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4.2
4.2.1
CHAPTER 4. VECTOR SPACES
Subspaces
De…nitions and Examples
Often, we work with vector spaces which consists of an appropriate subset of
vectors from a larger vector space. We might expect that most of the properties
of the larger space would be passed to that subset of vectors. Only two of them,
closure for addition and scalar multiplication, should be veri…ed. This is indeed
the case. We formalize this with a de…nition and a theorem.
De…nition 261 (Subspace) A subset W of a vector space V is called a subspace of V if it is itself a vector space under the addition and scalar multiplication de…ned on V .
So, in theory, to prove W is a subspace of V , we would need to verify the
ten axioms. However, since the elements of W are also in V , we know that most
axioms will hold. The only two which must be checked are closure under both
operations. For example, if we take two elements of W , we are guaranteed their
sum is in V since V is a vector space. But we don’t know it is in W . We have
the following theorem.
Theorem 262 If W is a non-empty subset of a vector space V , then W is a
subspace of V if and only if the two conditions below are satis…ed:
1. W is closed under vector addition that is if u 2 W and v 2 W then
u + v 2 W.
2. W is closed under scalar multiplication that is if u 2 W and c is any
scalar, then cu 2 W .
Proof. We prove both directions.
1. Suppose that W is a subspace of V . Then, by de…nition, it is a vector
space. All ten axioms hold, in particular the two conditions of the theorem.
2. Suppose that W V , W 6= ; and the two conditions hold. We must show
the ten axioms hold. Axioms 1 and 6 will hold by our assumption. If
u 2 W then u 2 W by assumption. We know that u + ( u) = 0 2 W .
So, 0 2 W . The remaining axioms are automatically satis…ed since every
element of W is also an element of V and V is a vector space.
Remark 263 It is easier (fewer axioms to check) to show that a set is a subspace of another vector space than proving it is a vector space. We will use this
whenever possible.
Remark 264 Any subset which does not contain the zero vector cannot be a
subspace because it won’t be a vector space.
4.2. SUBSPACES
123
Remark 265 Do not forget to prove that the subset you are studying is not
empty.
Example 266 (polynomials) Earlier, we established that Pn was a vector
space by proving it directly. We can also prove it by showing it is a subspace of
F ( 1; 1). Since polynomials are functions with domain ( 1; 1), we see that
Pn F ( 1; 1). In addition, it is not empty. As we veri…ed when we proved it
was a vector space, it is closed under addition and scalar multiplication. Hence,
it is a subspace of F ( 1; 1).
Example 267 (symmetric matrices) Let V = fA 2 Mnn : A is symmetricg.
Then, (V; +; :) with standard matrix addition and scalar multiplication is a vector space. We prove it by showing it is a subspace of Mnn . First, V is not
empty, it contains the zero n n matrix. Since the sum of two symmetric matrices is symmetric (why?), it is closed under addition. Also, the product of a
symmetric matrix by a scalar is also a symmetric matrix (why?), so it is closed
under scalar multiplication.
Example 268 R2 is not a subspace of R3 . However, the set H = f(x; y; 0) : x 2 R and y 2 Rg
is a subspace of R3 , it acts like R2 .
Example 269 We saw earlier that the set of function de…ned on an interval
[a; b], denoted F [a; b] (a or b can be in…nite) was a vector space. This space has
many subspaces of interest.
1. C [a; b], the space of continuous functions. This is a subspace of F [a; b].
It contain the zero function. We know from calculus that if f and g are
two continuous functions, so is f + g and cf for any scalar c. Thus, it is
closed under addition and scalar multiplication.
2. C 1 [a; b], the space of functions whose …rst derivative is continuous. It is
a subspace of F [a; b] for the same reasons. If f and g are two functions
0
whose …rst derivative is continuous, then (f + g) = f 0 + g 0 will also be
0
0
continuous and (cf ) = cf will be continuous as well. In fact, remembering from calculus that if a function is di¤ erentiable it is also continuous,
we see that C 1 [a; b] is a subspace of F [a; b].
3. Similarly, we can de…ne for every positive integer n, C n [a; b] to be the
space of function whose nth derivative is continuous. This is a subspace
of F [a; b] as well as a subspace of C n 1 [a; b].
4. Pn is a subspace of C m [a; b] for every positive integer m.
Example 270 (solution space of a homogeneous system) If A is m n,
then the solutions of the homogeneous system Ax = 0 form a vector space. This
is an important result, we give it as a theorem.
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CHAPTER 4. VECTOR SPACES
Theorem 271 If Ax = 0 is a homogeneous system of m equations and n unknowns, then the set of solutions is a subspace of Rn .
Proof. Let W be the set of solutions. First, it we must establish that W 6= ;.
We know from a theorem studied in class that 0 2 W. Also, W
Rn because
the solutions of Ax = 0 are n 1 vectors. Next, we must establish closure under
addition. If x and x0 are solutions, we must show x + x0 is also a solution.
A (x + x0 ) = Ax + Ax0 = 0 + 0. Similarly, A (cx) = cAx = c0 = 0 for any
scalar c. Thus we see that W is also closed under scalar multiplication.
