### Chapter 5 Orthogonality

```5.1
Orthogonality
Definitions
 A set of vectors is called an orthogonal set if all pairs of
distinct vectors in the set are orthogonal.
 An orthonormal set is an orthogonal set of unit vectors.
 An orthogonal (orthonormal) basis for a subspace W of
n
R is a basis for W that is an orthogonal (orthonormal) set.
 An orthogonal matrix is a square matrix whose columns
form an orthonormal set.
Examples
1) Is the following set of vectors orthogonal? orthonormal?


a)  1

 2

3   2  1  
    
, 4 , 1 
    
 1   2  

b) { e1 , e 2 ,..., e n }
2) Find an orthogonal basis and an orthonormal basis
n
for the subspace W of R
x 

 

W   y : x  y  2 z  0
 
z 

 

Theorems
All vectors in an orthogonal set are linearly independent.
Let {v1, v2,…, vk } be an orthogonal basis for a subspace
n
W of R and w be any vector in W. Then the unique
scalars c1 ,c2 , …, ck such that w = c1v1 + c2v2 + …+ ckvk
are given by
w  vi
ci 
for i  1,..., k
vi  vi
Proof: To find ci we take the dot product with vi
w vi = (c1v1 + c2v2 + …+ ckvk ) vi
Examples
3) The orthogonal basis for the subspace W in previous
example is  1    1 
  
 1  , 1
  0  1
  


 

Pick a vector in W and express it in terms of the vectors
in the basis.
4) Is the following matrix orthogonal?
 3

A 
1

 2
2
4
1
1 

1

2 
0

B  1

 0
0
0
1
1

0

0 
 cos 
C  
 sin 
 sin  

cos  
If it is orthogonal, find its inverse and its transpose.
Theorems on Orthogonal Matrix
The following statements are equivalent for a matrix A :
1) A is orthogonal
-1
T
2) A = A
n
3) ||Av|| = ||v|| for every v in R
n
4) Av1∙ Av2 = v1∙ v2 for every v1 ,v2 in R
Let A be an orthogonal matrix. Then
1) its rows form an orthonormal set.
-1
2) A is also orthogonal.
3) |det(A)| = 1
4) |λ| = 1 where λ is an eigenvalue of A
5) If A and B are orthogonal matrices, then so is AB
5.2
Orthogonal Complements
and Orthogonal Projections
Orthogonal Complements
 Recall: A normal vector n to a plane is orthogonal to
every vector in that plane. If the plane passes through
the origin, then it is a subspace W of R3 .
 Also, span(n) is also a subspace of R3
 Note that every vector in span(n) is orthogonal to
every vector in subspace W . Then span(n) is called
orthogonal complement of W.
Definition:
 A vector v is said to be orthogonal to a subspace W of
n
R if it is orthogonal to all vectors in W.
 The set of all vectors that are orthogonal to W is called
the orthogonal complement of W, denoted W ┴ . That is
W perp
W

 {v  R
n
: v  w  0  w  W}
http://www.math.tamu.edu/~yvorobet/MATH304-2011C/Lect3-02web.pdf
Example
3
1) Find the orthogonal complements for W of R .
 1  
 
a) W  span  2 
 
  3 
 
b) W  plane with direction
c) (subspace
x 

 

W   y : x  y  2 z  0
 
z 

 

spanned
by) vectors
1 
0
 
 
1 and 1
 
 
 0 
 1 
Theorems
n
Let W be a subspace of R .
n
┴
1) W is a subspace of R .
2) (W ┴)┴ = W
3) W ∩ W ┴ = {0}
4) If W = span(w1,w2,…,wk), then v is in W ┴ iff v∙wi = 0
for all i =1,…,k.
Let A be an m x n matrix. Then
(row(A))┴ = null(A)
and
(col(A))┴ = null(AT)
Proof?
Example
2) Use previous theorem to find the orthogonal complements
3
for W of R .
1 
0
 
 
a) W  plane w ith direction (subspace spann ed by) vectors 1 and 1
 
 
 0 
 1 
b)
 3 
 
2
 
W  subspace spanned by vectors  0  ,
 
  1
 4 
 1 
 
2
 
  2  an d
 
 0 
 1 
 3 
 
2
 
 6 
 
2 
 5 
Orthogonal Projections
 u v
w 1  proj v u  
 v 2

u
w2
w1

 u v
v  

v v



v

w 2 = perp v u  u - w 1
v
Let u and v be nonzero vectors.

w1 is called the vector component of u along v
(or projection of u onto v), and is denoted by projvu

w2 is called the vector component of u orthogonal to v
Orthogonal Projections
n
 Let W be a subspace of R with an orthogonal basis
{u1, u2,…, uk }, the orthogonal projection of v onto W is
defined as:
projW v = proju v + proju v + … + proju v
1
2
k
 The component of v orthogonal to W is the vector
perpW v = v – projw v
n
n
Let W be a subspace of R and v be any vector in R .
