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0
...
Now, if Span{v1,
, vb Wk+d
V, then it is a basis for V
...
, vb Wk+!• Wk+2},
=
=
· · ·
...
=
which is linearly independent
...
EXAMPLE 10
Let C
=
{[� �], [-� -�]}
Extend C to a basis for M(2, 2)
...
Consider
Row reducing the augmented matrix of the associated system gives
-2
-1
1
1
0
1
b1
b2
b3
b4
1
0
0
0
-2
1
0
0
b1
b2 - b1
b1 - b2 + b3
2b1 - 3b2 + b4
[�� �� ]
[� �]
Hence, '13 is not a spanning set of M(2, 2) since any matrix
(or 2b1 - 3b2 + b4 :/= 0) is not in Span'B
...
Hence, by the procedure above, we should add this matrix to the set
...
Consider
Row reducing the augmented matrix of the associated system gives
1
1
0
So, any matrix
-2
-1
1
[:� :�]
0
0
1
0
0
0
-2
1
0
0
0
0
1
0
bi
b2 - bi
b1 - b2 + b3
2b1 - 3b2 + b4
with 2b1 - 3b2 + b4 :/= 0 is not in Span '131• For example,
is not in the span of '131 and thus '131 is not a basis for M(2, 2)
...
Moreover, we can show that it spans
M(2, 2)
...
EXERCISE 4
Extend the set 'T
=
{[; ]}
to a basis for
R3
...
Theorem 4
Let Vbe an n-dimensional vector space
...
(2) A set of fewer than n vectors cannot span V
...
Proof:
(1) This is Lemma 2 above
...
If Vis n-dimensional,
then it has a basis containing n vectors
...
(3) If 13 is a linearly independent set of n vectors that does not span V, then it can
be extended to a basis for V by the procedure above
...
Similarly, if 13 is a spanning set for V that is not linearly independent, then
by the procedure above, there is a linearly independent proper subset of 13 that
spans V
...
EXAMPLE 11
1
...
Extend the basis 13 to obtain a basis C for
x1 +
2x2
-
3
x
=
0
...
Solution: (a) We know that a plane in JR3 has dimension 2
...
Observe that ii 1
and
v,
=
m
=
[�I
both satisfy the equation of the plane and neither is a scalar multiple
of the other
...
(b) From the procedure above, we need to add a vector that is not in the span of {v 1,
v2}
...
Observe that
V3
=
[�j
does not satisfy the equation of the plane and hence is not in the
{V 1, v2, v3} is a linearly independent
fore is a basis for JR3, according to Theorem 4
...
Thus,
set of three vectors in JR3 and there
EXERCISE 5
Produce a basis for the hyperplane in IR4 with equation
extend the basis to obtain a basis for IR4
...
3
Practice Problems
Al Determine whether each set is a basis for
(a)
{[il-l=: l [;]}
W�lHl}
minrnlHl}
wirnrnJ}
[-:]
...
·
(b)
(c)
(d)
(e)
{l � 1}
AS Determine the dimension of the vector space of
(a)
(b)
(c)
A2 Ut
B=
,
' 1 '
3
basis for IR4
...
Prove that
B
is a
l
{HJ
...
AS Obtain a basis for each of the following vector
spaces and determine the dimension
...
B={[-�
[� -�]}
11, determine a
x1 - x2 +X - X4 = 0
3
A 7 (a) Using the method in Example
B and determine the dimension of Span B
...
3
A6 (a) Using the method in Example
2
3
B
...
{[� �]
=
P2 a= -c}
I
M(2,2) a,b,c
I
E
�:]
[{ [�:l [i] o}
(d) S ={p(x)
(e) S
E
E
E
R3
E
P2 p(2) = O}
{[� !]
E
I
=
I
M(2,2) a= -c}
I
IR}
Homework Problems
Bl Determine whether each set is a basis forJR3
...
{[� �]·[=� =�]·[� -�]·[� �]}
{[ � �] ,[� �] [� �] ,
rn _;J
...
nrnrnJ}
{[�] HJ
...
Ul
}
-
(b) 3=
B6 Determine the dimension of the vector space of
polynomials spanned by3
...
(a)
(b)
B
=
=
B
{ [�l
...
ui Hll
_
+
determine a
+
=0 inJR3
...
BS (a) Using the method in Example
basis for the hyperplane
...
![ -�n
...
...
inJR4
...
B9 Obtain a basis for each of the following vector
S {[� �]EM(2,2) Ia,bE }
S { x2)p(x) p(x)EP2}
;;{[ ] R3 I[;;] Ul o}
S={[; :]EM(2,2)1a-b=O,
= EJR}
=
b
a
S {p(x) EPs Ip(-x) p(x) x}
spaces and determine the dimension
...
" Explain why this makes sense in terms
sional vector space V is a maximal (largest pos
sible) linearly independent set in V
...
(b) It may be said that "a basis for a finite dimen
sional vector space V is a minimal spanning set
of statements in this section
...
Prove
that if
=V
...
(b) Show that if {v1, Vz, v3} is a basis for a vector
space V, then for any s, t E JR, {v1, V2, v3 + tv1+
sv2} is also a basis for V
...
2
...