Download Linear Dependence and Linear Independence and more Exams Calculus in PDF only on Docsity!
4.5 Linear Dependence and Linear Independence 267
32. { v 1 , v 2 }, where v 1 , v 2 are collinear vectors in R^3. 33. Prove that if S and S′^ are subsets of a vector space V such that S is a subset of S′, then span(S) is a subset of span(S′). 34. Prove that
span{ v 1 , v 2 , v 3 } = span{ v 1 , v 2 }
if and only if v 3 can be written as a linear combination of v 1 and v 2.
4.5 Linear Dependence and Linear Independence
As indicated in the previous section, in analyzing a vector space we will be interested in determining a spanning set. The reader has perhaps already noticed that a vector space V can have many such spanning sets.
Example 4.5.1 Observe that {( 1 , 0 ), ( 0 , 1 )}, {( 1 , 0 ), ( 1 , 1 )}, and {( 1 , 0 ), ( 0 , 1 ), ( 1 , 2 )} are all spanning sets for R^2.
As another illustration, two different spanning sets for V = M 2 (R) were given in Exam- ple 4.4.5 and the remark that followed. Given the abundance of spanning sets available for a given vector space V , we are faced with a natural question: Is there a “best class of” spanning sets to use? The answer, to a large degree, is “yes”. For instance, in Exam- ple 4.5.1, the spanning set {( 1 , 0 ), ( 0 , 1 ), ( 1 , 2 )} contains an “extra” vector, ( 1 , 2 ), which seems to be unnecessary for spanning R^2 , since {( 1 , 0 ), ( 0 , 1 )} is already a spanning set. In some sense, {( 1 , 0 ), ( 0 , 1 )} is a more efficient spanning set. It is what we call a mini- mal spanning set , since it contains the minimum number of vectors needed to span the vector space. 3 But how will we know if we have found a minimal spanning set (assuming one exists)? Returning to the example above, we have seen that
span{( 1 , 0 ), ( 0 , 1 )} = span{( 1 , 0 ), ( 0 , 1 ), ( 1 , 2 )} = R^2.
Observe that the vector ( 1 , 2 ) is already a linear combination of ( 1 , 0 ) and ( 0 , 1 ), and therefore it does not add any new vectors to the linear span of {( 1 , 0 ), ( 0 , 1 )}. As a second example, consider the vectors v 1 = ( 1 , 1 , 1 ), v 2 = ( 3 , − 2 , 1 ), and v 3 = 4 v 1 + v 2 = ( 7 , 2 , 5 ). It is easily verified that det([ v 1 , v 2 , v 3 ]) = 0. Consequently, the three vectors lie in a plane (see Figure 4.5.1) and therefore, since they are not collinear, the linear span of these three vectors is the whole of this plane. Furthermore, the same plane is generated if we consider the linear span of v 1 and v 2 alone. As in the previous example, the reason that v 3 does not add any new vectors to the linear span of { v 1 , v 2 } is that it is already a linear combination of v 1 and v 2. It is not possible, however, to generate all vectors in the plane by taking linear combinations of just one vector, as we could generate only a line lying in the plane in that case. Consequently, { v 1 , v 2 } is a minimal spanning set for the subspace of R^3 consisting of all points lying on the plane. As a final example, recall from Example 1.2.16 that the solution space to the differ- ential equation y′′^ + y = 0
(^3) Since a single (nonzero) vector in R (^2) spans only the line through the origin along which it points, it cannot span all of R^2 ; hence, the minimum number of vectors required to span R^2 is 2.
