The following definition captures the essential idea that a subset of a vector space be a "linear" subset.
DEFINITION: A nonempty subset'l1ofa vector space 'l! is calleda linear subspace, or vector subspace,of'l! ifandonly ifthefollowing twoconditionsaresatisfied:
(1) ifA E'l1andB E'l1thenA+B E'l1, and (2) ifA E'l1andrem then rAE'l1.
These two conditions assert that applying the two basic vector opera- tions to elements of the collection'l1give again elements of the collec- tion 'l1. Ifthe vector space 'l!is complex then condition (2) should be replaced by
(2CC) ifA E'l1and c ECC then cAE'l1,
and likewise in the sequel. Note that 'l!is a linear subspace of itself, and we may refer to'l!as being alinear space.
PROPOSITION4.1.1: If'l1 isalinear subspaceofthevector space 'l!
then 'l1 is itselfa vector space ifwedefine vector addition andscalar multiplicationasin 'l!.
35 L. Smith, Linear Algebra
© Springer Science+Business Media New York 1998
PROOF: Notice that conditions (1), (2) in the definition of subspace assure us that we have operations on'l.1,i.e., ifA andB belong to 'l.1,so do A+B, and ifr belongs to m,rA also belongs to 'l.1. The properties expressed by Axioms 1, 2, 5, 6, 7, and 8 are valid for vectors in'11and hence for vectors in the smaller set'l.1. To verify Axiom 3, we first show that 0 E'l.1. Since'l.1is nonempty, there exists at least one vector AE'l.1.
By (2.1) and condition (2) for a subspace, 0 =OAE'l.1. Axiom 3 is now immediate, since it holds in'11. To verify Axiom 4, suppose that AE'l.1.
Then (-I)AE'l.1, but by Proposition 2.1.2, (-I)A=-A, and therefore -A E'l.1,and Axiom 4 holds. 0
There are some trivial examples of subspaces, to wit: '11 is always a subspace of'11 and the set consisting of the zero vector alone {O} is always a subspace of'11. We often abuse notation and write0 for this subspace. There is a also trivial example that isnota subspace, namely the empty set0, which the definition of subspace explicitly excludes.
ExAMPLE1: Here is a less trivial example of a linear subspace. We consider inmnthe subset 'l.1consisting of all vectors A=(a1, ... , an)
with the property that a1 + ... + an=O. If A=(a1, . .. , an),B= (b1 , .•• , bn ) E'l.1andcis a number, then
A+B=(a1+ b1, . .. , an +bn), c .A =(cal, ... , can)
and
a1+b1+ ... +an +bn=(a1 + +an)+(b1+ ... +bn)=,()+ 0=0, cal+ ...can =c(a1+ an)=c(O)=0,
so 'l.1is a linear subspace ofmn.
ExAMPLE2: The solution space to a linear homogeneous equation is a generalization of the preceding example. A linear homogeneous equation in the variablesXl, ••• , Xnis an equation of the form
A solution to this equation is a sequence ofn numbers(81, ... , 8n )such that
a181+ . " +a n8n=O.
IfA =(81, ... , 8n )and B=(t1,' .. , tn )are solutions to(*),define their sum by
A + B =(81+t1, . .. , 8n +tn ).
4.1 Basic Properties of Vector Subspaces 37 We claim that A+B is again a solution to(*). For we have
=alSl+altl+a2S2+a2t2+ ... +anSn+an t n
=alSl+a2S2+ ... +anSn+altl+a2t2+ ... +antn
=0 +0=0,
as we claimed. Next, defineaAfor a numbera to be
aA=(asl, ... , asn).
Simple manipulation shows that
al(aSl)+ ... +an(asn)= aalSl+ +aanSn
=a(alsl+ +ansn)
=a(O)=0,
so aA is again a solution to (*). We define a vector space 0/ by the interpretation
vector +-+solutionto(*), vector addition +-+as defined above, scalar multiplication +-+as defined above.
To show that 0/ is a vector space we will show that it is actually a linear subspace ofmn• For by definition the vectors of0/ are sequences
(SI," ., sn) of numbers and hence are vectors in mn. The process of adding solutions and multiplying solutions by scalars is exactly the process of adding vectors in mn and multiplying a vector ofmn by a number. In our preceding discussion we checked the following:
(1) IfA, B E0/, then A+BE0/.
(2) If A E0/, thenaAEo/for any numbera.
Thus we may apply Proposition 4.1.1 to conclude that 0/ is a vector space. But wait! In order to apply 4.1.1 to 0/, we must know that 0/ is nonempty, that is, that(*) has at least one solution. Happily this is a simple point, because (0, ... , 0) is a solution to(*),as one easily sees, since
alO+a20+ ... +anO=0+ ... +0 =O.
Thus 0/ is a linear subspace ofmn.
The preceding example may be extended from one equation to many, but this is a topic for further study. (See chapter 13.)
