Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
Bases and Dimensions
4
...
11• For example, the dot product of two vectors a and b was defined in terms of
the standard components of the vectors
...
Therefore, it would be useful to define the same concept for any vector space
...
2 that the two important properties of the standard basis in IR
...
11 and it was linearly independent
...
Why would it be important that the
set 13 be linearly independent? The following theorem answers this question
...
,v11} be a spanning set for a vector space V
...
=
Proof: Let x be any vector in V
...
Assume that there are linear combinations
=
This gives
which implies
If 13 is linearly independent, then we must have a; - b;
Hence, x has a unique representation
...
has a solution where at least one of the coefficients is non-zero
...
+ Ovn
Hence, 0 can be expressed as a linear combination of the vectors in 13 in multiple
ways
...
Definition
Basis
A set 13 of vectors in a vector space Vis a basis if it is a linearly independent spanning
set for V
...
However, we would like every vector space to have a basis, so we define the empty set
to be a basis for the trivial vector space
...
1
...
It is called the standard basis for M(2, 2)
...
1
...
, �}is called the standard basis for P11•
EXAMPLE2
�ove that the set C
{[i], m, [m
JR
...
JR
...
3
[t3 1] [ti t1 t2 t2 t3]
t2 t3
1] [1 1 l
Solution: We need to show that Span C
prove that Span C
P3
...
To
=
we need to show that every vector 1 E
can be written as a
linear combination of the vectors in C
...
3
...
3
...
2
...
3
...
Therefore, we have a unique solution when we take
1
=
0, so C is also linearly independent
...
Consider the equation
o[ o] [ 2] [o
i
0
ti
=
-1
0
1
+t2
3
]
[ ][
2
+t3
1
1
1
s
ti +2t3
=
-ti+3t2+t3
3
2t1 +t2+St3
ti+ 2+3t3
]
Row reducing the coefficient matrix of the corresponding system gives
1
0
2
1
0
2
2
1
5
0
1
1
-1
3
1
0
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+!
In this case, (4
...
Thus, t1 v1 +
+
tkvk + tk+! Wk+I
tk+I
0 implies that t1
, vb wk+!} is
0, and hence {v1,
linearly independent
...
If
not, we repeat the procedure to add another vector Wk+2 to get {v1,
...
• •
=
=
=
·
· ·
=
tk
=
· ·
·
• •
...
In this fashion, we must eventually get a basis, since
according to Lemma 2, there cannot be more than n linearly independent vectors in an
n-dimensional vector space
...
Solution: We first want to determine whether C is a spanning set for M(2, 2)
...
In particular,
with b1 -b2+b3 :/= 0
is not in the span of '13
...
We let
and repeat the procedure
...
Adding
get
rn �]
[� �]
to '131 we
By construction '132 is a linearly independent
...
Thus, it is a basis for M(2, 2)
...
Knowing the dimension of a finite dimensional vector space Vis very useful when
trying to invent a basis for V, as the next theorem demonstrates
...
Then
(1) A set of more than n vectors in V must be linearly dependent
...
(3) A set with n elements of Vis a spanning set for V if and only if it is linearly
independent
...
(2) Suppose that Vcan be spanned by a set of k
< n
vectors
...
This means we have a set of n linearly
independent vectors in a set spanned by k
<
n vectors which contradicts (1)
...
But, this would give a
linearly independent set of more than n vectors in V, which contradicts (1)
...
But, then V would be spanned by a set of fewer that n vectors, which
contradicts (2)
...
Produce a basis 13 for the plane P in
IR3
with equation
2
...
JR3
...
By Theorem 4, we just need
to pick two linearly independent vectors that lie in the plane
...
Thus they are linearly independent, and 13
=
{v 1, v2} is a basis for the
plane P
...
But, Span{v 1, v2} is in the plane, so we need to pick any vector not in the plane
...
plane
...
x1 - x2 + X - 2x4 = 0
3
and
PROBLEMS 4
...
[J
�
{ [ ]}
JR3
...
basis for the plane
(a)
for
(b)
B=
JR3
...
r!HH lm
{[�l r=�i flrnHm
basis for the hyperplane
in IR4•
(b) Extend the basis of part (a) to obtain a basis
for IR4
...
(a) S ={a+bx+cx2
(b) S =
(c) s =
·
A4 Select a basis for Span B from each of the following
sets
(a)
B and determine the dimension of Span B
...
B=
a
(b) Extend the basis of part (a) to obtain a basis
A3 Select a basis for Span B from each of the following
sets
11, determine
2x1 - x2 - X = 0 in JR3
...
B={l+x,l+x+x2,l+x3}
B= (1+x,1 - x, 1+x3,1 - x3}
B = {1 + x + x2,1 - x3,1 - 2x + 2x2 - x3,
1 - x2+2x3,x2+x3}
polynomials spanned by
�]·[� -�]·[� =�J
...
(a)
{[ il Ul Ul}
BS Select a basis for Span3 from each of the following
sets3 and determine the dimension of Span3
...
r; -�n
1 x+x2,x+x2+x3, 1-x3}
x+ x x2+x3, 1 x2+x3}
11,
x1 3x2 4x3
11,
x1 x2+2x3+ =
(a) 3=
·
(b)
(c)
(d)
(e)
{[�J
...
lm
mHm
[
{[J J
...
(a) 3= { +
(b) 3= {l+
(c) {
+
(d) {
+
+
B4 Select a basis for Span3 from each of the following
sets3 and determine the dimension of Span3
...
UJ rm
{[ :i [-irnl
...
for JR3
...
B3 Determine whether each set is a basis for
+
+
(b) Extend the basis of part (a) to obtain a basis
Prove that3 is a basis for
(b) {
+
basis for the plane
{[� �]·[� !] [! -�J
...
M(2, 2)
...
P
2
+x+x2, 1 -x2}
11++x2,
x2, 21-xx 2x2,-2 2x}
x2, -x-x2, 1+2x2}
3+2x+2x2,1+x+x2,1 -x - x2}
(a) {1
x2,
B7 (a) Using the method in Example
B2 Let3=
-
,
_
...
determine a
+
X
4
0
(b) Extend the basis of part (a) to obtain a basis
for JR4
...
(a)
(b)
=
JR
= Cl+
(c) s =
(d)
(e)
=
I
=
e
c
0, c
0
=
for all
Conceptual Problems
Dl (a) It may be said that "a basis for a finite dimen
for V
...
" Explain
why this makes sense in terms of statements in
this section
...
D2 Let V be an n-dimensional vector space
...
S
S
is a subspace of V and dim
S
= n, then
D3 (a) Show that if {v1, v2} is a basis for a vector space
V, then for any real number t, {v1, v2+ tvi} is
also a basis for V
...
D4 In Problem 4
...
D2 you proved that V
=
{(a, b) I
a, b E JR, b > O}, with addition defined by (a, b) EB
(c, d) = (ad+ be, bd) and scalar multiplication de
fined by t O (a, b) = (taY-1, b1), was a vector space
over R Find, with justification, a basis for V and
hence determine the dimension of V