268 CHAPTER 4 Vector Spaces
z
y
x
(7, 2, 5)
(3,2, 1)
(3,2, 0) (1, 1, 0)
(1, 1, 1)
(7, 2, 0)
v 3 4 v 1 v 2
v 2^ v^1
Figure 4.5.1: v 3 = 4 v 1 + v 2 lies in the plane through the origin containing v 1 and v 2 , and so, span{ v 1 , v 2 , v 3 } = span{ v 1 , v 2 }.
can be written as span{y 1 , y 2 }, where y 1 (x) = cos x and y 2 (x) = sin x. However, if we let y 3 (x) = 3 cos x − 2 sin x, for instance, then {y 1 , y 2 , y 3 } is also a spanning set for the solution space of the differential equation, since
span{y 1 , y 2 , y 3 } = {c 1 cos x + c 2 sin x + c 3 (3 cos x − 2 sin x) : c 1 , c 2 , c 3 ∈ R} = {(c 1 + 3 c 3 ) cos x + (c 2 − 2 c 3 ) sin x : c 1 , c 2 , c 3 ∈ R} = {d 1 cos x + d 2 sin x : d 1 , d 2 ∈ R} = span{y 1 , y 2 }.
The reason that {y 1 , y 2 , y 3 } is not a minimal spanning set for the solution space is that y 3 is a linear combination of y 1 and y 2 , and therefore, as we have just shown, it does not add any new vectors to the linear span of {cos x, sin x}. More generally, it is not too difficult to extend the argument used in the preceding examples to establish the following general result.
Theorem 4.5.2 Let { v 1 , v 2 ,... , v k } be a set of at least two vectors in a vector space V. If one of the vectors in the set is a linear combination of the other vectors in the set, then that vector can be deleted from the given set of vectors and the linear span of the resulting set of vectors will be the same as the linear span of { v 1 , v 2 ,... , v k }.
Proof The proof of this result is left for the exercises (Problem 48).
For instance, if v 1 is a linear combination of v 2 , v 3 ,... , v k , then Theorem 4.5.2 says that span{ v 1 , v 2 ,... , v k } = span{ v 2 , v 3 ,... , v k }.
In this case, the set { v 1 , v 2 ,... , v k } is not a minimal spanning set. To determine a minimal spanning set, the problem we face in view of Theorem 4.5. is that of determining when a vector in { v 1 , v 2 ,... , v k } can be expressed as a linear combination of the remaining vectors in the set. The correct formulation for solving this problem requires the concepts of linear dependence and linear independence, which we are now ready to introduce. First we consider linear dependence.
270 CHAPTER 4 Vector Spaces
Theorem 4.5.6 Let { v 1 , v 2 ,... , v k } be a set of at least two vectors in a vector space V. Then { v 1 , v 2 ,... , v k } is linearly dependent if and only if at least one of the vectors in the set can be expressed as a linear combination of the others.
Proof If { v 1 , v 2 ,... , v k } is linearly dependent, then according to Definition 4.5.3, there exist scalars c 1 , c 2 ,... , c (^) k , not all zero, such that
c 1 v 1 + c 2 v 2 + · · · + ck v k = 0.
Suppose that ci = 0. Then we can express v i as a linear combination of the other vectors as follows:
v i = −
ci
(c 1 v 1 + c 2 v 2 + · · · + ci− 1 v i− 1 + ci+ 1 v i+ 1 + · · · + ck v k ).
Conversely, suppose that one of the vectors, say, v j , can be expressed as a linear combi- nation of the remaining vectors. That is,
v j = c 1 v 1 + c 2 v 2 + · · · + cj − 1 v j − 1 + cj + 1 v j + 1 + · · · + ck v k.
Adding (− 1 ) v j to both sides of this equation yields
c 1 v 1 + c 2 v 2 + · · · + cj − 1 v j − 1 − v j + cj + 1 v j + 1 + · · · + ck v k = 0.
