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Unit 21
Subspaces of
<
n
Defnition 23.1.
A
subspace
of
<
n
is a nonempty set ,
S
, of vectors from
<
n
such that both of the following conditions are met:
1. if
~v
1
and
2
are in
S
then (
1
+
2
)
∈
S
(closed under addition)
2. if
1
∈
S
and
c
is a scalar then
c~v
1
∈
S
(closed under scalar multiplica
tion)
Example 1.
Istheset
S
of all vectors in
<
4
with Frst and last components
being 0 a subspace of
<
4
?
Solution:
Let
1
and
2
be in
S
,say
1
=(0
,a,b,
0)
2
,c,d,
0)
This is an extremely important stepchoosing two arbitrary vectors from
the set. We must use the deFnition of S in order to decide what form the
above vectors must have. Having done this, we must show the two closure
conditions are met:
1. Closure under addition:
Show
1
+
2
is in
S
1
+
2
=(
0
0) + (0
0)
0
,a
+
c, b
+
d,
0)
both Frst and last components are 0 and therefore
1
+
2
is in
S
,and
since
1
and
2
were arbitrary we have S closed under addition.
1
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View Full Document 2. Closure under scalar multiplication:
Show
c~v
1
is in
S
where
c
is a scalar
1
=
c
(0
,a,b,
0)
=(
0
,ca,cb,
0)
both frst and last components are 0 and thereFore
1
is in
S
.S
ince
~v
1
was arbitrary we have S closed under scalar multiplication
ThereFore,
S
is a subspace.
Example 2.
Let
A
be an
m
×
n
matrix and
S
the subset oF
<
n
consisting oF
all vectors
~x
satisFying
A~x
=
~
0. Show that S is a subspace oF
<
n
Solution:
Suppose
~x
1
and
2
are arbitrary vectors From
S
, so by the defni
tion oF S we have
A~x
1
=
~
0and
2
=
~
0
We must show the two closure conditions are met:
1. Closure under addition:
Show
1
+
2
is in
S
,
i.e.
A
(
1
+
2
)=
~
0:
A
(
1
+
2
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This note was uploaded on 04/15/2010 for the course MATH 20C taught by Professor Lit during the Spring '10 term at UCLA.
 Spring '10
 lit
 Calculus, Addition, Vectors, Scalar

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