Section 4.2: The Vector Space Rnand Subspaces 235
so we must conclude that u and v are linearly dependent vectors. Since 0,u it
40. Since the vectors u, v, w are linearly dependent , there exist scalars p, q, r not all zero
such that .pqr uvw 0 If r = 0, then p and q are scalars not both zero such that
.pquv 0 But this contradicts the given fact that u and v are linearly independent.
Hence r 0, so we can solve for
SECTION 4.2
THE VECTOR SPACE Rn AND SUBSPACES
The main objective in this section is for the student to understand what types of subsets of the vector
space Rn of n-tuples of real numbers are subspaces — playing the role in Rn of lines and planes
through the origin in R3. Our first reason for studying subspaces is the fact that the solution space
of any homogeneous linear system Ax = 0 is a subspace of Rn.
1. If 12 12
( , ,0) and ( , ,0)xx yyxy are vectors in W, then their sum
2. Suppose 123 12 3
(, , )and (, , )
x
xx yyyxy are vectors in W, so 12
5andxx 12
5.yy
Then their sum 112 23 3 123
(, , )(,,)
x
yx yx y sss sxy satisfies the same
condition
236 Chapter 4: Vector Spaces
3. The typical vector in W is of the form 13
(,1, )
x
xx with second coordinate 1. But the
particular scalar multiple 13
2(2,2,2)
x
xx of such a vector has second coordinate 21,
4. The typical vector 123
(, , )
x
xxx in W has coordinate sum 123
x
xx equal to 1. But
then the particular scalar multiple 123
2(2,2,2)
x
xxx of such a vector has coordinate
sum
5. Suppose 1234 1 2 3 4
(, , , )and (, , , )
x
xxx yyyyxy are vectors in W, so
1234 123 4
234 0and 234 0.xxxx yy y y
Then their sum 112 23 34 4 1234
(, , , )(,,,)
x
yx yx yx y sssssxy satisfies the
6. Suppose 1234 1 2 3 4
(, , , )and (, , , )
x
xxx yyyyxy are vectors in W, so
132 4 1 32 4
3, 4 and 3, 4.
x
xx x y yy y
Section 4.2: The Vector Space Rnand Subspaces 237
x
7. The vectors (1, 1) and (1, 1)xy are in W, but their sum (2, 0)xy is not,
because 20. Hence W is not a subspace of R2.
8. W is simply the zero subspace {}0 of R2.
11. Suppose 1234 1 2 3 4
(, , , )and (, , , )
x
xxx yyyyxy are vectors in W, so
12 34 12 3 4
and .
x
xxx yyyy
Then their sum 112 23 34 4 1234
(, , , )(,,,)
x
yx yx yx y sssssxy satisfies the
x
238 Chapter 4: Vector Spaces
12. The vectors (1, 0,1, 0) and (0, 2, 0,3)xy are in W (because both products are 0 in
13. The vectors (1, 0,1, 0) and (0,1, 0,1)xy are in W (because the product of the 4
components is 0 in each case) but their sum (1,1,1,1)sxy is not, because
1234 10.ss ss Hence W is not a subspace of R4.
In Problems 15–22, we first reduce the coefficient matrix A to echelon form E in order to
solve the given homogeneous system Ax 0 .
15.
1414 1014
1218 0102
x
x
AE
16.
1437 1015
2117 0113
1 2 3 11 0000
AE
Thus 34
and
x
sxt are free variables. We solve for 12
5and 3,
x
st x st
so
x
Section 4.2: The Vector Space Rnand Subspaces 239
17.
13 8 1 10 12
13105 013 1
14 11 2 000 0
AE
18.
132 5 1 100 2 3
27411 2 010 1 4
26512 7 001 2 5
AE
19.
1356 1010
21 4 4 0120
13 7 1 0001
AE
20.
151 8 100 5
250 5 010 3
271 9 001 2
AE
240 Chapter 4: Vector Spaces
21.
17 2 3 100 3
27 1 4 010 2
35 1 5 001 4
AE
22.
133 3 100 6
275 1 010 4
274 4 001 3
AE
23. Let u be a vector in W. Then 0u is also in W. But 0u = (0+0)u = 0u + 0u, so upon
subtracting 0u from each side, we see that 0u = 0, the zero vector.
24. (a) Problem 23 shows that 0u = 0 for every vector u.
25. If W is a subspace, then it contains the scalar multiples au and bv, and hence contains
26. The sum of any two scalar multiples of u is a scalar multiple of u, as is any scalar
multiple of a scalar multiple of u.