Since the coefficient of v j is − 1 = 0, the set of vectors { v 1 , v 2 ,... , v k } is linearly dependent. As far as the minimal-spanning-set idea is concerned, Theorems 4.5.6 and 4.5.2 tell us that a linearly dependent spanning set for a (nontrivial) vector space V cannot be a minimal spanning set. On the other hand, we will see in the next section that a linearly independent spanning set for V must be a minimal spanning set for V. For the remainder of this section, however, we focus more on the mechanics of determining whether a given set of vectors is linearly independent or linearly dependent. Sometimes this can be done by inspection. For example, Figure 4.5.2 illustrates that any set of three vectors in R^2 is linearly dependent.
x
y
v 1
v 2
v 3
Figure 4.5.2: The set of vectors { v 1 , v 2 , v 3 } is linearly dependent in R^2 , since v 3 is a linear combination of v 1 and v 2.
As another example, let V be the vector space of all functions defined on an interval I. If f 1 (x) = 1 , f 2 (x) = 2 sin 2 x, f 3 (x) = −5 cos 2 x, then {f 1 , f 2 , f 3 } is linearly dependent in V , since the identity sin 2 x + cos 2 x = 1 implies that for all x ∈ I ,
f 1 (x) = 12 f 2 (x) − 15 f 3 (x).
We can therefore conclude from Theorem 4.5.2 that
span{ 1 , 2 sin 2 x, −5 cos 2 x} = span{2 sin 2 x, −5 cos 2 x}.
In relatively simple examples, the following general results can be applied. They are a direct consequence of the definition of linearly dependent vectors and are left for the exercises (Problem 49).
Proposition 4.5.7 Let V be a vector space.
1. Any set of two vectors in V is linearly dependent if and only if the vectors are proportional.
4.5 Linear Dependence and Linear Independence 271
2. Any set of vectors in V containing the zero vector is linearly dependent.
Remark We emphasize that the first result in Proposition 4.5.7 holds only for the case of two vectors. It cannot be applied to sets containing more than two vectors.
Example 4.5.8 If v 1 = ( 1 , 2 , − 9 ) and v 2 = (− 2 , − 4 , 18 ), then { v 1 , v 2 } is linearly dependent in R^3 , since v 2 = − 2 v 1. Geometrically, v 1 and v 2 lie on the same line.
Example 4.5.9 If
A 1 =
[
]
, A 2 =
[
]
, A 3 =
[
]
then {A 1 , A 2 , A 3 } is linearly dependent in M 2 (R), since it contains the zero vector from M 2 (R). For more complicated situations, we must resort to Definitions 4.5.3 and 4.5.4, although conceptually it is always helpful to keep in mind that the essence of the problem we are solving is to determine whether a vector in a given set can be expressed as a linear combination of the remaining vectors in the set. We now give some examples to illustrate the use of Definitions 4.5.3 and 4.5.4.
Example 4.5.10 If v 1 = ( 1 , 2 , − 1 ) v 2 = ( 2 , − 1 , 1 ), and v 3 = ( 8 , 1 , 1 ), show that { v 1 , v 2 , v 3 } is linearly
dependent in R^3 , and determine the linear dependency relationship. Solution: We must first establish that there are values of the scalars c 1 , c 2 , c 3 , not all zero, such that
c 1 v 1 + c 2 v 2 + c 3 v 3 = 0. (4.5.1)
Substituting for the given vectors yields
c 1 ( 1 , 2 , − 1 ) + c 2 ( 2 , − 1 , 1 ) + c 3 ( 8 , 1 , 1 ) = ( 0 , 0 , 0 ).
That is, (c 1 + 2 c 2 + 8 c 3 , 2 c 1 − c 2 + c 3 , −c 1 + c 2 + c 3 ) = ( 0 , 0 , 0 ). Equating corresponding components on either side of this equation yields
c 1 + 2 c 2 + 8 c 3 = 0 , 2 c 1 − c 2 + c 3 = 0 , −c 1 + c 2 + c 3 = 0.
