SUBSPACES

Definition: Let S be a nonempty subset of

a vector space V. S is said to be a

subspace of V, if S is a vector space

under the same operations of addition

and scalar multiplication as in V.

Theorem A nonempty subset S of a vector

space V is a subspace of V, if the

following conditions are satisfied:

(i)If u, vS, then u+vS.

(ii)If uS and any scalar, then u S.

In other words, a subset S of a vector

space V is a subspace of V, if it is closed

under addition and scalar

multiplication defined in V.

Proof: Let S be a subspace of V. Then S is a vector space under the same operations as those of V. Hence, S satisfies (i) and (ii). Conversely, if (i) and (ii) are satisfied, then we have to prove that S satisfies all the axioms of a vector space. 1 and 2 are satisfied for S, because this is exactly what (i) and (ii) say. We shall now show that V3 for S follows from 1 and 2 for S.

0u=0 for any uV and therefore for any u0

S. Taking any u0 S,

We see that, from (i), 0 =0u0 S. Similarly,

(-1) u =-u for any u V and therefore

also for a given u0 S. Hence,

-u0 S for every u0 S. Thus, S2 and S3

hold in S.

Example :1 Let V=R3 the vector space as

usual further Let

U={u=(x1x2,x3) V/x1+x2+x3=0}

Now to show that U is a subspace of V.

Let U= (x1,x2,x3), v=(y1,y2,y3) U, then

x1+x2+x3=0, y1+y2+y3=0 by definition of

U. u+v=(x1+y2, x1+y2, x3+y3)

Now (x1+y1) + (x2 +y2)+(x3+y3)

=(x1+x2+x3) + (y1+y2,+y3)=0+0=0 from

above,this shows that u+vU.Now

u= (x1,x2,x3) = (x1, x2, x3)

x1+ x2+ x3 = (x1+ x2+ x3)= 0=0

This shows uU, U is a subspace of V=R3

Example :2 Let V3 = {u =(x1,x2,x3 ) }

with usual addition and multiplication

U={u=(x1,x2,x3) V3 /x1x2+x2x3=0}

Let u,v U, u=(x1,x2,x3), v = (y1,y2,y3 ))

Then x1x2 + x2x3=0 , y1y2 + y2y3=0

Now consider u+v=(x1+y1, x3+y2, x3+y3)

Now to check (x1+y1) (x2+y2)+(x2+y2)(x3+y3)

= (x2+y2) (x1+y1+ x3+y3)=(x2+y2)(x1+x3+

y1+y3)= (x2+y2)(x1+x3)+(x2+y2)(y1+y3)

= x2(x1+x3)+y2(x1+x3)+x2(y1+y3)+0

0 may not be zero in general in

particular

In particular (x1,x2,x3)= u = (1,0,3),

v=( y1,y2,y3 ) = (3 ,2,-2 ) U

since 0(1+3) =0 ,3(2-2)=0 But

y2(x1+x3)+x2(y1+y3)=3(1+3)+0(2-2)=120

Hence U is not a subspace of V.

Theorem 2.2: Let V be a vector space. If U and W be subspaces of V, then U W is also a subspace of V. Proof: Clearly U W is a subset of V, since U and W are subsets of V. Let u,vUW, then u,v U and u,vW, u+v U, u U, u+vW, u W ,since

U and W are subspaces of V, therefore u+vUW,uUW by definition of intersection.

Hence U W is a subspace of V. Any

intersection of subspaces of a vector space

V is a subspace of V.

V0 = { 0} is the smallest subspace of every

vector space.

Remark:However in general union of two

subspaces need not be a subspace of vector

space V i.e. in general U W need not be

a subspace of V .

Example 3: Let U={(x,0,0)x-real number}

and w ={(0,y,0)y-real number} be two

subspaces of vector space V3 ={(x,y,z)x,y,z

any real numbers}

In this UW is set of vectors, which does

not contain a vector, which is sum of

a vector of U and a vector of W.

For all scalars , ,

(x,0,0)+(0,y,0)=(x,y,0) is not in UW

hence UW is not a subspace of V.

Example:4 Let V = P4 be a vector space of

polynomials of degree 4 with usual

addition and scalar multiplication

Let U={p(x) p4 / =2p’(1) } prove

that U is a subspace of P4 .

Let p(x), q(x) U, then by definition

)1(q)1(p)1()qp(and)x(q)x(p)x()qp(

),x(q)x(p)x)(qp(Now

)1(q2)1(qand)1(p2)1(p

)1(p

4P of subspce a is UHence

U (x) p)( Therefore

(1))p2( (1)p (1) p also

U (x) q)(p Hence

(1))q2(p

(1)q(1)p2( (1)q2 (1)p2

(1)q (1)p (1))q(p

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