Problem Set 2  Solutions
J. Scholtz
October 15, 2006
Question 1: 1.5/4
In order to show that the set
{
e
1
, . . . , e
n
}
is linearly independent we need to
show that the following equation is satisﬁed only if
c
1
=
. . .
=
c
n
= 0:
c
1
e
1
+
c
2
e
2
+
. . .
+
c
n
e
n
= 0
But since
e
1
= (1
,
0
, . . . ,
0) and so on, then the LHS is equal to (
c
1
, c
2
, . . . , c
n
).
Hence the last equation reads: (
c
1
, c
2
, . . . , c
n
) = (0
,
0
, . . . ,
0) Which can only be
satisﬁed if all
c
i
s are zero.
Question 2: 1.5/13
a) Two vectors
Claim:
If
u
and
v
are linearly independent then so are
u
+
v
and
u

v
.
Proof:
The equation
a
(
u
+
v
)+
b
(
u

v
) = 0 expands as (
a
+
b
)
u
+(
a

b
)
v
= 0,
but since
u, v
are linearly independent by our assumption, then
a
+
b
= 0 and
a

b
= 0, as
F
is not of charcteristic 2 then
a
=

b
and
a
=
b
imply
a
=
b
= 0
and so our vectors are linearly independent.
Claim:
If
u
+
v
and
u

v
are linearly independent then so are
u
and
u
.
Proof:
If
u
+
v
and
u

v
are linearly independent then so are
x
=
u
+
v
2
and
y
=
u

v
2
. (Dividing by 2 is posssible since Char(
F
)
6
= 2). Then observe that
u
=
x
+
y
and
v
=
x

y
and use the previous part.
b) Three vectors
This is almost the same as the previous part. Please see me if you need more
explanation.
1
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View Full DocumentQuestion 3: 1.5/17
An upper triangular matrix is of form:
A
=
c
11
c
21
. . .
c
n
1
0
.
.
.
. . .
c
n
2
.
.
.
0
.
.
.
.
.
.
0
0
0
c
nn
Then the collumns form vectors of form
v
i
= (0
, . . . ,
0
, k
ii
, . . . , k
in
) where
the ﬁrst
i

1 elements are zeroes. This means that their linear combination
gives:
a
1
v
1
+
. . .
+
a
n
v
n
= (
a
1
k
11
, a
1
k
12
+
a
2
k
22
, . . . ,
X
a
j
k
jn
) = 0
But
k
11
6
= 0 since the diagonal entries are nonzero, hence
a
1
= 0. Then the
second entry reads
a
2
k
22
= 0, and agains since
k
22
6
= 0,then
a
2
= 0. This process
can be repeated ﬁnitely many times (ntimes) to show that
a
1
=
. . .
=
a
n
= 0
and hence the set is linearly independent.
Question 4: 1.6/3(de)
part d)
We can check, the only solution to:
a
(

1 + 2
x
+ 4
x
2
) +
b
(3

4
x

10
x
2
) +
c
(

2

5
x

6
x
2
) = 0
is
a
=
b
=
c
= 0. Hence these three polynomials are linearly independent. And
since we know that dim(
P
2
(
R
)) = 3. Then we know these three polynomials
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 Spring '08
 GUREVITCH
 Linear Algebra, Vector Space, basis

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