Section 4.3: Linear Combinations and Independence of Vectors 241
27. Let a1u + b1v and a2u + b2v be two vectors in
.Wabuv Then the sum
28. If u and v are vectors in W, then Au = ku and Av = kv. It follows that
29. If Ax0 = b and y = x – x0, then
30. Let W denote the intersection of the subspaces U and V. If u and v are vectors in W,
then these two vectors are both in U and in V. Hence the linear combination au + bv is
both in U and in V, and hence is in the intersection W, which therefore is a subspace. If
U and V are non-coincident planes through the origin if R3, then their intersection W is a
line through the origin.
SECTION 4.3
LINEAR COMBINATIONS AND
INDEPENDENCE OF VECTORS
In this section we use two types of computational problems as aids in understanding linear
independence and dependence. The first of these problems is that of expressing a vector w as a
linear combination of k given vectors 12
,,,
k
vv v (if possible). The second is that of
determining whether k given vectors 12
,,,
k
vv v are linearly independent. For vectors in Rn,
242 Chapter 4: Vector Spaces
1. 3
21
2,vv so the two vectors v1 and v2 are linearly dependent.
3. The three vectors v1, v2, and v3 are linearly dependent, as are any 3 vectors in R2. The
4. The four vectors v1, v2, v3, and v4 are linearly dependent, as are any 4 vectors in R3. The
5. The equation 11 2 2 3 3
cc c vvv0 yields
6. The equation 11 2 2 3 3
cc c vvv0 yields
1 2 3 123233
(1,0,0) (1,1,0) (1,1,1) ( , , ) (0,0,0).c c c cccccc
7. The equation 11 2 2 3 3
cc c vvv0 yields
Section 4.3: Linear Combinations and Independence of Vectors 243
8. Here inspection of the three given vectors reveals that 312
,vvv so the vectors
v1, v2, and v3 are linearly dependent.
9. 11 2 2
ccvvw
53 1 10 2
10. 11 2 2
ccvvw
36 3 107
11. 11 2 2
ccvvw
731 101
244 Chapter 4: Vector Spaces
12. 11 2 2
ccvvw
724 102
13. 11 2 2
ccvvw
155 100
14. 11 2 2 3 3
cc c vvvw
10 0 2 1000
15. 11 2 2 3 3
cc c vvvw
2314 1003
Section 4.3: Linear Combinations and Independence of Vectors 245
16. 11 2 2 3 3
cc c vvvw
24 1 7 100 6
12 3
In Problems 17–22,
123
Avvv is the coefficient matrix of the homogeneous linear
system corresponding to the vector equation 11 2 2 3 3 .cc c vvv0 Inspection of the indicated
reduced echelon form E of A then reveals whether or not a nontrivial solution exists.
123 100
18.
24 2 103/5
051 011/5
AE
25 2 100
04 1 010
246 Chapter 4: Vector Spaces
20.
123 100
111 010
AE
311 101
012 012
335 107/9
907 015/9
23. Because v1 and v2 are linearly independent, the vector equation
11 2 2 1 1 2 2 1 2
()()cc c c uu vv vv 0
yields the homogeneous linear system
24. Because v1 and v2 are linearly independent, the vector equation
yields the homogeneous linear system
Section 4.3: Linear Combinations and Independence of Vectors 247
25. Because the vectors 123
,,vvv are linearly independent, the vector equation
11 2 2 3 3 1 1 2 1 2 3 1 2 3
() ( 2) ( 2 3)cc c c c c u u u v vv vvv 0
yields the homogeneous linear system
26. Because the vectors 123
,,vvv are linearly independent, the vector equation
11 2 2 3 3 1 2 3 2 1 3 3 1 2
()()()cccccc uuu vv vv vv 0
yields the homogeneous linear system
then shows that 123
0,cc c and therefore that the vectors 123
,,uu u are linearly
independent.
27. If the elements of S are 12
,,,
k
vv v with 1,v0 then we can take 11c and
20.
k
cc This choice gives coefficients 12
,, ,
k
cc c not all zero such that
248 Chapter 4: Vector Spaces
28. Because the set S of vectors 12
,,,
k
vv v is linearly dependent, there exist scalars
m
30. Let W be the subspace of V spanned by the vectors 12
,,,.
k
vv v Because U is a
31. If S is contained in span(T), then every vector in S is a linear combination of vectors in
T. Hence every vector in span(S) is a linear combination of linear combinations of
vectors in T. Therefore every vector in span(S) is a linear combination of vectors in T,
k
33. The determinant of the kk identity matrix is nonzero, so it follows immediately from
Theorem 3 in this section that the vectors 12
,,,
k
vv v are linearly independent.
