Notes on Homework 12
1.
(a) Consider the function g : V → V given by
m
g(v) = v − ∑ hv, ei iei ;
i =1
we’ll show that it’s a linear transformation. As usual, there are two things to show:
First, suppose v1 , v2 ∈ V; we’ll show g(v1 + v2 ) = g(v1 ) + g(v2 ). Well,
m
g ( v 1 + v 2 ) = v 1 + v 2 − ∑ h v 1 + v 2 , ei i ei
i =1
m
= v1 + v2 − ∑ (hv1 , ei i + hv2 , ei i)ei additivity of h·, ·i
i =1
m
= v1 + v2 − ∑ (hv1 , ei iei + hv2 , ei iei ) distributivity of vector mult
i =1
m
m
= v1 + v2 − ∑ (hv1 , ei iei ) − ∑ hv2 , ei iei
i =1
m
i =1
m
= v1 − ∑ (hv1 , ei iei ) + v2 − ∑ hv2 , ei iei
i =1
i =1
= g(v1 ) + g(v2 ).
Second, suppose v ∈ V, λ ∈ R; we’ll show g(λv) = λg(v), by computing:
m
g(λv) = λv − ∑ hλv, ei iei
i =1
m
= λv − ∑ λ hv, ei iei
i =1
m
= λ (v − ∑ λ hv, ei iei )
i =1
= λg(v).
(b) It need not betruethat f (u) = u. Many examples are possible,
but, consider the subx
2
space U = {
: x ∈ R} ⊂ R2 ; as a basis, try e1 =
. (Note that has length
0
0
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
two, and thus {e1 } is not an orthonormal basis for U.) Suppose v =
a
0
∈ U. Then
f (v) = hv, e1 ie1
= ( a · 2 + 0 · 0)
4a
=
6= v
0
2.
2
0
(a) From class, we know that trying to find a polynomial a0 + a1 x which goes through the
four points is the same as solving the equation
 


4
1 1 1
1 2 a0



=
3
1 3 a1
1
1 4
Ax = b
This probably isn’t possible, but we can find the best approximate solution,
x̂ = ( A T A)−1 A T b
4
=
7
− 10
(computation omitted); the best-fitting line is
y=−
7
x + 4.
10
(b) To find the best-fitting degree two polynomial, we apply the same method, except using
the matrix


1 1 12
1 2 22 

B=
1 3 32 
1 4 42
we calculate
ŷ = ( B T B)−1 B T b
 21 
4

= − 39
20
1
4
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
and the best-fitting quadratic is
y=
1
39
21
x−
x+
4
20
4
(c) Now, use

1
1
C=
1
1
1
2
3
4
12
22
32
42

13
23 

33 
43
we calculate
ẑ = (C T C )−1 C T b


21
−27

=
 23 
2
− 32
and the best-fitting cubic is
3
2
y = − x3 + 3x − 27x + 21
2
2
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008

3.

 
2
1
(a) There are many different ways to do this. For instance, let z1 =  0 , z2 =  2 ,
1
0
 
0
z3 =  1 , and let
1


2 1 0
A = 0 2 1 .
1 0 1
To find numbers x1 , x2 , x3 such that x1 z1 + x2 z2 + x3 z3 = v is to solve the matrix equation

  
x1
1
A  x2  =  4  .
7
x3
(Why?) So, solve this equation any way you like. For example
 2
1 
− 15
5
5
2
2
A−1 =  51
−
5
5
2
1
4
−5
5
5
(calculation omitted)

1
sothatx = A−1  4 
7


1
=  −1 
6

Similarly, to find numbers y1 , y2 , y3 such that y1 z1 + y2 z2 + y3 z3 = w, use
 
3
−1 
3 
y=A
2
 
1
= 1 
1
(b) Let B = { z1 , z2 , z3 }. By definition, the Gram matrix is given by Gi j = h zi , z j i. Thus, for
example,
G1,2 = h z1 , z2 i
= 2·1+0·2+1·0
=2
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008
Note that this is calculated by multiplying the first row of A T against the second column
of A. More generally,
G = AT · A


5 2 1
= 2 5 2
1 2 2
(Of course, you can also just compute each h zi , z j i directly and then assemble the information at the end.)
(c) From class, we know that
hv, wi = [v]BT G [w]B

 
5 2 1
1



1 
= (1 − 1 6) 2 5 2
1 2 2
1
= 29
(d) We could also calculate directly:
hv, wi = 1 · 3 + 4 · 3 + 7 · 2
= 29
Professor Dan Bates
Colorado State University
M369 Linear Algebra
Fall 2008