The reduced row-echelon form of the augmented matrix of this system is
Consequently, the system has an infinite number of solutions for c 1 , c 2 , c 3 , so the vectors are linearly dependent. In order to determine a specific linear dependency relationship, we proceed to find c 1 , c 2 , and c 3. Setting c 3 = t, we have c 2 = − 3 t and c 1 = − 2 t. Taking t = 1 and
4.5 Linear Dependence and Linear Independence 273
Proof Since the given set is linearly dependent, at least one of the vectors in the set is a linear combination of the remaining vectors, by Theorem 4.5.6. Thus, by Theorem 4.5.2, we can delete that vector from the set, and the resulting set of vectors will span the same subspace of V as the original set. If the resulting set is linearly independent, then we are done. If not, then we can repeat the procedure to eliminate another vector in the set. Continuing in this manner (with a finite number of iterations), we will obtain a linearly independent set that spans the same subspace of V as the subspace spanned by the original set of vectors.
Remark Corollary 4.5.12 is actually true even if the set of vectors in question is infinite, but we shall not need to consider that case in this text. In the case of an infinite set of vectors, other techniques are required for the proof. Note that the linearly independent set obtained using the procedure given in the previous theorem is not unique, and therefore the question arises whether the number of vectors in any resulting linearly independent set is independent of the manner in which the procedure is applied. We will give an affirmative answer to this question in Section 4.6.
Example 4.5.13 Let v 1 = ( 1 , 2 , 3 ), v 2 = (− 1 , 1 , 4 ), v 3 = ( 3 , 3 , 2 ), and v 4 = (− 2 , − 4 , − 6 ). De-
termine a linearly independent set of vectors that spans the same subspace of R^3 as span{ v 1 , v 2 , v 3 , v 4 }. Solution: Setting c 1 v 1 + c 2 v 2 + c 3 v 3 + c 4 v 4 = 0 requires that
c 1 ( 1 , 2 , 3 ) + c 2 (− 1 , 1 , 4 ) + c 3 ( 3 , 3 , 2 ) + c 4 (− 2 , − 4 , − 6 ) = ( 0 , 0 , 0 ),
leading to the linear system
c 1 − c 2 + 3 c 3 − 2 c 4 = 0 , 2 c 1 + c 2 + 3 c 3 − 4 c 4 = 0 , 3 c 1 + 4 c 2 + 2 c 3 − 6 c 4 = 0.
The augmented matrix of this system is
and the reduced row-echelon form of the augmented matrix of this system is
The system has two free variables, c 3 = s and c 4 = t, and so { v 1 , v 2 , v 3 , v 4 } is linearly dependent. Then c 2 = s and c 1 = 2 t − 2 s. So the general form of the solution is
( 2 t − 2 s, s, s, t) = s(− 2 , 1 , 1 , 0 ) + t ( 2 , 0 , 0 , 1 ).
Setting s = 1 and t = 0 yields the linear combination
− 2 v 1 + v 2 + v 3 = 0 , (4.5.2)
274 CHAPTER 4 Vector Spaces
and setting s = 0 and t = 1 yields the linear combination
2 v 1 + v 4 = 0. (4.5.3)
We can solve (4.5.2) for v 3 in terms of v 1 and v 2 , and we can solve (4.5.3) for v 4 in terms of v 1. Hence, according to Theorem 4.5.2, we have
span{ v 1 , v 2 , v 3 , v 4 } = span{ v 1 , v 2 }.
By Proposition 4.5.7, v 1 and v 2 are linearly independent, so { v 1 , v 2 } is the linearly independent set we are seeking. Geometrically, the subspace of R^3 spanned by v 1 and v 2 is a plane, and the vectors v 3 and v 4 lie in this plane.
