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Orthogonality. When the angle between two vectors is π/2(900)π/2(900), we
say that the vectors are orthogonal. A quick look at the definition of angle
(Equation 12 from "Linear Algebra: Direction Cosines") leads to this equivalent definition for orthogonality:
(x,y)=0⇔x
and
y
are orthogonal.
(x,y)=0⇔xandy are orthogonal.(1)
For example, in Figure 1(a), the vectors x=31x=31 and y=-26y=-26 are clearly
orthogonal, and their inner product is zero:
(
x
,
y
)
=
3
(
-
2
)
+
1
(
6
)
=
0
.
(
x
,
y
)
=
3
(
-
2
)
+
1
(
6
)
=
0
.
(2)
In Figure 1(b), the vectors x=310,y=-260x=310,y=-260, and z=004z=004 are
mutually orthogonal, and the inner product between each pair is zero:
(
x
,
y
)
=
3
(
-
2
)
+
1
(
6
)
+
0
(
0
)
=
0
(
x
,
y
)
=
3
(
-
2
)
+
1
(
6
)
+
0
(
0
)
=
0
(3)
(
x
,
z
)
=
3
(
0
)
+
1
(
0
)
+
0
(
4
)
=
0
(
x
,
z
)
=
3
(
0
)
+
1
(
0
)
+
0
(
4
)
=
0
(4)
(
y
,
z
)
=
-
2
(
0
)
+
6
(
0
)
+
0
(
4
)
=
0
.
(
y
,
z
)
=
-
2
(
0
)
+
6
(
0
)
+
0
(
4
)
=
0
.
(5)
Which of the following pairs of vectors is orthogonal:
- x=101,y=010x=101,y=010 ;
- x=110,y=111x=110,y=111 ;
- x=e1,y=e3x=e1,y=e3 ;
- x=abx=ab , y=-ba?y=-ba?
We can use the inner product to find the projection of
one vector onto another as illustrated in Figure 2. Geometrically we find
the projection of
x
x onto
y
y by dropping a perpendicular from the head of
x
x
onto the line containing
y
y. The perpendicular is the dashed line in the figure.
The point where the perpendicular intersects
y
y (or an extension of
y
y) is the
projection of
x
x onto
y
y, or the component of
x
x along
y
y. Let's call it
z
z.
Draw a figure like Figure 2 showing the projection of
yy onto
x.x.
The vector
z
z lies along
y
y, so we may write it as the product of its norm
||z||||z|| and its direction vector
u
y
u
y
:
z=||z||uy=||z||y||y||.z=||z||uy=||z||y||y||.
(6)
But what is norm ||z||||z||? From Figure 2 we see that the vector
xx is just
zz,
plus a vector
vv that is orthogonal to
yy:
x
=
z
+
v
,
(
v
,
y
)
=
0
.
x
=
z
+
v
,
(
v
,
y
)
=
0
.
(7)
Therefore we may write the inner product between
xx and
yy as
(x,y)=(z+v,y)=(z,y)+(v,y)=(z,y).(x,y)=(z+v,y)=(z,y)+(v,y)=(z,y).
(8)
But because
z
z and
y
y both lie along
y
y, we may write the inner product (x,y)(x,y)
as
(
x
,
y
)
=
(
z
,
y
)
=
(
|
|
z
|
|
u
y
,
|
|
y
|
|
u
y
)
=
|
|
z
|
|
|
|
y
|
|
(
u
y
,
u
y
)
=
|
|
z
|
|
|
|
y
|
|
|
|
u
y
|
|
2
=
|
|
z
|
|
|
|
y
|
|
.
(
x
,
y
)
=
(
z
,
y
)
=
(
|
|
z
|
|
u
y
,
|
|
y
|
|
u
y
)
=
|
|
z
|
|
|
|
y
|
|
(
u
y
,
u
y
)
=
|
|
z
|
|
|
|
y
|
|
|
|
u
y
|
|
2
=
|
|
z
|
|
|
|
y
|
|
.
(9)
From this equation we may solve for ||z||=(x,y)||y||||z||=(x,y)||y|| and substitute ||z||||z|| into
Equation 6 to write
z
z as
z
=
|
|
z
|
|
y
|
|
y
|
|
=
(
x
,
y
)
|
|
y
|
|
y
|
|
y
|
|
=
(
x
,
y
)
(
y
,
y
)
y
.
z
=
|
|
z
|
|
y
|
|
y
|
|
=
(
x
,
y
)
|
|
y
|
|
y
|
|
y
|
|
=
(
x
,
y
)
(
y
,
y
)
y
.
(10)
Equation 10 is what we wanted–an expression for the projection of
x
x onto
y
y in terms of
x
x and
y
y.
Show that ||z||||z|| and
z
z may be written in terms of cosθcosθ for
θ
θ as
illustrated in Figure 2:
|
|
z
|
|
=
|
|
x
|
|
c
o
s
θ
|
|
z
|
|
=
|
|
x
|
|
c
o
s
θ
(11)
z
=
|
|
x
|
|
cos
θ
|
|
y
|
|
y
.
z
=
|
|
x
|
|
cos
θ
|
|
y
|
|
y
.
