k numbers c 1 , c 2 ,... , ck , not all zero, such that
c 1 v 1 + c 2 v 2 + · · · + ck vk = 0 ;
0 is a non-trivial linear combination of v 1 , v 2 ,... , vk. The set of vectors is linearly independent
if it is not linearly dependent.
NOTE: If there exists one set of k numbers c 1 , c 2 ,... , ck, not all zero, then there exist infinitely
many such sets. For example, 2 c 1 , 2 c 2 ,... , 2 ck is another such set; and
1 3 c 1 ,
1 3 c 2 ,... ,
1 3 ck is
another such set; and so on.
The definition of linear dependence can also be stated as: The vectors v 1 , v 2 ,... , vk are
linearly dependent if one of the vectors can be written as a linear combination of the others. For
example if
c 1 v 1 + c 2 v 2 + · · · + ckvk = 0
and c 1 6 = 0, then
v 1 = −
c 2
c 1
v 2 −
c 3
c 1
v 3 − · · · −
ck
c 1
vk = λ 2 v 2 + λ 3 v 3 + · · · + λk vk.
This form of the definition parallels the definition of the linear dependence of two vectors.
Stated in terms of linear independence, the definition can be stated equivalently as:
The set of vectors {v 1 , v 2 ,... , vk} is linearly independent if
c 1 v 1 + c 2 v 2 + · · · + ck vk = 0 implies c 1 = c 2 = · · · = ck = 0;
i.e., the only linear combination of v 1 , v 2 ,... , vk which equals 0 is the trivial linear
combination. The set of vectors is linearly dependent if it is not linearly independent.
Writing the augmented matrix and reducing to row echelon form, we get
The row echelon form implies that system (A) has infinitely nontrivial solutions. Thus, we can find
three numbers c 1 , c 2 , c 3 , not all zero, such that
c 1 v 1 + c 2 v 2 + c 3 v 3 = 0.
In fact, we can find infinitely many such sets c 1 , c 2 , c 3. Note that the vectors v 1 , v 2 , v 3 appear
as the columns of the coefficient matrix in system (A).
Method 2 Form the matrix with rows v 1 , v 2 , v 3 and reduce to echelon form:
The row of zeros indicates that the zero vector is a nontrivial linear combination of v 1 , v 2 , v 3.
Thus the vectors are linearly dependent.
Method 3 Calculate the determinant of the matrix of coefficients:
∣ ∣ ∣ ∣ ∣ ∣ ∣
1 2 7
Therefore, as we saw in the preceding section, system (A) has infinitely many nontrivial solutions.
(d) Determine whether the vectors v 1 = ( 1, 2 , −3 ), v 2 = ( 1, − 3 , 2 ), v 3 = ( 2, − 1 , 5 ) are
linearly dependent or linearly independent.
SOLUTION In part (c) we illustrated three methods for determining whether or not a set of vectors
is linearly dependent or linearly independent. We could use any one of the three methods here. The
determinant method is probably the easiest. Since a determinant can be evaluated by expanding
across any row or down any column, it does not make any difference whether we write the vectors
as rows or columns. We’ll write them as rows.
∣ ∣ ∣ ∣ ∣ ∣ ∣
1 2 − 3
Since the determinant is nonzero, the only solution to the vector equation
c 1 v 1 + c 2 v 2 + c 3 v 3 = 0.
is the trivial solution; the vectors are linearly independent.
Example 3. Determine whether the vectors v 1 = ( 1, 2 , −4 ), v 2 = ( 2, 0 , 5 ), v 3 = ( 1, − 1 , 7 ), v 4 =
( 2, − 2 , −6 ) are linearly dependent or linearly independent.
SOLUTION In this case we need to determine whether or not there are four numbers c 1 , c 2 , c 3 , c 4 ,
not all zero such that
c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = 0.
Equating the components of the vector on the left and the vector on the right, we get the homoge-
neous system of equations
c 1 + 2c 2 + c 3 + 2c 4 = 0
2 c 1 − c 3 − 2 c 4 = 0
− 4 c 1 + 5c 2 + 7c 3 − 6 c 4 = 0
We know that this system has nontrivial solutions since it is a homogeneous system with more
unknowns than equations. Thus the vectors must be linearly dependent.
Example 4. Vectors in R
4
Let v 1 = ( 2, 0 , −1 4), v 2 = ( 2, − 1 , 0 2), v 3 = ( − 2 , 4 , −3 4), v 4 = ( 1, − 1 , 3 0), v 5 =
(a) Determine whether the vectors v 1 , v 2 , v 3 , v 4 , v 5 are linearly dependent of linearly indepen-
dent.
SOLUTION The vectors are linearly dependent because the vector equation
c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 + c 5 v 5 = 0.
leads to a homogeneous system with more unknowns than equations.