Linear Dependence and Linear Independence in Rn
Let { v 1 , v 2 ,... , v k } be a set of vectors in Rn, and let A denote the matrix that has v 1 , v 2 ,... , v k as column vectors. Thus,
A = [ v 1 , v 2 ,... , v k ]. (4.5.4)
Since each of the given vectors is in Rn, it follows that A has n rows and is therefore an n × k matrix. The linear combination c 1 v 1 + c 2 v 2 + · · · + ck v k = 0 can be written in matrix form as (see Theorem 2.2.9)
A c = 0 , (4.5.5)
where A is given in Equation (4.5.4) and c = [c 1 c 2... ck ]T^. Consequently, we can state the following theorem and corollary:
Theorem 4.5.14 Let v 1 , v 2 ,... , v k be vectors in Rn^ and A = [ v 1 , v 2 ,... , v k ]. Then { v 1 , v 2 ,... , v k } is linearly dependent if and only if the linear system A c = 0 has a nontrivial solution.
Corollary 4.5.15 Let v 1 , v 2 ,... , v k be vectors in Rn^ and A = [ v 1 , v 2 ,... , v k ].
1. If k > n, then { v 1 , v 2 ,... , v k } is linearly dependent. 2. If k = n, then { v 1 , v 2 ,... , v k } is linearly dependent if and only if det(A) = 0.
Proof If k > n, the system (4.5.5) has an infinite number of solutions (see Corol- lary 2.5.11), hence the vectors are linearly dependent by Theorem 4.5.14. On the other hand, if k = n, the system (4.5.5) is n × n, and hence, from Corollary 3.2.5, it has an infinite number of solutions if and only if det(A) = 0.
Example 4.5.16 Determine whether the given vectors are linearly dependent or linearly independent in R^4.
1. v 1 = ( 1 , 3 , − 1 , 0 ), v 2 = ( 2 , 9 , − 1 , 3 ), v 3 = ( 4 , 5 , 6 , 11 ), v 4 = ( 1 , − 1 , 2 , 5 ), v 5 = ( 3 , − 2 , 6 , 7 ). 2. v 1 = ( 1 , 4 , 1 , 7 ), v 2 = ( 3 , − 5 , 2 , 3 ), v 3 = ( 2 , − 1 , 6 , 9 ), v 4 = (− 2 , 3 , 1 , 6 ).
276 CHAPTER 4 Vector Spaces
Example 4.5.19 If f 1 (x) = sin x and f 2 (x) = cos x on (−∞, ∞), then
W f 1 , f 2 = sin x cos x cos x − sin x = (sin x)(− sin x) − (cos x)(cos x)
= −(sin 2 x + cos 2 x) = − 1.
Example 4.5.20 If f 1 (x) = x, f 2 (x) = x^2 , and f 3 (x) = x^3 on (−∞, ∞), then
W f 1 , f 2 , f 3 =
x x^2 x^3 1 2x 3 x^2 0 2 6 x
= x( 12 x^2 − 6 x^2 ) − ( 6 x^3 − 2 x^3 ) = 2 x^3.
We can now state and prove the main result about the Wronskian.
Theorem 4.5.21 Let f 1 , f 2 ,... , f (^) k be functions in C k−^1 (I ). If W [f 1 , f 2 ,... , f (^) k ] is nonzero at some point x 0 in I , then {f 1 , f 2 ,... , f (^) k } is linearly independent on I.
Proof To apply Definition 4.5.17, assume that
c 1 f 1 (x) + c 2 f 2 (x) + · · · + ck fk (x) = 0 ,
for all x in I. Then, differentiating k − 1 times yields the linear system
c 1 f 1 (x) + c 2 f 2 (x) + · · · + ck fk (x) = 0 , c 1 f 1 ′(x) + c 2 f 2 ′(x) + · · · + ck f (^) k′ (x) = 0 , .. . c 1 f 1 (k −^1 )(x) + c 2 f 2 (k −^1 )(x) + · · · + ck f (^) k(k −^1 )(x) = 0 ,
where the unknowns in the system are c 1 , c 2 ,... , c (^) k. We wish to show that c 1 = c 2 = · · · = ck = 0. The determinant of the matrix of coefficients of this system is just W f 1 , f 2 ,... , f (^) k . Consequently, if W [f 1 , f 2 ,... , f (^) k ](x 0 ) = 0 for some x 0 in I , then the determinant of the matrix of coefficients of the system is nonzero at that point, and therefore the only solution to the system is the trivial solution c 1 = c 2 = · · · = ck = 0. That is, the given set of functions is linearly independent on I.