(12)
Orthogonal Decomposition. You already know how to decompose
a vector in terms of the unit coordinate vectors,
x
=
(
x
,
e
1
)
e
1
+
(
x
,
e
2
)
e
2
+
⋯
+
(
x
,
e
n
)
e
n
.
x
=
(
x
,
e
1
)
e
1
+
(
x
,
e
2
)
e
2
+
⋯
+
(
x
,
e
n
)
e
n
.
(13)
In this equation, (x,ek)ek(x,ek)ek is the component of
x
x along
e
k
e
k
, or the projection
of
x
x onto
e
k
e
k
, but the set of unit coordinate vectors is not the only possible
basis for decomposing a vector. Let's consider an arbitrary pair of orthogonal
vectors
x
x and
y
y:
(
x
,
y
)
=
0
.
(
x
,
y
)
=
0
.
(14)
The sum of
x
x and
y
y produces a new vector
w
w, illustrated in Figure 3, where
we have used a two-dimensional drawing to represent
n
n dimensions. The norm
squared of
w
w is
|
|
w
|
|
2
=
(
w
,
w
)
=
[
(
x
+
y
)
,
(
x
+
y
)
]
=
(
x
,
x
)
+
(
x
,
y
)
+
(
y
,
x
)
+
(
y
,
y
)
=
|
|
x
|
|
2
+
|
|
y
|
|
2
.
|
|
w
|
|
2
=
(
w
,
w
)
=
[
(
x
+
y
)
,
(
x
+
y
)
]
=
(
x
,
x
)
+
(
x
,
y
)
+
(
y
,
x
)
+
(
y
,
y
)
=
|
|
x
|
|
2
+
|
|
y
|
|
2
.
This is the Pythagorean theorem in
n
n dimensions! The length squared of w
w is
just the sum of the squares of the lengths of its two orthogonal components.
The projection of
w
w onto
x
x is
x
x, and the projection of
w
w onto
y
y is
y
y:
w
=
(
1
)
x
+
(
1
)
y
.
w
=
(
1
)
x
+
(
1
)
y
.
(15)
If we scale
w
w by
a
a to produce the vector
z=awz=aw, the orthogonal decomposition
of
z
z is
z
=
a
w
=
(
a
)
x
+
(
a
)
y
.
z
=
a
w
=
(
a
)
x
+
(
a
)
y
.
(16)
Let's turn this argument around. Instead of building ww from orthogonal vectors
x
x and
y
y, let's begin with arbitrary
w
w and x
x and see whether we can compute an orthogonal decomposition. The projection of
w
w onto
x
x is found from Equation 10:
wx=(w,x)(x,x)x.wx=(w,x)(x,x)x.(17)
But there must be another component of
w
w such that
w
w is equal to the sum
of the components. Let's call the unknown component
w
y
w
y
. Then
w
=
w
x
+
w
y
.
w
=
w
x
+
w
y
.
(18)
Now, since we know
w
w and
w
x
w
x
already, we find
w
y
w
y
to be
w
y
=
w
-
w
x
=
w
-
(
w
,
x
)
(
x
,
x
)
x
.
w
y
=
w
-
w
x
=
w
-
(
w
,
x
)
(
x
,
x
)
x
.
(19)
Interestingly, the way we have decomposed
w
w will always produce
w
x
w
x
and
w
>y
w
>y
orthogonal to each other. Let's check this:
(
w
x
,
w
y
)
=
(
(
w
,
x
)
(
x
,
x
)
x
,
w
-
(
w
,
x
)
(
x
,
x
)
x
)
=
(
w
,
x
)
(
x
,
x
)
(
x
,
w
)
-
(
w
,
x
)
2
(
x
,
x
)
2
(
x
,
x
)
=
(
w
,
x
)
2
(
x
,
x
)
-
(
w
,
x
)
2
(
x
,
x
)
=
0
.
(
w
x
,
w
y
)
=
(
(
w
,
x
)
(
x
,
x
)
x
,
w
-
(
w
,
x
)
(
x
,
x
)
x
)
=
(
w
,
x
)
(
x
,
x
)
(
x
,
w
)
-
(
w
,
x
)
2
(
x
,
x
)
2
(
x
,
x
)
=
(
w
,
x
)
2
(
x
,
x
)
-
(
w
,
x
)
2
(
x
,
x
)
=
0
.
(20)
To summarize, we have taken two arbitrary vectors,
w
w and
x
x, and decomposed
w
w into a component in the direction of
x
x and a component orthogonal to
x
x.
(What does "Endorsed by" mean?)
"Reviewer's Comments: 'I recommend this book as a "required primary textbook." This text attempts to lower the barriers for students that take courses such as Principles of Electrical Engineering, […]"