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DIMENSION
Math 21b, O. Knill
Homework: Section 3.3: 22,24,32,38,52,36*,56*
LINEAR SPACE.
X
is a
linear space
if
~
0
∈
X
and
if
X
is closed under addition and scalar multiplica
tion.
Examples:
R
n
, X
= ker(
A
)
, X
= im(
A
) are linear
spaces.
REVIEW BASIS.
B
=
{
~v
1
, . . . , ~v
n
} ⊂
X
B
linear independent
:
c
1
~v
1
+
...
+
c
n
~v
n
= 0 implies
c
1
=
...
=
c
n
= 0.
B
span
X
:
~v
∈
X
then
~v
=
a
1
~v
1
+
...
+
a
n
~v
n
.
B
basis
: both linear independent and span.
BASIS: ENOUGH BUT NOT TOO MUCH. The spanning
condition for a basis assures that there are
enough
vectors
to represent any other vector, the linear independence condi
tion assures that there are
not too many
vectors. A basis is,
where J.Lo meets A.Hi:
Left: J.Lopez in
”Enough”
, right ”The
man who new
too much
” by A.Hitchcock
AN
UNUSUAL
EXAMPLE.
Let
X
be
the
space
of
polynomials
up
to
degree
4.
For
example
p
(
x
)
=
3
x
4
+ 2
x
3
+
x
+ 5
is
an
element
in
this
space.
It
is
straightforward
to
check
that
X
is
a
linear
space.
The
”zero
vector” is
the
function
f
(
x
)
=
0
which
is
zero
everywhere.
We
claim
that
e
1
(
x
) = 1
, e
2
(
x
) =
x, e
3
(
x
) =
x
2
, e
4
(
x
) =
x
3
and
e
5
(
x
) =
x
4
form a basis in
X
.
PROOF. The vectors span the space:
every polynomial
f
(
x
) =
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
is a sum
f
=
c
0
e
1
+
c
1
e
2
+
c
2
e
3
+
c
3
e
4
+
c
4
e
5
of basis elements.
The vectors are linearly independent: a nontrivial relation 0 =
c
0
e
1
+
c
1
e
2
+
c
2
e
3
+
c
3
e
4
+
c
4
e
5
would mean that
c
0
+
c
1
x
+
c
2
x
2
+
c
3
x
3
+
c
4
x
4
= 0 for all
x
which is not possible unless all
c
j
are zero.
DIMENSION. The number of elements in a basis of
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 Spring '03
 JUDSON
 Linear Algebra, Algebra, Addition, Multiplication, Scalar

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