MAT1341A: LECTURE NOTES FOR FALL 2008
MONICA NEVINS (HEAVILY BORROWING FROM HANDWRITTEN NOTES OF BARRY JESSUP)
5.
Subspaces and Spanning Sets
Last time, we established the concept of a
vector space
, which is a set
V
on which we can
perform two operations (addition and scalar multiplication) such that 10 axioms are satisfied.
The result is: a vector space is algebraically indistinguishable from things like
R
n
(as far as
addition and scalar multiplication go).
We met some particularly nice vector spaces:
•
R
n
,
n
≥
1
• E
, a space of linear equations in 3 variables
•
F
[
a, b
], functions on the interval [
a, b
]
•
F
(
R
), functions on the real line
•
M
m
×
n
(
R
),
m
×
n
real matrices
And for each of these, we had to check 10 axioms. But we noticed that sometimes, the last
6 axioms (the ones dealing purely with algebraic properties) were “obvious”.
Let’s consider this more carefully.
5.1.
Subsets of Vectors Spaces.
[N, 5.1.2] Suppose
V
is a vector space, and
W
⊂
V
is a
subset
of
V
.
Example:
Let
V
=
R
2
(we know it’s a vector space), and let
W
=
{
(
x,
2
x
)

x
∈
R
}
. Use the
same
operations of addition and scalar multiplication on
W
as we use in
R
2
. What do we
really need to verify to decide if it’s a vector space?
closure:
(1) closure under addition: if (
x,
2
x
) and (
y,
2
y
) are in
W
, then (
x,
2
x
) + (
y,
2
y
) =
(
x
+
y,
2(
x
+
y
)), which is in
W
because it has the correct form (
z,
2
z
) with
z
=
x
+
y
. YES.
(2) closure under scalar multiplication: if (
x,
2
x
)
∈
W
and
c
∈
R
then
c
(
x,
2
x
) = (
cx,
2(
cx
))
∈
W
as well. YES.
existence:
(3) the zero vector is (0
,
0), which is in
W
. But we don’t need to bother checking
if 0 +
u
=
u
for each
u
∈
W
, because we know this is true for all
u
∈
R
2
already (since
V
is
a vector space and so satisfies axiom (3)). So it’s enough to know that 0
∈
W
.
Date
: September 18, 2008.
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MONICA NEVINS (HEAVILY BORROWING FROM HANDWRITTEN NOTES OF BARRY JESSUP)
(4) if
u
∈
W
, then we know that

u
= (

1)
u
∈
W
because (2) is true. And again, we know
that the negative has the correct property because we checked it when we decided that
V
was a vector space.
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 Winter '10
 DAIGLE
 Vector Space, Sets, Complex number, Monica Nevins

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