SOLUTION TO ASSIGNMENT 8  MATH 251, WINTER 2007
In this assignment
F
=
R
or
C
.
1. Consider results of an experiment given by a series of points:
(
x
1
, y
1
)
,
(
x
2
, y
2
)
, . . . ,
(
x
n
, y
n
)
,
where
x
1
< x
2
<
· · ·
< x
n
and the
x
i
, y
i
are real numbers.
We assume that the actual law governing this data is linear. Namely, that there is an equation of the
form
f
A,B
(
x
) =
Ax
+
B
that fits the data up to experimental errors.
Therefore, we look for such an
equation
Ax
+
B
that fits the data best. Our measure for that is “the method of list squares”. Namely,
given a line
Ax
+
B
, let
d
i
=

y
i

(
Ax
i
+
B
)

(the distance between the theoretical
y
and the observed
y
).
Then we seek to minimize
d
2
1
+
d
2
2
+
· · ·
+
d
2
n
.
Let
T
:
R
2
→
R
n
,
be the map
T
(
A, B
) = (
f
A,B
(
x
1
)
, . . . , f
A,B
(
x
n
))
.
Prove that
T
is a linear map:
We can represent
T
by a matrix
(
A, B
)
7→
x
1
1
x
2
1
.
.
.
.
.
.
x
n
1
A
B
¶
.
So clearly it is linear.
and that the problem we seek to solve is to minimize
k
T
(
A, B
)

(
y
1
, . . . , y
n
)
k
2
.
Let us calculate:
k
T
(
A, B
)

(
y
1
, . . . , y
n
)
k
2
=
k
(
Ax
1
+
B, . . . , Ax
n
+
B
)

(
y
1
, . . . , y
n
)
k
2
=
∑
n
i
=1
(
Ax
i
+
B

y
i
)
2
=
∑
n
i
=1
d
2
i
.
Let
W
be the subspace of
R
n
which is the image of
T
.
Prove that
W
is two dimensional and that
{
s
1
, s
2
}
is a basis for
W
, where
s
1
= (1
,
1
, . . . ,
1)
, s
2
= (
x
1
, x
2
, . . . , x
n
).
W
is spanned by
T
(0
,
1)
and
T
(1
,
0)
, which are just
s
1
and
s
2
.
W
is two dimensional, unless
x
1
=
x
2
=
· · ·
=
x
n
, which is not the case.
Assume for simplicity that
∑
n
i
=1
x
i
= 0 (this can always be achieved by shifting the data). Find an
orthonormal basis for
W
and use it to find the vector in
W
closest to (
y
1
, . . . , y
n
).
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
2
SOLUTION TO ASSIGNMENT 8  MATH 251, WINTER 2007
The vector closest to
(
y
1
, . . . , y
n
)
is its orthogonal projection on
W
. Let
v
1
=
n

1
/
2
(1
,
1
, . . . ,
1)
,
v
2
= (
n
X
i
=1
x
2
i
)

1
/
2
(
x
1
, . . . , x
n
)
.
Then
{
v
1
, v
2
}
are an orthonormal basis for
W
.
The projection is the function
v
7→ h
v, v
1
i
v
1
+
h
v, v
2
i
v
2
.
Taking
v
= (
y
1
, . . . , y
n
)
we get that the projection is
∑
n
i
=1
y
i
n
(1
, . . . ,
1) +
h
x
,y
i
k
x
k
2
x
, where we have let
x
=
(
x
1
, . . . , x
n
)
, y
= (
y
1
, . . . , y
n
)
.
Put now everything together to get explicit formulas for
A, B
such that
f
A,B
(
x
) is the best linear
approximation to the data
(
x
1
, y
1
)
,
(
x
2
, y
2
)
, . . . ,
(
x
n
, y
n
)
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Goren
 Linear Algebra, Real Numbers, Ker

Click to edit the document details