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0
0
3
0
0
0
Observe that this implies that there are non-trivial solutions to the system
...
EXAMPLE4
Is the set C
=
2
2
2
{3+2x+2x , 1 +x , 1 +x+x } a basis for P2?
Solution: Consider the equation
2
2
2
2
ao +aix+a2x = ti(3+2x+2x )+t2( 1 +x )+t3( 1 +x+x )
2
= (3ti+t2+t3)+(2ti+t3)X+(2t1 +t2+t3)X
Row reducing the coefficient matrix of the corresponding system gives
[ i l [ o ol
3
2
2
1
0
1
1
1 � 0
1
0
1
0
0
l
Observe that this implies that the system is consistent and has a unique solution for all
ao +aix+a2x
EXERCISE 1
2
E
P2
...
2
2
Prove that the set 23 = {l +2x+x , 1 +x , 1 +x} is a basis for P2•
EXAMPLES
Determine a basis for the subspace§
=
{p(x)
E
P I p(l)
2
=
O} of P
...
By the Factor Theorem, if p(l) = 0, then (x- 1) is a factor of p(x)
...
Consider
Thus, we see that§ =
The only solution is
(x - l)(ax +b)
=
t1
=
t
2
=
0
...
Thus,
{x2 - x, x - 1} is a linearly independent spanning set of§ and hence a basis
...
One standard way of doing this is to first determine a spanning set for the
vector space and then to remove vectors from the spanning set until we have a basis
...
, vk} is a spanning set for a non-trivial vector space V
...
If'T is linearly independent, then
Tis a basis for V, and we are done
...
Then, we
can solve the equation for v; to get
Suppose that
T
{v1,
=
•
•
•
·
·
·
=
So, for any x E V we have
x =
=
a1v1 +
·
·
·
+a;-1 V;-1 +a;v; +a;+1V;+1 +
l�
·
·
·
+akvk
a1V1 +
...
+t;-1Vi-J +t;+1V;+1 +
...
This
T\{v;} is a spanning set for V
...
Otherwise, we repeat the procedure to omit
a second vector, say
v1, and get T\{v;, v1}, which still spans V
...
(Certainly, if there is only one non-zero
vector left, it forms a linearly independent set
...
EXAMPLE6
If T
={[_il [-:J
...
,
0
1, we get
t1
t2
t3
4
=
l
Ul-Hl+�H� Hl UJ+:I
[ �]
Tl{[-�I}=
1
, which
0
or
Thus, we can omit -
from T and consider
{fJHJ
...
{[_iJ
...
[i]}
is
EXERCISE 2
Let 'B
{1
=
-
x, 2 + 2x + x2, x + x2, 1 + x2}
...
Dimension
We saw in Section 2
...
We now prove that this result holds for general vector spaces
...
3
...
If {u1 ,
...
Proof: Since each ui, 1 $ i $ k, is a vector in V, it can be written as a linear combi
nation of the v/s
...
If k > n, then this system would have a non
trivial solution, which would imply that {u1,
...
But, we
of n equations in k unknowns t1,
• • • ,
assumed that {u1,
...
, v,,} and C
=
{u1,
• • •
n
...
Proof: On one hand, 'Bis a basis for V, so it is linearly independent
...
Thus, by Lemma 2, we get that
independent as it is a basis for V, and Span 'B
gives
n
�
k
...
=
n
$
k
...
So Lemma 2
•
As in Section 2
...
Definition
Dimension
If a vector space V has a basis with n vectors, then we say that the dimension of V is
n and write
dimV = n
0
...
The dimension of the trivial vector space is defined to be
Remark
Properties of infinite-dimensional spaces are beyond the scope of this book
...
11 is n-dimensional because the standard basis contains n vectors
...
(c) The vector space Pn is (n + 1 )-dimensional as it has the standard basis
{ 1 , X, x2,
•
...
(d) The vector space C(a, b) is infinite-dimensional as it contains all polynomials
(along with many other types of functions)
...
EXAMPLES
Ut S =Span
ml
...
nl}
Show that ilim S
=
2
...
[=�]}
We row reduce the corresponding coefficent matrix
-1
3
2
-2
2
3
Observe that this implies that
of the first two vectors Thus, S
5
1
and
=
Span
3
-
5
can be written as linear combinations
Moreom, B =
]
...
Thus, dim§
EXAMPLE9
Let§
=
{[: �]
E
Solution: Since d
=
2
...
Determine the dimension of§
...
Thus, dim§
EXERCISE 3
Find the dimension of§
=
{a+ bx+ cx2 + dx3
E
=
for S is also linearly
3
...
Extending a Linearly Independent Subset to a Basis
{vi,
...
If Span T * V,
then there exists some vector wk+! that is in V but not in Span T
...
1)
If tk+I * 0, then we have
Wk+I
=
t1
--V1 tk+I
' ' ·
tk
- -Vk
tk+!
and so Wk+! can be written as a linear combination of the vectors in T, which cannot
be since Wk+!