Math 115A: Linear Algebra
Homework 6: Final version: Due Friday November 4
Homework from text:
§
2.4 4, 6, 7, 17,
§
2.5 2d, 4, 5, 7.
§
5.1 3, 8, 17, 20.
Double star 11: Let
P
n
be all real polynomials of degree less than or equal
to
n
. Let
c
0
,c
1
,...c
n
be distinct numbers. Let
T
:
P
n
→
R
n
+1
be deﬁned by
T
(
p
(
x
)) = (
p
(
c
0
)
,p
(
c
1
)
,...,p
(
c
n
))
.
Show
T
is an isomorphism. You may use that a polynomial
p
(
x
)
∈
P
n
has
at most
n
roots unless
p
(
x
) = 0. The fact that
T
is onto is called Lagrange
interpolation, famous in applied math.
Double star 12: Suppose
V
has dimension 2 and
T
:
V
→
V
is linear.
Suppose
T
2
=
T
. Suppose
T
is not 0 and that
T
is not an isomorphism.
Show
V
has a basis
{
α
1
,α
2
}
=
B
so that
[
T
]
B
=
1 0
0 0
!
Hint: Show there are
α
1
∈
R
(
T
),
α
2
∈
N
(
T
) which form a basis.
Double star 13: Suppose dim
V
= 3 and
T
3
= 0, but
T
2
6
= 0. Show that we
can ﬁnd a basis
B
so that
[
T
]
B
=
0 0 0
1 0 0
0 1 0
Hint: If
T
2
(
α
)
6
= 0, consider
B
=
{
α,T
(
α
)
,T
2
(
α
)
}
.
You have to show
B
is a
basis.
Problem: In this problem, the scalar ﬁeld is
C
. Let
A
=
cos(
θ
)

sin(
θ
)
sin(
θ
)
cos(
θ
)
!
Find a matrix
Q
so that
Q

1
AQ
is diagonal. What is
Q

1
AQ
?
Challenge problems: Text
§
2.5: 13.
Challenge 6.1 A variant of double star 12: Suppose
V
has dimension 2 and
T
:
V
→
V
is linear. Suppose
T
2
=
T
. Suppose
T
is not 0 and that
T
is not
the identity. Show
V
has a basis
{
α
1
,α
2
}
=
B
so that
[
T
]
B
=
1 0
0 0
!
Challenge 6.2 Suppose dim
V
= 3 and
T
2
= 0, but
T
6
= 0. Show there is a
basis
B
so that
[
T
]
B
=
0 0 0
1 0 0
0 0 0