4.2.2
Linear Combinations and Spanning Sets
De…nition 272 Given n vectors v1 , v2 , :::, vn from a vector space V , a linear
combination of these vectors is an expression of the form
k1 v1 + k2 v2 + ::: + kn vn
where k1 , k2 , :::, kn are scalars.
Remark 273 Note that since V is a vector space, k1 v1 + k2 v2 + ::: + kn vn is
also an element of V . So, a linear combination of vectors in a vector space is a
vector in the same vector space.
Remark 274 A linear combination of one vector is simply a scalar multiple of
that vector.
Example 275 Earlier, we saw that in R3 , every vector !
u = (x; y; z) could
!
!
!
!
be written as u = x i + y j + z k thus every vector in R3 is some linear
!
! !
combination of i , j and k .
An important question is given a set of vectors v1 , v2 , :::, vn in a vector
space V , how do we know if another vector w of V can be written as a linear
combination of v1 , v2 , :::, vn . This amounts to solving a system. We know
that this will happen if one can …nd scalars k1 , k2 , :::, kn such that k1 v1 +
k2 v2 + ::: + kn vn = w. 2
This is3the same as solving the system V k = w where
k1
6 k2 7
6
7
V = [v1 ; v2 ; :::; vn ], k = 6 . 7. We illustrate this with an example.
4 .. 5
kn
Example 276 Consider the vectors in R3 u = (1; 2; 1) and v = (6; 4; 2).
1. Is w = (9; 2; 7) a linear combination of u and v?
We need to solve k1 u+k2 v = w that is (9; 2; 7) = k1 (1; 2; 1)+k2 (6; 4; 2).
This gives us the system
8
< k1 + 6k2 = 9
2k1 + 4k2 = 2
:
k1 + 2k2 = 7
Solving using Gauss-Jordan gives k1 =
3 and k2 = 2, thus w =
3u+2v.
4.2. SUBSPACES
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2. Is w0 = (4; 1; 8) a linear combination of u and v?
We need to solve k1 u+k2 v = wthat is (4; 1; 8) = k1 (1; 2; 1)+k2 (6; 4; 2).
This gives us the system
8
< k1 + 6k2 = 4
2k1 + 4k2 = 1
:
k1 + 2k2 = 2
The reader will verify that this system has no solutions.
This examples illustrates the fact that if we take some vectors v1 ; v2 ; :::; vn
from a vector space V , some vectors of V maybe a linear combination of
v1 ; v2 ; :::; vn , but others may not. If we construct the set W consisting of
all the possible linear combinations of v1 ; v2 ; :::; vn , then it is a vector space.
We state this as a theorem.
Theorem 277 If v1 ; v2 ; :::; vn are vectors in a vector space V , then
1. The set W of all linear combinations of v1 ; v2 ; :::; vn is a subspace of V .
2. It is the smallest subspace of V which contains v1 ; v2 ; :::; vn .
Proof. We prove each part separately.
1. We must prove three things: 1) W 6= ;, 2) W is closed under addition and
3) W is closed under scalar multiplication.
1) W 6= ;: 0 2 W since 0 = 0v1 + 0v2 + ::: + 0vn .
2) W is closed under addition: Let u and v be two vectors in W . This
means that u = k1 v1 +k2 v2 +:::+kn vn and v = l1 v1 +l2 v2 +:::+ln vn .
So,
u+v
= k1 v1 + k2 v2 + ::: + kn vn + l1 v1 + l2 v2 + ::: + ln vn
= (k1 + l1 ) v1 + (k2 + l2 ) v2 + ::: + (kn + ln ) vn
2 W
and
3) W is closed under scalar multiplication:
cu = c (k1 v1 + k2 v2 + ::: + ckn vn )
= ck1 v1 + ck2 v2 + ::: + ckn vn
2 W
2. Each of the vectors v1 ; v2 ; :::; vn is an element of W since vi = 0v1 +
0v2 + :::1vi + ::: + kn vn . Suppose that W 0 is another subspace containing
v1 ; v2 ; :::; vn . Since it is closed under addition and scalar multiplication,
it contains every linear combination of v1 ; v2 ; :::; vn that is vectors of W .
Thus W W 0 .
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CHAPTER 4. VECTOR SPACES
De…nition 278 We de…ne Span (v1 ; v2 ; :::; vn ) to be the subspace of all linear
combinations of v1 ; v2 ; :::; vn .
n! ! !o
Example 279 Earlier we saw that Span i ; j ; k = R3 .
Example 280 The polynomials 1; x; x2 ; :::; xn span Pn since every polynomial
can be written as a0 + a1 x + a2 x2 + ::: + an xn .
4.2.3
Problems
1. Do # 1, 2, 3, 4, 5, 7, 8, 9, 11, 22, 23 on pages 238-240.
2. Show that H as de…ned in example 268 is a subspace of R3 .
3. Any plane P in space which goes through the origin has equation ax +
by + cz = 0 where a, b, c are constants. Prove that such a plane is a vector
space by proving it is a subspace of R3 .