MATH 225 - LECTURE 20
1. Inner Product, Length and Distance




v1
u1




 v2 
 u2 



Defintion. The Inner product u = 
 ...  and v =  ...  is defined as




vn
un
u·v =



3



 −5 

 and v = 
Example. Let u = 
 2




6
1
2
0
1



. Find u · v.


Solution:
Theorem 1. Let u, v and w be vectors in Rn , and let c be any scalar.
Then
(a) u · v = v · u
(b) (u + v) · w = u · w + v · w
1
(c) (cu) · v =c (u · v) = u · (cv)
(d) u · u ≥ 0, and u · u = 0 if and only if u = 0.


v1


 v2 

Defintion. For v = 
 ... , the length or norm of v is the nonnegative


vn
scalar kvk defined by
"
Example. If v =
#
a
, then
b
kvk =
That is the distance between 0 and v):
2


1


 −2 

Example. Let v = 
 2 . Then


0
kvk =
Defintion. The distance between u and v in Rn is defined as
Example. This agrees with the usual formulas for R2 . Let u, v in R2 .
Then
u−v =
and
dist (u, v) =
3
2. Orthogonal Vectors
Figure 1. Orthogonal vectors
By Pythagorean Theorem,
[dist (u, v)]2 = kuk2 + kvk2 .
But we also get,
[dist (u, v)]2 = ku − vk2 = (u − v) · (u − v)
So if u and v are orthogonal,
4
Defintion. Two vectors u and v are said to be orthogonal (to each other)
if u · v = 0.
Example. Are these vector orthogonal?
"
# "
#
−1
,
,
1
  

3
−1
  

•  1  ,  3 .
2
1
•
1
1
5
3. Orthogonal Complements
Defintion. If a vector z is orthogonal to every vector in a subspace W
of Rn , then z is said to be orthogonal to W . The set of vectors z that
are orthogonal to W is called the orthogonal complement of W and is
denoted by W ⊥ .
Figure 2. Orthogonal complements



 

1
−1
0
 

 

Example. Let z =  1 . Then z is orthogonal to span{ 1  ,  1 }.
1
0
−1
6
4. Motivation
Not all linear systems have solutions.
"
1 2
2 4
#"
x1
x2
#
"
#
3
=
exists. Why?
2
(" #)
1
Ax is a point on the line spanned by
and b is not on the line.
2
So Ax 6= b for all x.
Example. No solution to
Figure 3. No solutions
b so that Ab
Instead find x
x lies ”closest” to b.
Figure 4. The closest solution!
7
"
b=
Using information we will learn in this chapter, we will find that x
"
1.4
0
#
#
1. 4
.
2. 8
Segment joining Ab
x and b is perpendicular ( or orthogonal) to the set of
solutions to Ax = b.
so that Ab
x=
That’s why we developed fundamental ideas of length, orthogonality
and orthogonal projections.
This is crucial for any empirical science!
8
,