1.
FALSE
correct
2.
TRUE
Explanation:
A nonzero vector
x
is an eigenvector of
A
if and only if
A
x
=
λ
x
for some scalar
λ
. Now
A
x
=
3
6
7
3
2
7
5
6
4
1
−
2
2
=
5
13
1
negationslash
=
λ
1
−
2
2
for any choice of scalar
λ
.
Consequently, the statement is
FALSE
.
011
10.0points
If the matrix
A
=
bracketleftbigg
2
−
1
1
4
bracketrightbigg
is diagonalizable,
i.e.
,
A
=
PDP

1
with
P
invertible and
D
diagonal, which of the fol
lowing is a choice for
D
?
1.
D
=
bracketleftbigg
3
0
0
3
bracketrightbigg
2.
D
=
bracketleftbigg
−
3
0
0
2
bracketrightbigg
3.
D
=
bracketleftbigg
3
0
0
2
bracketrightbigg
4.
D
=
bracketleftbigg
−
3
0
0
−
3
bracketrightbigg
5.
A
is not diagonalizable
correct
Explanation:
Since
det [
A
−
λI
] =
bracketleftbigg
2
−
λ
−
1
1
4
−
λ
bracketrightbigg
= 1 + (2
−
λ
)(4
−
λ
) = 9
−
6
λ
+
λ
2
,
the eigenvalues of
A
are the solutions of
9
−
6
λ
+
λ
2
= (3
−
λ
)
2
= 0
,
i.e.
,
λ
= 3
,
3.
On the other hand, when
λ
= 3,
rref(
A
−
λI
) = rref
bracketleftbigg
−
1
−
1
1
1
bracketrightbigg
=
bracketleftbigg
1
1
0
0
bracketrightbigg
,
so
x
2
is the only free variable.
Thus the
eigenspace Nul(
A
−
3
I
) has dimension 1. But
then, when
λ
= 3,
geo mult
A
(
λ
)
< alg mult
A
(
λ
)
.
Consequently,
A
is not diagonalizable
.
012
10.0points
If the matrix
A
=
bracketleftbigg
1
4
1
−
2
bracketrightbigg
is diagonalizable,
i.e.
,
A
=
PDP

1
with
P
invertible and
D
diagonal, which of the fol
lowing is a choice for
D
?
1.
D
=
bracketleftbigg
1
0
0
−
2
bracketrightbigg
2.
A
is not diagonalizable
3.
D
=
bracketleftbigg
−
1
0
0
2
bracketrightbigg
4.
D
=
bracketleftbigg
3
0
0
−
2
bracketrightbigg
5.
D
=
bracketleftbigg
−
3
0
0
2
bracketrightbigg
correct
Explanation:
huynh (lth436) – HW09 – gilbert – (57245)
6
Since
det [
A
−
λI
] =
bracketleftbigg
1
−
λ
4
1
−
2
−
λ
bracketrightbigg
=
−
4
−
(1
−
λ
)(2 +
λ
) =
λ
2
+
λ
−
6
,
the eigenvalues of
A
are the solutions of
λ
2
+
λ
−
6 = (
λ
+ 3)(
λ
−
2) = 0
,
i.e.
,
λ
=
−
3
,
2.
Thus
A
is diagonalizable
because the eigenvalues of
A
are distinct, and
A
=
PDP

1
with
D
=
bracketleftbigg
λ
1
0
0
λ
2
bracketrightbigg
=
bracketleftbigg
−
3
0
0
2
bracketrightbigg
.
013
10.0points
If the matrix
A
=
bracketleftbigg
1
4
1
−
2
bracketrightbigg
is diagonalizable,
i.e.
,
A
=
PDP

1
with
P
invertible and
D
diagonal, which of the fol
lowing is a choice for
P
?
1.
A
is not diagonalizable
2.
P
=
bracketleftbigg
−
1
1
1
4
bracketrightbigg
3.
P
=
bracketleftbigg
−
1
4
1
1
bracketrightbigg
correct
4.
P
=
bracketleftbigg
−
1
4
2
1
bracketrightbigg
5.
P
=
bracketleftbigg
2
4
−
1
1
bracketrightbigg
Explanation:
Since
det [
A
−
λI
] =
bracketleftbigg
1
−
λ
4
1
−
2
−
λ
bracketrightbigg
=
−
4
−
(1
−
λ
)(2 +
λ
) =
λ
2
+
λ
−
6
,
the eigenvalues of
A
are the solutions of
λ
2
+
λ
−
6 = (
λ
+ 3)(
λ
−
2) = 0
,
i.e.
,
λ
=
−
3
,
2.
Thus
A
is diagonalizable
because the eigenvalues of
A
are distinct, and
A
=
PDP

1
with
P
= [
v
1
v
2
]
where
v
1
and
v
2
are eigenvectors correspond
ing to
λ
1
and
λ
2
respectively.
To determine
v
1
and
v
2
we solve the equation
A
x
=
λ
x
.