DEFINITION:IfAI, ... ,An arevectors of'll, thena linear combi- nation ofAI, ... , Anisa vectorofthe form
whereaI, ... , an arenumbers.
DEFINITION: lfthe vectorsAI, ... , An are fixed, thelinear span ofAI, ... , An,denoted byL(Al, ... , An),or Span {AI,"" An},is the setofallvectors of'll thatarelinear combinationsofAI, ... , An.
PRoPOSITION4.1.2: Suppose that AI, ... , An are vectors of'll.
ThenL(AI , ... , An)isalinearsubspaceof'll.
PROOF: We must verify that the two conditions of the definition of a linear subspace are satisfied by the set of all linear combinations of AI, ... , An. So suppose
Then
for suitable numbers a~,... , a~, a~, ... , a~. Then using the general- ized associative and commutative laws, we find
A' A"'A+ = al I + ... +an n'A + al I + ... +"A an"An
'A "A 'A"A
=al I + al I + +an n+an n
=(a~+a~)AI+ +(a~+a~)An,
which shows that A' + A" is again a linear combination of AI, ... , An, that is, A' +A" eL(Al, ... ,An). SimilarlyifAeL(AI, ... ,An)and rem., then
A= alAI + ... +anAn for suitable numbersal,... , an,so
rA=r(alAI + +anAn)
=ralAI + +ranAn
showing that rA eL(Al, ... , An). Therefore L(AI, ... , An) is a sub- spaceof'll. D
The idea ofthe linear span is not restricted to finite sets of vectors, but may be extended to arbitrary sets of vectors as follows.
4.1 Basic Properties of Vector Subspaces 39 DEFINITION: Let0/ bea vector space and E c 0/; that is, E isa collection ofvectors in 0/. A linear combination ofvectors inE is a vector in0/ofthe form
where At, ... , An EE. The linear span ofE denoted by £(E), or Span {E},is the setofall vectors that are linear combinationsofvectors ofE.
PROPOSITION4.1.3: Let 0/ be a vector space and E c 0/. Then
£(E) isa linear subspaceofo/.
The proof of Proposition 4.1.3 follows closely the proof of Proposition 4.1.2 and will be left to the diligent student. Note that the linear span allows us to assign to each subset E of a vector space 0/ a subspace
£(E)of0/,and thatE C £(E).
PRoPOsmON4.1.4: Let 0/ be a vector space and E C 0/. Then E =£(E) ifand only ifE isalinear subspaceof0/.
PROOF: Suppose that E is a linear subspace of0/. Ifthe vectors AI, ... ,An belong to E and aI, ... , an are numbers, then the vector alAI+ ... +anAnbelongs toE becauseEis closed under the operations of scalar multiplication and vector addition. Therefore, £(E) c E.
SinceE c £(E),we conclude thatE =£(E).
Conversely, suppose thatE =£(E). IfA,BEE, then A+B is certainly a linear combination of vectors inE and hence A+B belongs to £(E), which, since £(E) =E,leads us to conclude A+BEE. Likewise, aA is a linear combination of vectors ofE and hence belongs to £(E)=E.
Therefore,E is closed under vector addition and scalar multiplication, and hence E is a linear subspace of0/. 0
The preceding propositions show that in general a vector space has an abundance of subspaces.
EXAMPLE 3: In lR3 consider the subspace spanned by the two vectors A =(-1, 0,1) and B =(0, 1,0), see Figure 4.1.1. Note that this is just the plane through the origin with equation x +z=O. That is, the vectors in£(A,B) are those vectors(x, y,z) ElR3whose coordinates
satisfy the equationx - z =O.
y
Figure 4.1.1
PRoPOSITION4.1.6: LetSandf['besubspacesofo/. ThenS ("\f['is alsoa subspaceofo/.
PRooF: 0eS and0ef['since Sand f['are subspaces. Therefore
oeS ("\f['.
Suppose that A eS ("\f['and B eS ("\f['. Then A eS and B eS. SinceS is a subspa~,A +B eS. Likewise, A ef['and B ef[',and sincef['is a subspace, A + B e'l'. Therefore, A + B eS ("\ 'l'.
Ifr is a number, then since Sandf['aresubspaces, rA eS and rA ef[', so rAeS("\rt,showing thatS ("\f['is again a subspace of0/. D
DEFINITION: IfS and 'l' aresubspaces of0/, theirsum, denoted byS +f[',is definedtobe thesetofall vectors Cin0/ oftheform
C=A+B, wbereAeS andB ef['.
PRoPOSITION4.1.6: IfS andf[' aresubspaces of the vector space 0/, then so isS + 'l'.
PRooF: Suppose that C... C2 eS + 'l'. Write
Then
Cl =A1+B1,
C2 =A2 +B2,
Al eS,B1e'l', A2 eS;B2 ef['.