122 CHAPTER 4. VECTOR SPACES

4.2 Subspaces

4.2.1 De�nitions and Examples

Often, we work with vector spaces which consists of an appropriate subset ofvectors from a larger vector space. We might expect that most of the propertiesof 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 indeedthe 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 sub-space of V if it is itself a vector space under the addition and scalar multipli-cation de�ned on V .

So, in theory, to prove W is a subspace of V , we would need to verify theten axioms. However, since the elements ofW are also in V , we know that mostaxioms will hold. The only two which must be checked are closure under bothoperations. For example, if we take two elements of W , we are guaranteed theirsum is in V since V is a vector space. But we don�t know it is in W . We havethe following theorem.

Theorem 262 If W is a non-empty subset of a vector space V , then W is asubspace 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 thenu+ v 2W .

2. W is closed under scalar multiplication that is if u 2 W and c is anyscalar, then cu 2W .

Proof. We prove both directions.

1. Suppose that W is a subspace of V . Then, by de�nition, it is a vectorspace. 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 showthe ten axioms hold. Axioms 1 and 6 will hold by our assumption. Ifu 2 W then �u 2 W by assumption. We know that u + (�u) = 0 2 W .So, 0 2W . The remaining axioms are automatically satis�ed since everyelement 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 sub-space of another vector space than proving it is a vector space. We will use thiswhenever possible.

Remark 264 Any subset which does not contain the zero vector cannot be asubspace 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 notempty.

Example 266 (polynomials) Earlier, we established that Pn was a vectorspace by proving it directly. We can also prove it by showing it is a subspace ofF (�1;1). Since polynomials are functions with domain (�1;1), we see thatPn � F (�1;1). In addition, it is not empty. As we veri�ed when we proved itwas 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 2Mnn : A is symmetricg.Then, (V;+; :) with standard matrix addition and scalar multiplication is a vec-tor space. We prove it by showing it is a subspace of Mnn. First, V is notempty, it contains the zero n� n matrix. Since the sum of two symmetric ma-trices is symmetric (why?), it is closed under addition. Also, the product of asymmetric matrix by a scalar is also a symmetric matrix (why?), so it is closedunder 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 Rgis 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 hasmany 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 aretwo continuous functions, so is f + g and cf for any scalar c. Thus, it isclosed under addition and scalar multiplication.

2. C1 [a; b], the space of functions whose �rst derivative is continuous. It isa subspace of F [a; b] for the same reasons. If f and g are two functionswhose �rst derivative is continuous, then (f + g)0 = f 0 + g0 will also becontinuous and (cf)0 = cf 0 will be continuous as well. In fact, remember-ing from calculus that if a function is di¤erentiable it is also continuous,we see that C1 [a; b] is a subspace of F [a; b].

3. Similarly, we can de�ne for every positive integer n, Cn [a; b] to be thespace of function whose nth derivative is continuous. This is a subspaceof F [a; b] as well as a subspace of Cn�1 [a; b].

4. Pn is a subspace of Cm [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. Thisis an important result, we give it as a theorem.

124 CHAPTER 4. VECTOR SPACES

Theorem 271 If Ax = 0 is a homogeneous system of m equations and n un-knowns, 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 2W. Also, W � Rn becausethe solutions of Ax = 0 are n�1 vectors. Next, we must establish closure underaddition. 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 anyscalar 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 linearcombination of these vectors is an expression of the form

k1v1 + k2v2 + :::+ knvn

where k1, k2, :::, kn are scalars.

Remark 273 Note that since V is a vector space, k1v1 + k2v2 + :::+ knvn isalso an element of V . So, a linear combination of vectors in a vector space is avector in the same vector space.

Remark 274 A linear combination of one vector is simply a scalar multiple ofthat vector.

Example 275 Earlier, we saw that in R3, every vector �!u = (x; y; z) couldbe 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 vectorspace V , how do we know if another vector w of V can be written as a linearcombination of v1, v2, :::, vn. This amounts to solving a system. We knowthat this will happen if one can �nd scalars k1, k2, :::, kn such that k1v1 +k2v2 + ::: + knvn = w. This is the same as solving the system V k = w where

V = [v1;v2; :::;vn], k =

26664k1k2...kn

37775. We illustrate this with an example.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 k1u+k2v = 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 125

2. Is w0 = (4;�1; 8) a linear combination of u and v?We need to solve k1u+k2v = 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; :::;vnfrom a vector space V , some vectors of V maybe a linear combination ofv1;v2; :::;vn, but others may not. If we construct the set W consisting ofall 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 and3) W is closed under scalar multiplication.

� 1) W 6= ;: 0 2W since 0 = 0v1 + 0v2 + :::+ 0vn.

� 2)W is closed under addition: Let u and v be two vectors inW . Thismeans that u = k1v1+k2v2+:::+knvn and v = l1v1+l2v2+:::+lnvn.So,

u+ v = k1v1 + k2v2 + :::+ knvn + l1v1 + l2v2 + :::+ lnvn

= (k1 + l1)v1 + (k2 + l2)v2 + :::+ (kn + ln)vn

2 W

and

� 3) W is closed under scalar multiplication:

cu = c (k1v1 + k2v2 + :::+ cknvn)

= ck1v1 + ck2v2 + :::+ cknvn

2 W

2. Each of the vectors v1;v2; :::;vn is an element of W since vi = 0v1 +0v2 + :::1vi + :::+ knvn. Suppose that W 0 is another subspace containingv1;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.

126 CHAPTER 4. VECTOR SPACES

De�nition 278 We de�ne Span (v1;v2; :::;vn) to be the subspace of all linearcombinations of v1;v2; :::;vn.

Example 279 Earlier we saw that Spann�!i ;�!j ;�!ko= R3.

Example 280 The polynomials 1; x; x2; :::; xn span Pn since every polynomialcan be written as a0 + a1x+ a2x2 + :::+ anxn.

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 vectorspace by proving it is a subspace of R3.