MATH 245 Linear Algebra 2, Solutions to Assignment 3
1:
(a) Find the leastsquares best ﬁt quadratic
f
∈
P
2
for the following data points.
x
i

1 0 1 2
3
y
i
0
2 3 2

2
Solution: Let
A
=
1
x
1
x
1
2
1
x
2
x
2
2
1
x
3
x
3
2
1
x
4
x
4
2
1
x
5
x
5
2
=
1

1 1
1
0
0
1
1
1
1
2
4
1
3
9
and
y
=
y
1
y
2
y
3
y
4
y
5
=
0
2
3
2

2
. Then we have
A
t
A
=
1
1 1 1 1

1 0 1 2 3
1
0 1 4 9
1

1 1
1
0
0
1
1
1
1
2
4
1
3
9
=
5
5
15
5
15 35
15 35 99
.
A
t
y
=
1
1 1 1 1

1 0 1 2 3
1
0 1 4 9
0
2
3
2

2
=
5
1

7
, and
(
A
t
A
±
±
A
t
y
)
=
5
5
15
5
15 35
15 35 99
±
±
±
±
±
±
5
1

7
∼
1
1
3
0 10 20
0 20 54
±
±
±
±
±
±
1

4

22
∼
1 1
3
0 1
2
0 0 14
±
±
±
±
±
±
1

2
5

14
∼
1 0 1
0 1 2
0 0 1
±
±
±
±
±
±
7
5

2
5

1
∼
1 0 0
0 1 0
0 0 1
±
±
±
±
±
±
12
5
8
5

1
.
Thus the best ﬁt quadratic is
f
(
x
) =
12
5
+
8
5
x

x
2
.
(b) Let
{
p
1
,p
2
,
···
,p
l
}
be a linearly independent set of polynomials in
P
m
and let (
x
1
,y
1
)
,
(
x
2
,y
2
)
,
···
,
(
x
n
,y
n
)
be points in
R
2
such that at least
m
+1 of the
x
coordinates
x
i
are distinct. Show that there exists a unique
polynomial
f
∈
Span
{
p
1
,p
2
,
···
,p
l
}
which minimizes the sum
n
∑
i
=1
(
y
i

f
(
x
i
)
)
2
.
Solution: For
p
(
t
) =
a
0
+
a
1
t
+
···
+
a
m
t
m
∈
P
m
, we write
p
(
x
) =
(
p
(
x
1
)
,p
(
x
2
)
,
···
,p
(
x
n
)
)
t
∈
R
n
. Note that
p
(
x
) =
p
(
x
1
)
p
(
x
2
)
.
.
.
p
(
x
n
)
=
a
0
+
a
1
x
1
+
···
+
a
m
x
1
m
a
0
+
a
1
x
2
+
···
+
a
m
x
2
m
.
.
.
a
0
+
a
1
x
n
+
···
+
a
m
x
n
m
=
A
[
p
]
,
where
A
=
1
x
1
x
1
2
···
x
1
m
1
x
2
x
2
2
···
x
2
m
.
.
.
1
x
n
x
n
2
···
x
n
m
∈
M
n
×
(
m
+1)
and [
p
]=
a
0
a
1
.
.
.
a
m
∈
R
m
+1
is the coordinate vector of
p
.
We need to ﬁnd
c
∈
R
l
such that the polynomial
f
=
c
1
p
1
+
c
2
p
2
+
···
+
c
l
p
l
minimizes
±
±
y

f
(
x
)
±
±
. Note that
f
(
x
) =
c
1
p
1
(
x
) +
···
+
c
l
p
l
(
x
) =
c
1
A
[
p
1
] +
···
+
c
l
A
[
p
l
] =
(
A
[
p
1
]
,
···
,A
[
p
l
]
)
c
=
APc,
where
P
=
(
[
p
1
]
,
[
p
2
]
,
···
,
[
p
l
]
)
∈
M
(
m
+1)
×
l
. Since some
m
+ 1 of the
x
i
are distinct, it follows from question
5 on assignment 2 that some set of
m
+ 1 rows of
A
is linearly independent, so
A
has rank
m
+ 1 and hence
A
is 1:1. Since
{
p
1
,p
2
,
···
,p
l
}
is linearly independent, it follows that
²
[
p
1
]
,
[
p
2
]
,
···
,
[
p
l
]
³
is also linearly