calles (dac3476) – HW11 – gilbert – (54425)
1
This printout should have 22 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
10.0 points
IF the matrix
A
=
b
2
−
1
1
4
B
is diagonalizable,
i.e.
,
A
=
PDP

1
with
P
invertible and
D
diagonal, which oF the Fol
lowing is a choice For
D
?
1.
A
is not diagonalizable
correct
2.
D
=
b
3
0
0
3
B
3.
D
=
b
−
3
0
0
−
3
B
4.
D
=
b
3
0
0
2
B
5.
D
=
b
−
3
0
0
2
B
Explanation:
Since
det [
A
−
λI
] =
b
2
−
λ
−
1
1
4
−
λ
B
= 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
b
−
1
−
1
1
1
B
=
b
1
1
0
0
B
,
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
.
002
10.0 points
An
n
×
n
matrix can be diagonalizable, but
not invertible.
True or ±alse?
1.
±ALSE
2.
TRUE
correct
Explanation:
Consider the 3
×
3 triangular matrix
A
=
5
−
8
1
0
0
7
0
0
−
2
Because
A
is triangular, its eigenvalues are the
entries along the diagonal,
i.e.
,
λ
= 5
,
0
,
−
2.
Since these are distinct,
A
is diagonalizable.
On the other hand, one oF its eigenvalues
is zero, so
A
is not invertible (or note that
det[
A
] = 0 because
det[
A
] = 5(0)(
−
2) = 0
is the product oF the diagonal values oF
A
, so
A
is not invertible). ThereFore, an
n
×
n
matrix
A
can be diagonalizable, but not invertible.
Consequently, the statement is
TRUE
.
003
10.0 points
IF the matrix
A
=
b
1
4
1
−
2
B
is diagonalizable,
i.e.
,
A
=
PDP

1
with
P
invertible and
D
diagonal, which oF the Fol
lowing is a choice For
P
?
calles (dac3476) – HW11 – gilbert – (54425)
2
1.
P
=
b
−
1
4
2
1
B
2.
P
=
b
−
1
1
1
4
B
3.
P
=
b
2
4
−
1
1
B
4.
P
=
b
−
1
4
1
1
B
correct
5.
A
is not diagonalizable
Explanation:
Since
det [
A
−
λI
] =
b
1
−
λ
4
1
−
2
−
λ
B
=
−
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
.
A
x
=
λ
x
, λ
=
−
3:
b
1
4
1
−
2
Bb
x
1
x
2
B
=
b
x
1
+ 4
x
2
x
1
−
2
x
2
B
=
−
3
b
x
1
x
2
B
,
which can be written as
x
1
+ 4
x
2
=
−
3
x
1
,
x
1
−
2
x
2
=
−
3
x
2
,
i.e.
,
x
1
=
−
x
2
. So one choice of
v
1
is
v
1
=
b
−
1
1
B
.
A
x
=
λ
x
, λ
= 2:
b
1
4
1
−
2
Bb
x
1
x
2
B
=
b
x
1
+ 4
x
2
x
1
−
2
x
2
B
= 2
b
x
1
x
2
B
,
which can be written as
x
1
+ 4
x
2
= 2
x
1
,
x
1
−
2
x
2
= 2
x
2
,
i.e.
,
x
1
= 4
x
2
. So one choice of
v
2
is
v
2
=
b
4
1
B
.
Consequently,
A
=
PDP

1
with
P
=
b
−
1
4
1
1
B
.
004
10.0 points
If an
n
×
n
matrix
A
is diagonalizable, then
A
has
n
distinct eigenvalues.