34. If the vectors 12
,,,
n
vv v are linearly independent, then by Theorem 2 the matrix
35. Because the vectors 12
,,,
k
vv v are linearly independent, Theorem 3 implies that some
kk submatrix A0 of A has nonzero determinant. Let A0 consist of the rows
12
,, ,
k
ii i of the matrix A, and let C0 denote the kk submatrix consisting of the
Section 4.4: Bases and Dimension for Vector Spaces 249
SECTION 4.4
BASES AND DIMENSION FOR VECTOR SPACES
A basis
12
,,,
k
vv v for a subspace W of Rn enables up to visualize W as a k-dimensional
plane (or “hyperplane”) through the origin in Rn. In case W is the solution space of a
homogeneous linear system, a basis for W is a maximal linearly independent set of solutions of
the system, and every other solution is a linear combination of these particular solutions.
2. We note that v2 = 2v1. Consequently the vectors 123
,,vv v are linearly dependent, and
therefore do not form a basis for R3.
5. The three given vectors 123
,,vv v all lie in the 2-dimensional subspace x1 = 0 of R3.
Therefore they are linearly dependent, and hence do not form a basis for R3.
9. The single equation 25 0xyz is already a system in reduced echelon form, with
free variables y and z. With ,,25ysztx s t we get the solution vector
250 Chapter 4: Vector Spaces
10. The single equation 0yz is already a system in reduced echelon form, with free
variables x and z. With ,
x
sy z t we get the solution vector
11. The line of intersection of the planes in Problems 9 and 11 is the solution space of the
system
12. The typical vector in R4 of the form ( , , , )abcd with abcd can be written as
13. The typical vector in R4 of the form (,,, )abcd with 3and 4ac bd can be written
as
14. The typical vector in R4 of the form (,,, )abcd with 2and 3ab cd can be
written as
( 2 , , 3 , ) ( 2,1, 0, 0) (0, 0, 3,1).bb d d b d v
Section 4.4: Bases and Dimension for Vector Spaces 251
In Problems 15–26, we show first the reduction of the coefficient matrix A to echelon form E.
Then we write the typical solution vector as a linear combination of basis vectors for the
subspace of the given system.
15. 123 1011
231 01 7
AE
With free variable 312
and 11 , 7
x
txtxt we get the solution vector
(11 , 7 , ) (11, 7,1).ttt tx Thus the solution space of the given system is 1-
dimensional with basis consisting of the vector 1(11, 7,1).v
x
17. 1 3 2 4 1 0 11 11
2573 0135
AE
With free variables 34 1 2
, and with 11 11 , 3 5
x
sx t x s tx s t we get the
solution vector
( 11 11 , 3 5 , , ) ( 11, 3,1, 0) ( 11, 5,0,1).ststst s t x
Thus the solution space of the given system is 2-dimensional with basis consisting of the
vectors 12
( 11, 3,1,0) and ( 11, 5,0,1). vv
x
252 Chapter 4: Vector Spaces
19.
1385 1034
21 411 0123
13 313 0000
AE
x
20.
13105 1012
14 11 2 013 1
13 8 1 000 0
AE
x
21.
1437 1015
2117 0113
1 2 3 11 0000
AE
With free variables 34 1 2
,andwith 5, 3
x
sx t x s tx s t we get the solution
vector
Section 4.4: Bases and Dimension for Vector Spaces 253
23.
1 5 13 14 1 0 2 0
2 5 11 12 0 1 3 0
2 7 17 19 0 0 0 1
AE
With free variable 3124
and with 2 , 3 , 0xs x sx sx we get the solution
vector (2 , 3 , ,0) (2, 3,1,0).sss s x Thus the solution space of the given system is
1-dimensional with basis consisting of the vector 1(2, 3,1,0).v
x
25.
127 9 31 120 23
247 1134 001 14
365 1129 000 0 0
AE
With free variables 245 1 3
, , and with 2 2 3 , 4
x
rx sx t x r s tx s t we get
the solution vector
254 Chapter 4: Vector Spaces
26.
31 31110 100 2 3
58 2 2 7 010 1 4
25 0 114 001 2 5
AE
27. If the vectors 12
,,,
n
vv v are linearly independent, and w is another vector in V, then
the vectors 12
,,,,
n
wv v v are linearly dependent (because no n+1 vectors in the n–
dimensional vector space V are linearly independent). Hence there exist scalars
28. If the n vectors in S were not linearly independent, then some one of them would be a
linear combination of the others. These remaining n–1 vectors would then span the n–
dimensional vector space V, which is impossible. Therefore the spanning set S is also
linearly independent, and therefore is a basis for V.
29. Suppose 11 2 2 .
kk
cc c c vv v v 0 Then c = 0 because, otherwise, we could
solve for v as a linear combination of the vectors 12
,,,.
k
vv v But this is impossible,
k
30. Let
12
,,,
k
Svv v be a linearly independent set of k < n vectors in V. If the