(b) Determine whether the vectors v 1 , v 2 , v 3 , v 4 are linearly dependent or linearly independent.
SOLUTION To test for dependence/independence in this case, we have three options.
- Solve the system of equations
c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = 0.
A nontrivial solution implies that the vectors are linearly dependent; if the trivial solution is
the only solution, then the vectors are linearly independent.
- Form the matrix A having v 1 , v 2 , v 3 , v 4 as the rows and reduce to row-echelon form. If
the row-echelon form has one or more rows of zeros, the vectors are linearly dependent; four
nonzero rows means the vectors are linearly independent.
- Calculate det A. det A = 0 implies that the vectors are linearly dependent; det A 6 = 0
implies that the vectors are linearly independent.
Options 1 and 2 are essentially equivalent; the difference being that in option 1 the vectors appear
as columns. Option 2 requires a little less writing so we’ll use it.
A =
(verify this).
a 11 x 1 + a 12 x 2 + a 13 x 3 + · · · + a 1 nxn = b 1
a 21 x 1 + a 22 x 2 + a 23 x 3 + · · · + a 2 nxn = b 2
a 31 x 1 + a 32 x 2 + a 33 x 3 + · · · + a 3 nxn = b 3
am 1 x 1 + am 2 x 2 + am 3 x 3 + · · · + amnxn = bm
Note that we can write this system as the vector equation
x 1
a 11
a 21
. . .
am 1
a 12
a 22
. . .
am 2
a 1 n
a 2 n
. . .
amn
b 1
b 2
. . .
bm
which is
x 1 v 1 + x 2 v 2 + · · · + xnvn = b
where
v 1 =
a 11
a 21
. . .
am 1
, v 2 =
a 12
a 22
. . .
am 2
,... , vn =
a 1 n
a 2 n
. . .
amn
and b =
b 1
b 2
. . .
bm
Written in this form, the question of solving the system of equations can be interpreted as asking
whether or not the vector b can be written as a linear combination of the vectors v 1 , v 2 ,... , vn.
As we know, b may be written uniquely as a linear combination of v 1 , v 2 ,... , vn (the system
has a unique solution); b may not be expressible as a linear combination of v 1 , v 2 ,... , vn
(the system has no solution); or it may be possible to represent b as a linear combination of
v 1 , v 2 ,... , vn in infinitely many different ways (the system has infinitely many solutions).
Linear Dependence and Linear Independence of Functions
As we saw in Chapter 3, two functions, f and g, are linearly dependent if one is a multiple of
the other; otherwise they are linearly independent.
DEFINITION 2. Let f 1 , f 2 , f 3 ,... , fn be functions defined on an interval I. The functions
are linearly dependent if there exist n real numbers c 1 , c 2 ,... , cn, not all zero, such that
c 1 f 1 (x) + c 2 f 2 (x) + c 3 f 3 (x) + · · · + cnfn(x) ≡ 0;
that is,
c 1 f 1 (x) + c 2 f 2 (x) + c 3 f 3 (x) + · · · + cnfn(x) = 0 for all x ∈ I.
Otherwise the functions are linearly independent.
Equivalently, the functions f 1 , f 2 , f 3 ,... , fn are linearly independent if
c 1 f 1 (x) + c 2 f 2 (x) + c 3 f 3 (x) + · · · + cnfn(x) ≡ 0
only when c 1 = c 2 = · · · = cn = 0.
Example 5. Let f 1 (x) = 1, f 2 (x) = x, f 3 (x) = x
2 on I = (−∞, ∞). Show that f 1 , f 2 , f 3 are
linearly independent.
SOLUTION Suppose that the functions are linearly dependent. Then there exist three numbers
c 1 , c 2 , c 3 , not all zero, such that
c 1 1 + c 2 x + c 3 x
2 ≡ 0. (B)
Method 1 Since (B) holds for all x, we’ll let x = 0. Then we have
c 1 + c 2 (0) + c 3 (0) = 0 which implies c 1 = 0.
Since c 1 = 0, (B) becomes
c 2 x + c 3 x
2 ≡ 0 or x [c 2 + c 3 x] ≡ 0.
Since x is not identically zero, we must have c 2 + c 3 x ≡ 0. Letting x = 0, we have c 2 = 0.
Finally, c 1 = c 2 = 0 implies c 3 = 0. This contradicts our assumption that f 1 , f 2 , f 3 are linearly
dependent. Thus, the functions are linearly independent.
Method 2 Since (B) holds for all x, we’ll evaluate at x = 0, x = 1, x = 2. This yields the
system of equations
c 1 = 0
c 1 + c 2 + c 3 = 0
c 1 + 2c 2 + 4c 3 = 0.
It is easy to verify that the only solution of this system is c 1 = c 2 = c 3 = 0. Thus, the functions
are linearly independent.