Remarks
1. Notice that it is only necessary for W f 1 , f 2 ,... , f (^) k to be nonzero at one point in I for {f 1 , f 2 ,... , f (^) k } to be linearly independent on I. 2. Theorem 4.5.21 does not say that if W f 1 , f 2 ,... , f (^) k = 0 for every x in I , then {f 1 , f 2 ,... , f (^) k } is linearly dependent on I. As we will see in the next example below, the Wronskian of a linearly independent set of functions on an interval I can be identically zero on I. Instead, the logical equivalent of the preceding theorem is: If {f 1 , f 2 ,... , f (^) k } is linearly dependent on I , then W f 1 , f 2 ,... , f (^) k = 0 at every point of I.
4.5 Linear Dependence and Linear Independence 277
If W f 1 , f 2 ,... , f (^) k = 0 for all x in I , Theorem 4.5.21 gives no information as to the linear dependence or independence of {f 1 , f 2 ,... , f (^) k } on I.
Example 4.5.22 Determine whether the following functions are linearly dependent or linearly indepen-
dent on I = (−∞, ∞).
(a) f 1 (x) = e x^ , f 2 (x) = x^2 e x^.
(b) f 1 (x) = x, f 2 (x) = x + x^2 , f 3 (x) = 2 x − x^2.
(c) f 1 (x) = x^2 , f 2 (x) =
2 x^2 , if x ≥ 0 , −x^2 , if x < 0.
Solution:
(a) W f 1 , f 2 = e x^ x^2 e x e x^ e x^ (x^2 + 2 x)
= e^2 x^ (x^2 + 2 x) − x^2 e^2 x^ = 2 xe^2 x^.
Since W f 1 , f 2 = 0 (except at x = 0), the functions are linearly independent on (−∞, ∞).
(b)
W f 1 , f 2 , f 3 =
x x + x^2 2 x − x^2 1 1 + 2 x 2 − 2 x 0 2 − 2 = x [(− 2 )( 1 + 2 x) − 2 ( 2 − 2 x)]
−
[
(− 2 )(x + x^2 ) − 2 ( 2 x − x^2 )
]
Thus, no conclusion can be drawn from Theorem 4.5.21. However, a closer in- spection of the functions reveals, for example, that
f 2 = 3 f 1 − f 3.
Consequently, the functions are linearly dependent on (−∞, ∞).
(c) If x ≥ 0, then W f 1 , f 2 = x^2 2 x^2 2 x 4 x
whereas if x < 0, then
W f 1 , f 2 = x^2 −x^2 2 x − 2 x
Thus, W f 1 , f 2 = 0 for all x in (−∞, ∞), so no conclusion can be drawn from Theorem 4.5.21. Again we take a closer look at the given functions. They are sketched in Figure 4.5.3. In this case, we see that on the interval (−∞, 0 ), the functions are linearly dependent, since
f 1 + f 2 = 0.
4.5 Linear Dependence and Linear Independence 279
- Be able to produce a linearly independent set of vec- tors that spans a given subspace of a vector space V.
- Be able to conclude immediately that a set of k vectors in Rn^ is linearly dependent if k > n, and know what can be said in the case where k = n as well.
- Know what information the Wronskian does (and does not) give about the linear dependence or linear inde- pendence of a set of functions on an interval I.
True-False Review
For Questions 1–9, decide if the given statement is true or false , and give a brief justification for your answer. If true, you can quote a relevant definition or theorem from the text. If false, provide an example, illustration, or brief explanation of why the statement is false.
