Homework 3, Due March 2
Adam Gamzon
Related reading: sections 2.4  3.2 from the text
1. Determine whether or not
1
0
0
,
1
2
0
,
1
2
3
are linearly independent.
Solution:
To do this, solve the system
a
1
0
0
+
b
1
2
0
+
c
1
2
3
= 0
.
Using GaussJordan elimination,
1
1
1

0
0
2
2

0
0
0
3

0
→
1
1
1

0
0
1
1

0
0
0
3

0
→
1
0
0

0
0
1
1

0
0
0
3

0
→
1
0
0

0
0
1
1

0
0
0
1

0
→
1
0
0

0
0
1
0

0
0
0
1

0
.
Hence the only solution is
a
=
b
=
c
= 0, so these vectors are linearly independent.
2. Find the inverse of the matrix
0
0
1
0
1
0
1
0
1
.
Solution:
0
0
1

1
0
0
0
1
0

0
1
0
1
0
1

0
0
1
→
1
0
1

0
0
1
0
1
0

0
1
0
0
0
1

1
0
0
→
1
0
0


1
0
1
0
1
0

0
1
0
0
0
1

1
0
0
1
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So the inverse is

1
0
1
0
1
0
1
0
0
.
3.
(a) Show that if
A
2
is invertible then
A
is invertible.
Solution:
If
A
2
is invertible then there is a matrix, (
A
2
)

1
, such that
A
2
(
A
2
)

1
=
I
n
.
Since
A
2
=
AA
, associativity of matrix multiplication gives
I
n
=
A
2
(
A
2
)

1
=
A
(
A
(
A
2
)

1
).
That is,
A
is invertible and its inverse is
A
(
A
2
)

1
.
(b) Let
T
:
R
n
→
R
m
be a linear transformation.
If vectors
v
and
w
are in ker(
T
), is the
vector
v
+ 3
w
in ker(
T
)? Why? What is
T
(
v
+ 3
w
)?
Solution:
The vector
v
+ 3
w
is in ker(
T
) because
T
(
v
+ 3
w
) =
T
(
v
) + 3
T
(
w
) = 0 + 0 = 0
.
(Alternatively, one could argue that the kernel is a subspace, so
v
+ 3
w
must be in ker(
T
)
because subspaces are closed under addition and scalar multiplication.)
4.
(a) For the matrix
A
=
0
1
0
0
0
1
0
0
0
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 Spring '10
 STAFF
 Math, Linear Algebra, Algebra, Vector Space, Ker

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