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Solutions to Homework 4.
Math 110, Fall 2006.
Prob 2.3.13.
Let
A
= [
a
ij
]. Then
A
t
= [
a
ji
], and
tr(
A
) =
n
X
i
=1
a
ii
=
n
X
j
=1
a
jj
= tr(
A
t
)
.
The elements of
A
= [
a
ij
],
B
= [
b
ij
],
AB
= [
c
ij
] and
BA
= [
d
ij
] are connected by the formulas
c
ij
=
n
X
k
=1
a
ik
b
kj
,
d
ij
=
n
X
k
=1
b
ik
a
kj
.
So,
tr(
AB
) =
n
X
i
=1
c
ii
=
n
X
i
=1
n
X
k
=1
a
ik
b
ki
=
n
X
i
=1
n
X
k
=1
b
ik
a
ki
=
n
X
i
=1
d
ii
= tr(
BA
)
.
The secondlast equality is obtained by interchanging indices
i
and
k
.
Prob 2.3.15.
Suppose
A
has
n
columns
{
A
i
:
i
= 1
,...,n
}
, and write
A
= [
A
1
,A
2
,...,A
n
]. Then
MA
= [
MA
1
,...,MA
n
], i.e., the columns of
MA
are columnvectors
MA
1
,...,MA
n
. So if
A
j
=
∑
i
6
=
j
a
i
A
i
,
then
MA
j
=
M
(
X
i
6
=
j
a
i
A
i
) =
X
i
6
=
j
a
i
MA
i
,
i.e., the
j
th column of
MA
is a linear combination of its other columns with the same coeﬃcients.
Prob 2.3.17.
For any vector
x
, we have
x
=
Tx
+ (
x

Tx
). Note that
Tx
is in the range
R
(
T
) of
T
and
x

Tx
is in the kernel
N
(
T
), since
T
(
x

Tx
) =
Tx

T
2
x
=
Tx

Tx
= 0
.
This shows that
V
=
R
(
T
) +
N
(
T
). By the RankNullity theorem, we can therefore conclude that this sum
is direct (see Prob. 2.1.35a). Also note that
T
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 Fall '06
 Holtz
 Linear Algebra, Algebra, Formulas

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