1.
Solution:
scores
(a).
True
The eigenspace of
A
corresponding to eigenvalue
λ
is the null space of
A
I
λ
−
.
Since
2 2
A
×
∈ℜ
and with an eigenspace of dimension 2, rank
(
)
0
A
I
λ
−
=
, or,
equivalently,
(
)
0
A
I
λ
−
=
⇒
A
I
λ
=
.
5
(a).
False
D
is a diagonal matrix and its diagonal entry
ii
d
is the eigenvalue of
A
corresponding to column
i
p
, eigenvector corresponding to
ii
d
. If we change the
permutation of the columns of
P
, the order of the diagonal entries of
D
will
change.
Ex.
1
1
1
3
2
3
1
0
2
3
3
2
2
0
3
2
0
2
0
1
0
2
0
1
1
0
0
1
1
0
A
−
−
⎡
⎤
⎡
⎤ ⎡
⎤ ⎡
⎤
⎡
⎤ ⎡
⎤ ⎡
⎤
=
=
=
⎢
⎥
⎢
⎥ ⎢
⎥ ⎢
⎥
⎢
⎥ ⎢
⎥ ⎢
⎥
⎣
⎦
⎣
⎦ ⎣
⎦ ⎣
⎦
⎣
⎦ ⎣
⎦ ⎣
⎦
5
2.
Solution:
scores
Assume that
n
x
∈
^
satisfies
0
Ax
=
, then
(
)
0
T
T
A
Ax
A Ax
=
=
. Thus
(
)
(
)
Null
Null
T
A
A A
⊆
(1)
Assume that
n
x
∈
^
satisfies
0
T
A Ax
=
, then
(
)
2
0
T
T
T
T
x
A Ax
x A Ax
Ax
=
=
=
⇒
0
Ax
=
.
Thus
(
)
(
)
Null
Null
T
A A
A
⊆
(2)
From (1) and (2), we have
(
)
(
)
Null
Null
T
A
A A
=
⇒
Nullity of
Nullity of
T
A
A A
=
⇒
(
)
(
)
(
)
rank
rank
Nullity of
T
A
A A
n
A
=
=
−
(3)
10

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