1. Every vector space V possesses a unique minimal spanning set. 2. The set of column vectors of a 5 × 7 matrix A must be linearly dependent. 3. The set of column vectors of a 7 × 5 matrix A must be linearly independent. 4. Any nonempty subset of a linearly independent set of vectors is linearly independent. 5. If the Wronskian of a set of functions is nonzero at some point x 0 in an interval I , then the set of func- tions is linearly independent. 6. If it is possible to express one of the vectors in a set S as a linear combination of the others, then S is a linearly dependent set. 7. If a set of vectors S in a vector space V contains a linearly dependent subset, then S is itself a linearly dependent set. 8. A set of three vectors in a vector space V is linearly de- pendent if and only if all three vectors are proportional to one another. 9. If the Wronskian of a set of functions is identically zero at every point of an interval I , then the set of functions is linearly dependent.
Problems
For Problems 1–9, determine whether the given set of vectors is linearly independent or linearly dependent in Rn. In the case of linear dependence, find a dependency relationship.
1. {( 1 , − 1 ), ( 1 , 1 )}. 2. {( 2 , − 1 ), ( 3 , 2 ), ( 0 , 1 )}. 3. {( 1 , − 1 , 0 ), ( 0 , 1 , − 1 ), ( 1 , 1 , 1 )}. 4. {( 1 , 2 , 3 ), ( 1 , − 1 , 2 ), ( 1 , − 4 , 1 )}. 5. {(− 2 , 4 , − 6 ), ( 3 , − 6 , 9 )}. 6. {( 1 , − 1 , 2 ), ( 2 , 1 , 0 )}. 7. {(− 1 , 1 , 2 ), ( 0 , 2 , − 1 ), ( 3 , 1 , 2 ), (− 1 , − 1 , 1 )}. 8. {( 1 , − 1 , 2 , 3 ), ( 2 , − 1 , 1 , − 1 ), (− 1 , 1 , 1 , 1 )}. 9. {( 2 , − 1 , 0 , 1 ), ( 1 , 0 , − 1 , 2 ), ( 0 , 3 , 1 , 2 ), (− 1 , 1 , 2 , 1 )}. 10. Let v 1 = ( 1 , 2 , 3 ), v 2 = ( 4 , 5 , 6 ), v 3 = ( 7 , 8 , 9 ). De- termine whether { v 1 , v 2 , v 3 } is linearly independent in R^3. Describe span{ v 1 , v 2 , v 3 } geometrically. 11. Consider the vectors v 1 = ( 2 , − 1 , 5 ), v 2 = ( 1 , 3 , − 4 ), v 3 = (− 3 , − 9 , 12 ) in R^3.
(a) Show that { v 1 , v 2 , v 3 } is linearly dependent. (b) Is v 1 ∈ span{ v 2 , v 3 }? Draw a picture illustrating your answer.
12. Determine all values of the constant k for which the vectors ( 1 , 1 , k), ( 0 , 2 , k), and ( 1 , k, 6 ) are linearly de- pendent in R^3.
For Problems 13–14, determine all values of the constant k for which the given set of vectors is linearly independent in R^4.
13. {( 1 , 0 , 1 , k), (− 1 , 0 , k, 1 ), ( 2 , 0 , 1 , 3 )}. 14. {( 1 , 1 , 0 , − 1 ), ( 1 , k, 1 , 1 ), ( 2 , 1 , k, 1 ), (− 1 , 1 , 1 , k)}.
For Problems 15–17, determine whether the given set of vec- tors is linearly independent in M 2 (R).
15. A 1 =
[
]
, A 2 =
[
]
, A 3 =
[
]
16. A 1 =
[
]
, A 2 =
[
]
280 CHAPTER 4 Vector Spaces
17. A 1 =
[
]
, A 2 =
[
]
, A 3 =
[
]
For Problems 18–19, determine whether the given set of vec- tors is linearly independent in P 1.
