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
get weadding,on conditions abovecheckTo
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