Method 3 Our functions are differentiable, so we’ll differentiate (B) twice to get
c 1 1 + c 2 x + c 3 x
2 ≡ 0
c 2 + 2c 3 x ≡ 0
2 c 3 ≡ 0
From the last equation, c 3 = 0. Substituting c 3 = 0 in the second equation gives c 2 = 0.
Substituting c 2 = 0 and c 3 = 0 in the first equation gives c 1 = 0. Thus (B) holds only when
c 1 = c 2 = c 3 = 0, which implies that the functions are linearly independent.
Example 6. Let f 1 (x) = sin x, f 2 (x) = cos x, f 3 (x) = sin
x − 1 6
π
, x ∈ (−∞, ∞). Are these
functions linearly dependent or linearly independent?
SOLUTION By the addition formula for the sine function
sin
x −
1 6 π
= sin x cos
1 6 π − cos x sin
1 6 π =
1 2
3 sin x −
1 2 cos x,
Since f 3 is a linear combination of f 1 and f 2 , we can conclude that f 1 , f 2 , f 3 are linearly
dependent on (−∞, ∞).
DEFINITION 3. Suppose that the functions f 1 , f 2 , f 3 ,... , fn are n − 1-times differentiable
on an interval I. The determinant
W (x) =
f 1 (x) f 2 (x)... fn(x)
f ′ 1 (x) f ′ 2 (x)... f ′ n (x)
f ′′ 1 (x) f ′′ 2 (x)... f ′′ n (x)
. . .
f
(n−1) 1 (x)^ f
(n−1) 2 (x)^...^ f
(n−1) n (x)
is called the Wronskian of f 1 , f 2 , f 3 ,... , fn.
Theorem 1 can be stated equivalently as:
COROLLARY Suppose that the functions f 1 , f 2 , f 3 ,... , fn are (n − 1)-times differentiable
on an interval I and let W (x) be their Wronskian. If W (x) 6 = 0 for at least one x ∈ I, then
the functions are linearly independent on I.
This is a useful test for determining the linear independence of a set of functions.
Example 7. Show that the functions f 1 (x) ≡ 1 , f 2 (x) = x, f 3 (x) = x
2 , f 4 (x) = x
3 are linearly
independent.
SOLUTION These functions are three-times differentiable on (−∞, ∞). Their Wronskian is
W (x) =
1 x x
2 x
3
0 1 2 x 3 x
2
0 0 2 6 x
Since W 6 = 0, the functions are linearly independent.
Note: You can use the Wronskian to show that any set of distinct powers of x is a linearly
independent set.
Caution Theorem 1 says that if a set of (sufficiently differentiable) functions is linearly dependent
on an interval I, then their Wronskian is identically zero on I. The theorem does not say that
if the Wronskian of a set of functions is identically zero on some interval, then the functions are
linearly dependent on that interval. Here is an example of a pair of functions which are linearly
independent and whose Wronskian is identically zero.
Example 8. Let f(x) = x
2 and let
g(x) =
−x 2 − 2 < x < 0
x 2 0 ≤ x < 2
on (− 2 , 2). The only question is whether g is differentiable at 0. You can verify that it is. Thus
we can form their Wronskian:
For x ≥ 0,
W (x) =
x
2 x
2
2 x 2 x
For x < 0,
W (x) =
x 2 −x 2
2 x − 2 x
Thus, W (x) ≡ 0 on (− 2 , 2).
We can state that f and g are linearly independent because neither is a constant multiple of
the other (f = g on [0, 2), f = −g on (− 2 , 0)). Another way to see this is: Suppose that f and
g are linearly dependent. Then there exist two numbers c 1 , c 2 , not both zero, such that
c 1 f(x) + c 2 g(x) ≡ 0 on (− 2 , 2).
If we evaluate this identity at x = 1 and x = −1, we get the pair of equations
c 1 + c 2 = 0
c 1 − c 2 = 0
The only solution of this pair of equations is c 1 = c 2 = 0. Thus, f and g are linearly independent.
Exercises 5.
- Show that any set of vectors in R
n that contains the zero vector is a linearly dependent set.
Hint: Show that the set { 0 , v 2 , v 3 ,... , vk} is linearly dependent.
- Show that if the set {v 1 , v 2 , v 3 ,... , vk } is linearly dependent, then one of the vectors can
be written as a linear combination of the others.
Determine whether the set of vectors is linearly dependent or linearly independent. If it is
linearly dependent, express one of the vectors as a linear combination of the others.
- For which values of b are the vectors ( 1, b ), ( 3, −1 ) linearly independent?
- For which values of b are the vectors ( 3, b ), ( 6, b − 1 ) linearly independent?
- For which values of b are the vectors ( 2, −b ), ( 2b + 6, 4 b ) linearly dependent?
- For which values of b are the vectors ( 1, b, 2 b ) ( 2, 1 , 4 ) linearly independent?