18. p 1 (x) = 1 − x, p 2 (x) = 1 + x. 19. p 1 (x) = 2 + 3 x, p 2 (x) = 4 + 6 x. 20. Show that the vectors
p 1 (x) = a + bx and p 2 (x) = c + dx
are linearly independent in P 1 if and only if the con- stants a, b, c, d satisfy ad − bc = 0.
21. If f 1 (x) = cos 2x, f 2 (x) = sin 2 x, f 3 (x) = cos 2 x, determine whether {f 1 , f 2 , f 3 } is linearly dependent or linearly independent in C∞(−∞, ∞).
For Problems 22–28, determine a linearly independent set of vectors that spans the same subspace of V as that spanned by the original set of vectors.
22. V = R^3 , {( 1 , 2 , 3 ), (− 3 , 4 , 5 ), ( 1 , − 43 , − 53 )}. 23. V = R^3 , {( 3 , 1 , 5 ), ( 0 , 0 , 0 ), ( 1 , 2 , − 1 ), (− 1 , 2 , 3 )}. 24. V = R^3 , {( 1 , 1 , 1 ), ( 1 , − 1 , 1 ), ( 1 , − 3 , 1 ), ( 3 , 1 , 2 )}. 25. V = R^4 , {( 1 , 1 , − 1 , 1 ), ( 2 , − 1 , 3 , 1 ), ( 1 , 1 , 2 , 1 ), ( 2 , − 1 , 2 , 1 )}. 26. V = M 2 (R), {[ 1 2 3 4
]
[
]
[
]}
27. V = P 1 , { 2 − 5 x, 3 + 7 x, 4 − x}. 28. V = P 2 , { 2 + x^2 , 4 − 2 x + 3 x^2 , 1 + x}.
For Problems 29–33, use the Wronskian to show that the given functions are linearly independent on the given inter- val I.
29. f 1 (x) = 1 , f 2 (x) = x, f 3 (x) = x^2 , I = (−∞, ∞). 30. f 1 (x) = sin x, f 2 (x) = cos x, f 3 (x) = tan x, I = (−π/ 2 , π/ 2 ). 31. f 1 (x) = 1 , f 2 (x) = 3 x, f 3 (x) = x^2 − 1 , I = (−∞, ∞). 32. f 1 (x) = e^2 x^ , f 2 (x) = e^3 x^ , f 3 (x) = e−x^ , I = (−∞, ∞).
f 1 (x) =
x^2 , if x ≥ 0 , 3 x^3 , if x < 0 ,
f 2 (x) = 7 x^2 , I = (−∞, ∞).
For Problems 34–36, show that the Wronskian of the given functions is identically zero on (−∞, ∞). Determine whether the functions are linearly independent or linearly dependent on that interval.
34. f 1 (x) = 1 , f 2 (x) = x, f 3 (x) = 2 x − 1. 35. f 1 (x) = e x^ , f 2 (x) = e−x^ , f 3 (x) = cosh x. 36. f 1 (x) = 2 x^3 ,
f 2 (x) =
5 x^3 , ifx ≥ 0 , − 3 x^3 , if x < 0.
37. Consider the functions f 1 (x) = x,
f 2 (x) =
x, if x ≥ 0 , −x, if x < 0.
(a) Show that f 2 is not in C^1 (−∞, ∞). (b) Show that {f 1 , f 2 } is linearly dependent on the in- tervals (−∞, 0 ) and [ 0 , ∞), while it is linearly in- dependent on the interval (−∞, ∞). Justify your results by making a sketch showing both of the functions.
38. Determine whether the functions f 1 (x) = x,
f 2 (x) =
x, if x = 0 , 1 , if x = 0.
are linearly dependent or linearly independent on I = (−∞, ∞).
39. Show that the functions
f 1 (x) =
x − 1 , if x ≥ 1 , 2 (x − 1 ), if x < 1 ,
f 2 (x) = 2 x, f 3 (x) = 3 form a linearly independent set on (−∞, ∞). Determine all intervals on which {f 1 , f 2 , f 3 } is linearly dependent.
