MAT223H1b.doc

Page 1 of 51

Lecture #11 – Tuesday, February 10, 2004

ORTHOGONALITY

Definition

Let X, Y be vectors in Rn. The dot product is

R

∈

=

⋅

Y

X

Y

X

T.

So, if

=

n

x

x

x

X

2

1

,

=

n

y

y

y

Y

2

1

,

=

=+++=⋅

n

j

jjnn yxyxyxyxYX

1

2211 .

Definition

The length of a vector X, denoted XXX ⋅=. So if

=

n

x

x

x

X

2

1

, 0

22

2

2

1≥+++= n

xxxX.

Example

Let

−

=

4

2

0

1

X,

−

−

=

2

3

2

1

Y. Find

Y

X

⋅

and X.

•

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

1524322011

−

=

−

+

−

+

+

−

=

⋅

YX.

•

( ) ( ) ( ) ( )

214201 2222 =+−++=X.

Theorem

Let X, Y, Z be vectors in Rn. Then:

1)

X

Y

Y

X

⋅

=

⋅

. (Proof:

==

=

n

j

jj

n

j

jj xyyx

11

)

2)

(

)

ZXYXZYX

⋅

+

⋅

=

+

⋅

.

3) For a ∈ R,

(

)

(

)

(

)

YXaaYXYaX

⋅

=

⋅

=

⋅

.

4) 0≥X, 00 =⇔=xX.

Example

Let X be a vector in Rn, X ≠ 0. Find all Y ∈ Rn such that Y is collinear to X and is unitary.

• Let

R

∈

=

aaXY,. XaaXY== . We want 1=Y, so X

a

X

aXa11

1±===.

• So, X

X

Y1

±= .

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MAT223H1b.doc

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Definition

Let X, Y be two vectors in Rn. X and Y are orthogonal iff 0

=

⋅

YX.

Let

{

}

k

XXX ,,, 21 be a set of vectors in Rn.

{

}

k

XXX ,,, 21 is orthogonal iff kixi,,2,1,0

=

≠

and jiXX ji ≠= ,0.

Moreover, if kjXj,,2,1,1== , then {X1, X2,…, Xk} is orthonormal.

Example

The standard basis of Rn is a orthonormal set.

=

0

0

1

1

E,

=

0

1

0

2

E,

=

1

0

0

3

E.

Definition

If

{

}

k

XXX ,,, 21 is an orthogonal set, then

{

}

kk XaXaXa,,, 2211 is also an orthogonal set if

kjaj,,2,1,0

=

≠

.

For

j

jX

a1

=, the set

{

}

kk XaXaXa,,, 2211 will be orthonormal.

Example

Let

=

1

1

2

1

X,

−=

3

3

0

2

X,

−

−=

1

1

1

3

X. Construct an orthonormal set.

•

{

}

321 ,, XXX is an orthogonal set because:

• 0

≠

i

X.

• 0

21

=

XX , 0

31

=

XX , 0

32

=

XX .

• Find 3,2,1,

=

jaj such that

{

}

332211 ,, XaXaXa is orthonormal:

( ) ( ) ( )

6

1

112

11

222

1

1=

++

== X

a,

23

1

2=a,

3

1

3=a.

• So,

3

,

23

,

6

3

21 X

XX is orthonormal.

Theorem (Pythagorean)

If X and Y are orthogonal, then 222 YXYX+=+ .

Proof

X and Y are orthogonal; it means 0

=

⋅

YX. So,

( ) ( )

222 00 YXYYXXYYXYYXXXYXYXYX+=⋅+++⋅=⋅+⋅+⋅+⋅=+⋅+=+ .

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MAT223H1b.doc

Page 3 of 51

Theorem

Every orthogonal set of vectors of Rn is independent.

Proof

Let

{

}

k

XXX ,,, 21 be an orthogonal set. Consider 0

2211

=

+

+

+

=

kk XtXtXtY. We want to prove

that 0

=

j

t.

(

)

00

00

2

22112211

==

=

+

+

⋅

+

+

⋅

+

⋅

=

+

+

+

⋅

=

⋅

jjj

kkjjjjjkkjj

tXt

XtXXtXXtXXtXtXtXtXYX

.

Theorem

If

{

}

n

EEE ,,, 21 is an orthogonal basis of Rn, then n

X

R

∈

∀

,

n

n

nE

E

EX

E

E

EX

E

E

EX

X2

2

2

2

2

1

2

1

1

⋅

++

⋅

+

⋅

=.

Proof

Let X ∈ Rn. Then nn EtEtEtX

+

+

+

=

2211 because is a

{

}

n

EEE ,,, 21 basis of Rn.

( )

2

2

112211

j

j

jjjjnnjjjjjnnj

E

EX

tEtEEtEEtEEtEEtEtEtEX ⋅

==⋅++⋅++⋅=⋅++=⋅ .

Example

Let

=

0

1

2

1

X,

−=

3

3

0

2

X,

−=

1

1

1

3

X. Show that

{

}

321 ,, XXX is an orthogonal basis of R3.

• It is enough to prove 0

21

=

⋅

XX , 0

32

=

⋅

XX , 0

31

=

⋅

XX because:

•

{

}

321 ,, XXX is an orthogonal set, so X1, X2, X3 are independent.

• 3dim3=R. So a set of 3 linearly independent vectors is a basis.

Example

Write

=

c

b

a

X as a linear combination of

=

0

1

2

1

X,

−=

3

3

0

2

X,

−=

1

1

1

3

X.

• Since

{

}

321 ,, XXX is orthogonal,

3213

2

3

3

2

2

2

2

1

2

1

1

36

3

5

2X

cba

X

cb

X

ba

X

X

XX

X

X

XX

X

X

XX

X+−

+

+−

+

+

=

⋅

+

⋅

+

⋅

=.

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