Math 416  Abstract Linear Algebra
Fall 2011, section E1
Homework 3 solutions
Section 1.6
6.1.
(2 pts)
We want to show that any
w
∈
W
can
be expressed
uniquely
as a linear
combination of
Av
1
, . . . , Av
n
. The equation
w
=
c
1
Av
1
+
. . .
+
c
n
Av
n
=
A
(
c
1
v
1
+
. . .
+
c
n
v
n
)
is equivalent to the equation
A

1
w
=
A

1
A
(
c
1
v
1
+
. . .
+
c
n
v
n
) =
c
1
v
1
+
. . .
+
c
n
v
n
which
has
a
unique
solution (
c
1
, . . . , c
n
) since
{
v
1
, . . . , v
n
}
is a basis of
V
.
6.2.
(1 pt check)
A right inverse to
A
=
1
1
is a 2
×
1 matrix
B
=
c
d
satisfying
AB
=
I
1
, that is
AB
=
1
1
c
d
=
c
+
d
=
1
.
The right inverses of
A
are all matrices of the form
c
1

c
for some
c
∈
R
.
In particular,
A
has distinct right inverses, therefore
A
has no left inverse (by thm 6.1).
6.8.
A
cannot be invertible. If
A
were invertible, the condition
AB
= 0 would imply
A

1
AB
=
A

1
0 = 0, that is
B
= 0.
6.9. (2 pts)
T
1
x
1
x
2
x
3
x
4
x
5
=
x
1
x
4
x
3
x
2
x
5
=
1
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
1
x
1
x
2
x
3
x
4
x
5
T
2
x
1
x
2
x
3
x
4
x
5
=
x
1
x
2
+
ax
4
x
3
x
4
x
5
=
1
0
0
0
0
0
1
0
a
0
0
0
1
0
0
0
0
0
1
0
0
0
0
0
1
x
1
x
2
x
3
x
4
x
5
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 Fall '08
 Staff
 Math, Linear Algebra, Algebra, 1 pt, Diagonal matrix, Triangular matrix, X1, Normal matrix

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