Then there are unique vectors w1 in W and w2 in W ┴
such that v = w1 + w2 .
Examples
3) Find the orthogonal projection of v = [ 1, -1, 2 ] onto W and
the component of v orthogonal to W.
 1  
 
a) W  span  2 
 
3
 
1 
 -1 
 
 
b) W  subspace spanned by 1 and
1
 
 
 0 
 1 
x 

 

c) W   y : x  y  2 z  0 
 
z 

 

5.3
The Gram-Schmidt Process
And the QR Factorization
The Gram-Schmidt Process
Goal: To construct an orthogonal (orthonormal) basis for
n
any subspace of R .
each vector vi in the basis one at a time by finding the
component of vi orthogonal to W = span(x1, x2,…, xi-1 ).
 Let {x1, x2,…, xk } be a basis for a subspace W. Then
choose the following vectors:
v1 = x1,
v2 = x2 – projv x2
1
v3 = x3 – projv x3 – projv x3
1
2
… and so on
 Then {v1, v2,…, vk } is orthogonal basis for W .
 We can normalize each vector in the basis to form an
orthonormal basis.
Examples
1) Use the following basis to find an orthonormal basis for R
  3  1  
  ,  ,
 1   2  
3
2) Find an orthogonal basis for R that contains the vector
  1
  1
1 
 
 
 
2 ,
1
, 0
 
 
 
  1
 1 
 1 
2
The QR Factorization
If A is an m x n matrix with linearly independent columns,
then A can be factored as A = QR where R is an invertible
upper triangular matrix and Q is an m x n orthogonal
n
matrix. In fact columns of Q form orthonormal basis for R
which can be constructed from columns of A by using
Gram-Schmidt process.
Note: Since Q is orthogonal, Q-1 = QT and we have R = QT A
Examples
3) Find a QR factorization for the following matrices.
3
A  
1
1

2
1

A  2

 1
-1
1
-1
- 1

0

1 
5.4
Orthogonal Diagonalization
of Symmetric Matrices
Example
1) Diagonalize the matrix.
3
A  
2
2

6
Recall:
 A square matrix A is symmetric if AT = A.
 A square matrix A is diagonalizable if there exists a
matrix P and a diagonal matrix D such that P-1AP = D.
Orthogonal Diagonalization
Definition:
A square matrix A is orthogonally diagonalizable if there
exists an orthogonal matrix Q and a diagonal matrix D
such that Q-1AQ = D.
 Note that Q-1 = QT
Theorems
1. If A is orthogonally diagonalizable, then A is symmetric.
2. If A is a real symmetric matrix, then the eigenvalues of A
are real.
3. If A is a symmetric matrix, then any two eigenvectors
corresponding to distinct eigenvalues of A are orthogonal.
A square matrix A is orthogonally diagonalizable
if and only if it is symmetric.
Example
2) Orthogonally diagonalize the matrix
0

A  1

 1
1
0
1
1

1

0 
and write A in terms of matrices Q and D.
Theorem
If A is orthogonally diagonalizable, and QTAQ = D
then A can written as
A   1 q1 q1   2 q 2 q 2  ...   n q n q n
T
T
T
where qi is the orthonormal column of Q, and λi is
the corresponding eigenvalue.
This fact will help us construct the matrix A given
eigenvalues and orthogonal eigenvectors.
Example
3) Find a 2 x 2 matrix that has eigenvalues 2 and 7, with
corresponding eigenvectors

v1  
1
2


v
2
1 
  
2
5.5
Applications
A quadratic form in x and y :
 a
T
2
2
ax  by  cxy  x  1
2 c
c
x
b 
1
2
A quadratic form in x,y and z:
 a
2
2
2

ax  by  cz  dxy  exz  fyz  x T 12 d

 12 e
where x is the variable (column) matrix.
1
2
d
b
1
2
f
e

1
f
x
2

c 
1
2
A quadratic form in n variables is a function
n
f : R  R of the form:
f (x)  x Ax
T
where A is a symmetric n x n matrix and x is in R
A is called the matrix associated with f.
z  f ( x , y )  x  y  8 xy
2
2
z  f ( x, y )  2 x  5 y
2
2
n
The Principal Axes Theorem
Every quadratic form can be diagonalized. In fact,
if A is a symmetric n x n matrix and if Q is an
orthogonal matrix so that QTAQ = D then the change
of variable x = Qy transforms the quadratic form into
x A x  y D y  1 y1   2 y 2  ...   n y n
T
T
2
2
2
Example: Find a change of variable that transforms the
Quadratic into one with no cross-product terms.
z  f ( x , y )  x  y  8 xy
2
2
z  f ( x, y )  2 x  5 y
2
2
```