1.
SOLUTIONS
(1)(20 points)
(
a
) If the sequence
a
n
satisfies lim
n
→∞
a
n
+1

a
n
= 0 then lim
n
→∞
a
n
exists and is finite.
F
Take
a
n
=
√
n.
(
b
) If
q
is a polynomial of degree 5 and
q
(0) =
q
(0) =
q
(4)
(0) = 0 and
q
(0) = 1
, q
(0) =

1
6
, q
(5)
(0) =
1
120
then
q
is the fifth degree Taylor polynomial for
f
(
x
) = sin
x
at
c
= 0
.
T
Since
f
(0) =
f
(0) =
f
(4)
(0) = 0 and
f
(0) = 1
, f
(0) =

1
6
, f
(5)
(0) =
1
120
.
(
c
) The matrix
A
=
0 1 0
0 0 1
0 0 0
is diagonalizable.
F
One way to see this is that the only eigenvalue is 0
,
it has multiplicity 3
,
and the eigenspace
N
(
A

0
I
) =
{
(
t,
0
,
0)
T
:
t
∈
R
}
so there are not three linearly independent eigenvectors.
Another way to see this is that if
A
were diagonalizable, then for some invertible matrix
X
we would have
A
=
XDX

1
and
D
is the diagonal matrix with the eigenvalues of
A
on its diagonal. But this would be the
0 matrix and as a result
A
is also the 0 matrix, but it is not, so
A
can not be diagonalizable.
1
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2
(
d
) If an eigenvalue of the matrix
A
has multiplicity greater than one, then
A
is defective.
F
Defective means not diagonalizable, but the identity matrix has sole eigenvalue 1 and is
already diagonal.
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 Spring '10
 DONALDSON,NEIL
 Linear Algebra, Matrices, Sin, degree Taylor polynomial

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