MA 265 LECTURE NOTES: FRIDAY, MARCH 7
Bases and Dimension
Basic Results.
Let (
V,
+
,
·
) be a real vector space of dimension
n
. We recall a few statements we made
without proof at the end of the previous lecture.
(1) If
T
is a maximal independent set of vectors in
V
, then
T
contains exactly
n
vectors.
(2) If
T
is a set of
n
linearly independent vectors, then it is a basis for
V
.
(3) Any collection
S
of
k > n
vectors must be linearly dependent.
(4) If
T
is a minimal spanning set for
V
, then
T
contains exactly
n
vectors.
(5) It
T
is a set of
n
vectors which spans
V
, then it is a basis for
V
.
(6) Any collection of
S
of
k < n
vectors cannot span
V
.
We explain why these are true.
For (1), let
T
=
{
v
1
,
v
2
, . . . ,
v
m
}
be an maximal independent set of vectors. We show that
T
is a basis
for
V
. If this were
not
true, then
T
would not span
V
, so that we could find some vector
v
∈
V
not in the
span of
T
. Consider
S
=
{
v
1
,
v
2
, . . . ,
v
m
,
v
}
. If
S
were linearly dependent, then we may write
a
1
v
1
+
a
2
v
2
+
· · ·
+
a
m
v
m
+
a
v
=
0
=
⇒
v
=

a
1
a
v
1
+

a
2
a
v
2
+
· · ·
+

a
m
a
v
m
so that
v
is in the span of
T
– a contradiction. Hence
S
must be a linearly independent set which properly
contains
T
. This contradicts the maximality of
T
– so that our original assumption on
T
must have been
incorrect. Hence
T
must be a basis for
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 Spring '10
 ...
 Linear Algebra, Vector Space, Wj, linearly independent set, V1

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