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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