True or False?
1.
TRUE
2.
FALSE
correct
Explanation:
If an
n
×
n
matrix has
n
distinct
eigenvalues,
it is diagonalizable, but the converse does not
necessarily have to be true.
For example, when
A
=
1
3
3
−
3
−
5
−
3
3
3
1
,
then
det[
A
−
λI
] =
−
(
λ
−
1)(
λ
+ 2)
2
= 0
.
Thus the eigenvalues of
A
are
λ
= 1
,
−
2
,
−
2.
Now
rref(
A
−
I
) =
1
0
−
1
0
1
1
0
0
0
,
so
Nul(
A
−
I
) =
s
1
−
1
1
:
s
in
R
.
moreno (jcm5423) – HW11 – gilbert – (54425)
3
On the other hand,
rref(
A
+ 2
I
) =
1
1
1
0
0
0
0
0
0
,
so
Nul(
A
+ 2
I
)
=
s
−
1
1
0
+
t
−
1
0
1
:
s, t
in
R
.
But then
A
is diagonalizable because it has 3
linearly independent eigenvectors
1
−
1
1
,
−
1
1
0
,
−
1
0
1
.
Since the eigenvalue
λ
=
−
2 is repeated, how
ever,
A
does not have distinct eigenvalues.
Consequently, the statement is
FALSE
.
005
10.0 points
The eigenvalues of the matrix
A
=
2
0
−
2
1
3
2
0
0
3
are
λ
= 2
,
3
,
3.
If
A
is diagonalizable,
i.e.
,
A
=
PDP

1
with
P
invertible and
D
diagonal, which of
the following is a choice for
P
?
1.
P
=
−
1
0
2
1
1
0
0
0
1
2.
P
=
1
0
2
1
1
0
0
0
1
3.
P
=
−
1
0
−
2
1
1
0
0
0
1
correct
4.
A
is not diagonalizable
5.
P
=
1
0
−
2
1
1
0
0
0
1
Explanation:
We ±rst determine the eigenspaces corre
sponding to
λ
= 2
,
3:
λ
= 2 : since
rref(
A
−
2
I
) = rref
0
0
−
2
1
1
2
0
0
1
=
1
1
0
0
0
1
0
0
0
,
there is only free variable and
Nul(
A
−
2
I
) =
s
−
1
1
0
:
s
in
R
.
Thus Nul(
A
−
2
I
) has dimension 1, and
v
1
=
−
1
1
0
is a basis for Nul(
A
−
2
I
).
λ
= 3 : since
rref(
A
−
3
I
) = rref
−
1
0
−
2
1
0
2
0
0
0
=
1
0
2
0
0
0
0
0
0
,
there are two free variables and
Nul(
A
−
3
I
) =
s
0
1
0
+
t
−
2
0
1
:
s, t
in
R
.
Thus Nul(
A
−
3
I
) has dimension 2, and
v
2
=
0
1
0
,
v
3
=
−
2
0
1
,
moreno (jcm5423) – HW11 – gilbert – (54425)
4
form a basis for Nul(
A
−
3
I
).
Consequently,
A
is diagonalizable because
v
1
,
v
2
,
v
3
are linearly independent, and
A
=
PDP

1
with
P
= [
v
1
v
2
v
2
] =
−
1
0
−
2
1
1
0
0
0
1
.
006
10.0 points
Every
n
×
n
matrix
A
having eigenvectors
v
1
,
v
2
, . . .,
v
n
can be diagonalized.
True or False?
1.
TRUE
2.
FALSE
correct
Explanation:
An
n
×
n
matrix
A
can be diagonalized,
i.e.
written as
A
=
PDP

1
for some invert
ible matrix
P
and diagonal matrix
D
, if and
only if
A
has
linearly independent
eigenvectors
v
1
,
v
2
, . . .,
v
n
.
Consequently, the statement is
FALSE
.
007
10.0 points
If
A
is a diagonalizable
n
×
n
matrix, then
A
is invertible.
True or False?
1.
FALSE
correct
2.
TRUE
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.