Partial Solution Set, Leon
§
3.3
3.3.1
Determine whether the following vectors are linearly independent in
R
2
.
(a)
±
2
1
²
,
±
3
2
²
. Yes. These are clearly not scalar multiples of one another, and when
testing two vectors that’s all that we need to show.
(c)
±

2
1
²
,
±
1
3
²
,
±
2
4
²
. No. This can be shown in two ways. First, the easy way:
If the span of the ﬁrst two vectors is all of
R
2
(it is; they are linearly independent),
all three cannot help being linearly dependent. Done.
The almostaseasy way: Set up a homogeneous system in which the three vectors
in question are the columns of a matrix
A
. Then apply Gaussian elimination to
show that there are nontrivial solutions to the homogeneous equation
A
x
=
0
.
(Recommended, if you feel that you need more practice.)
3.3.2
Same as (1), except that we are now in
R
3
. A pair of vectors is linearly independent
unless they are scalar multiples of one another, and that takes care of (e). In (b), even if
we can ﬁnd three vectors that are linearly independent (and we can), it is easy to show
that those three span
R
3
, so if we add any vector(s) we create a linearly dependent set.
In other words, a set of four vectors from
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, Scalar, Vector Space, ax, linearly independent vectors

Click to edit the document details