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Up to this point we projected the vector v along a single unit vector n. Suppose we

^

have a set of mutually orthogonal unit vectors ni . The fact that these unit vectors are

^

mutually orthogonal means that di erent unit vectors are perpendicular to each other:

(^i nj ) = 0 when i = j. We can project v on each of these unit vectors and add these

n^ 6

projections. This gives us the projection of v on the subspace spanned by the unit vectors

ni:

^ X

Pv = ni (^i v) :

^n (10.11)

i

When the unit vectors ni span the full space we work in, the projected vector is identical

^

to the original vector. To see this, consider for example a three-dimensional space. Any

CHAPTER 10. LINEAR ALGEBRA

106

vector can be decomposed in the components along the x, y and z-axis, this can be written

as:

v =vxx+vy y+vz^ = x (^ v) +^ (^ v) + ^ (^ v)

^ ^ z ^x yy zz (10.12)

note that this expression has the same form as (10.11). This implies that when we sum

in (10.11) over a set of unit vectors that completely spans the space we work in, the right

P^ n

hand side of (10.11) is identical to the original vector v, i.e. i ni (^i v) = v. The

operator of the left hand side of this equality is therefore identical to the identity operator

I:

X

N

ninT = I :

^ ^i (10.13)

i=1

Keep in mind that N is the dimension of the space we work in, if we sum over a smaller

number of unit vectors we project on a subspace of the N-dimensional space. Expression

(10.13) expresses that the vectors ni (with i = 1

^ N) can be used to give a complete

representation of any vector. Such a set of vectors is called a complete set, and expression

(10.13) is called the closure relation.

Problem h: Verify explicitly that when the unit vectors ni are chosen to be the unit

^

vectors x, y and ^ along the x, y and z-axis that the right hand side of (10.13) is

^^ z

given by the 3 3 identity matrix.

There are of course many di erent ways of choosing a set of three orthogonal unit vectors

in three dimensions. Expression (10.13) should hold for every choice of a complete set of

unit vectors.

Problem i: Verify explicitly that when the unit vectors ni are chosen to be the unit

^

vectors ^, ^ and ' de ned in equations (3.6) for a system of spherical coordinates

r ^

that the right hand side of (10.13) is given by the 3 3 identity matrix.

10.2 A projection on vectors that are not orthogonal

In the previous section we considered the projection on a set of orthogonal unit vectors.

In this section we consider an example of a projection on a set of vectors that is not

necessarily orthogonal. Consider two vectors a and b in a three-dimensional space. These

two vectors span a two-dimensional plane. In this section we determine the projection of

a vector v on the plane spanned by the vectors a and b, see gure (10.2) for the geometry

of the problem. The projection of v on the plane will be denoted by vP .

By de nition the projected vector vP lies in the plane spanned by a and b, this vector

can therefore be written as:

vP = a+ b : (10.14)

The task of nding the projection can therefore be reduced to nding the two coe cients

and . These constants follow from the requirement that the vector joining v with its

projection vP = Pv is perpendicular to both a and b, see gure (10.2).

10.2. A PROJECTION ON VECTORS THAT ARE NOT ORTHOGONAL 107

.

v

. Pv

b

. a

Figure 10.2: De nition of the geometric variables for the projection on a plane.

Problem a: Show that this requirement is equivalent with the following system of equa-

tions for and :

(a a) + (a b) = (a v) (10.15)

(a b) + (b b) = (b v)

Problem b: Show that the solution of this system is given by

;b2a; (a b) b v

= 22

a b (a b)2

;

;a2 b; (a b) a v

= 22 (10.16)

a b (a b)2

;

where a denotes the length of the vector a: a and a similar notation is used

jaj,

for the vector b.

Problem c: Show using (10.14) and (10.16) that the projection operator for the projec-

tion on the plane (Pv = vP ) is given by

P = 2 2 1 2 b2 aaT + a2bbT (a b) abT + baT : (10.17)

a b (a b)

;

;

This example shows that projection on a set of non-orthogonal basis vectors is much more

complex than projecting on a set of orthonormal basis vectors. A di erent way of nding

the projection operator of expression (10.17) is by rst nding two orthogonal unit vectors

in the plane spanned by a and b and then using expression (10.11). One unit vector can

be found by dividing a by its length to give the unit vector ^ = a=

a The second unit

jaj.

vector can be found by considering the component b? of b perpendicular to ^ and by

a

^

normalizing the resulting vector to form the unit vector b? that is perpendicular to ^,a

see gure (10.3).

CHAPTER 10. LINEAR ALGEBRA

108

^

Problem d: Use expression (10.3) to nd b? and show that the projection operator P

of expression (10.17) can also be written as

aa ^ ^?

P = ^^T + b?bT : (10.18)

Note that this expression is consistent with (10.11).

.

v

. Pv

^

n

.

b

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