Partial Solution Set, Leon
§
3.4
3.4.3
Given the vectors
x
1
= (2
,
1)
T
,
x
2
= (4
,
3)
T
, and
x
3
= (7
,

3)
T
,
(a) Show that
x
1
and
x
2
form a basis for
R
2
.
(b) Why must
x
1
,
x
2
, and
x
3
be linearly dependent?
(c) What is the dimension of Span(
x
1
,
x
2
,
x
3
)?
Solution
:
(a) This follows because they are (by inspection) linearly independent in
R
2
. Since the
dimension of
R
2
is 2,
x
1
and
x
2
form a basis for
R
2
.
(b) Since
R
2
is 2dimensional, any collection of more than two vectors from
R
2
must
be linearly dependent.
(c) Since
x
1
and
x
2
form a basis for
R
2
, the dimension of Span(
x
1
,
x
2
,
x
3
) is 2, i.e.,
Span(
x
1
,
x
2
,
x
3
) =
R
2
.
3.4.4
Given the vectors,
x
1
= (3
,

2
,
4)
T
,
x
2
= (

3
,
2
,

4)
T
, and
x
3
= (

6
,
4
,

8)
T
, what is
the dimension of Span(
x
1
,
x
2
,
x
3
)?
Solution
: Both
x
2
and
x
3
are scalar multiples of
x
1
, so the dimension of Span(
x
1
,
x
2
,
x
3
)
is 1.
3.4.5
Given the vectors,
x
1
= (2
,
1
,
3)
T
,
x
2
= (3
,

1
,
4)
T
, and
x
3
= (2
,
6
,
4)
T
,
(a) Show that
x
1
,
x
2
, and
x
3
are linearly dependent.
(b) Show that
x
1
and
x
2
are linearly independent.
(c) What is the dimension of Span(
x
1
,
x
2
,
x
3
)?
(d) Give a geometric description of Span(
x
1
,
x
2
,
x
3
).
Solution
:
(a) Letting
A
=
x
1
x
2
x
3
, we consider the solutions to
A
x
=
0
. (Yes, this is one
of those situations in which the matrix turns out to be square, so the determinant
is a possibility.
But it is not recommended.
Use Gaussian elimination instead.)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Anshelvich
 Linear Algebra, Algebra, Vectors, Vector Space, X1,

Click to edit the document details