8.2 Orthogonal Matrices
The fact that the eigenvectors of a symmetric matrix A are orthogonal implies the
coordinate system defined by the eigenvectors is an orthogonal coordinate system. In
particular, it makes the diagonalization A = TDT-1 of the matrix A particularly nice.
Recall T is the matrix whose columns are the eigenvectors v1,…vn of A and D is the
diagonal matrix with the eigenvalues 1,…,n of A on the main diagonal and zero's
elsewhere. It is convenient to normalize the eigenvectors before using them as the
columns of T. Recall, normalizing a vector v means dividing it by its length | v | so the
v
resulting vector | v | has unit length. If we normalize an eigenvector the resulting vector is
still an eigenvector, since if we multiply an eigenvector by a constant it is still an
eigenvector. If we use the normalized eigenvectors as the columns of T then the columns
of T are orthogonal and have length one. Such a matrix is called an orthogonal matrix.
Definition 1. An orthogonal matrix is a square matrix S whose columns are orthogonal
and have length one.
Example 1. Find the diagonalization of A =  - 6
17 - 6 
8
using the normalized eigenvectors
as the columns of T. From Example 3 in the previous section, the eigenvalues of A are
1
-2
1 = 5 and 2 = 20 and the eigenvectors are v1 =  2  and v2 =  1 . One has
1/ 5 
- 2/ 5 
| v1 | = | v2 | = 5 . So the normalized eigenvectors are u1 = 
and u2 = 
. So
 2/ 5 
 1/ 5 
1 1 -2
T =  2 1  and the diagonalization of A is
5
A = TDT-1 =
Note that T =
1 1 -2
52 1 
1 1 -2
52 1 
-1
 5 0   1  1 - 2 
 0 20   5  2 1 
is an orthogonal matrix.
cos  - sin  
 sin  cos   be the matrix for a rotation by an angle . Then is
1 0
orthogonal R. Let F =  0 - 1  be the matrix for a reflection across the y axix. Then F is
cos  - sin    1 0   cos 
sin  
orthogonal. Also RF = 
 sin  cos    0 - 1  =  sin  - cos   is orthogonal.
Example 2. Let R = 
a b
a
Proposition 1. If S =  c d  is orthogonal then S = R or S = RF where  is the angle  c 
makes with the x axis.
8.2 - 1
a
Proof. S is orthogonal   c  is a unit vector  a2 + c2 = 1  r = 1 and  are the polar
a
a
b
coordinates of  c   a = cos  and c = sin . S is orthogonal   c  and  d  are orthogonal
b
-c
b
  d  = t  a  for some number t. S is orthogonal   d  is a unit vector  t = 1 or t = - 1.
If t = 1 then S = R. If t = - 1 then S = RF. //
An orthogonal matrix is nice because it is easy to compute its inverse since its inverse
turns out to be equal to their transpose.
Theorem 2. S is orthogonal if and only if S-1 = ST.
Proof. It suffices to show S is orthogonal  STS = I, i.e. (STS)ij = Iij. One has (STS)ij
equal to the product of row i of ST and column j of S. However row i of ST is the
transpose of column i of S. So (STS)ij = (S●,i)T(S●,j) = S●,i . S●,j. Note that S is orthogonal
 S●,i . S●,j = 0 = Iij if i  j and S●,i . S●,j = 1 = Iij if i = j. So S is orthogonal  = Iij. //
Example 3. Find the inverse of T =
1 1 -2
.
52 1 
1  1 2
. In particular the diagonalization
5-2 1
1 1 -25 0 1  1 2 
.
5  2 1   0 20  5  - 2 1 
T is orthogonal so T-1 = TT =
17 - 6
Example 1 is  - 6 8  =
of A in
An application of this would be to compute An. One has
1  1 - 2   5n 0  1  1 2 
n
5  2 1   0 20  5  - 2 1 
1
5n - 4(20n)
2(5n) - 2(20n) 
=  2(5n) - 2(20n)
4(5n) + 20n 
5
An =
= 5  2(5n)
1
5n
- 2(20n)   1 2 
20n   - 2 1 
6 -2
Problem 2. Let A =  - 2 3  be the matrix in Problem 1 in section 6.1. (a) Normalize the
eigenvectors you found in that problem. (b) What is the matrix T whose columns are the
normalized eigenvectors. (c) Find the inverse of T. (d) Find An.
0.98 0.12
Problem 2. Let A =  0.12 1.08  be the matrix in Problem 2 in section 6.1. (a) Normalize the
eigenvectors you found in that problem. (b) What is the matrix T whose columns are the
normalized eigenvectors. (c) Find the inverse of T. (d) Find An. (e) Find the solution to the
difference equations
sn+1 = 0.98sn + 0.12gn
gn+1 = 0.12sn + 1.08gn
with the initial conditions s0 = 1 and g0 = 2.
8.2 - 2
1  3
1 2
1  3 2
and u1 =
. (b) T =
. (c) T is orthogoanl
13  - 2 
13  3 
13  - 2 3 
1 3 -2
1  9(0.9)n + 4(1.16)n - 6(0.9)n + 6(1.16)n 
n
so T-1 = TT =
.
(d)
A
=
13  - 6(0.9)n + 6(1.16)n 4(0.9)n + 9(1.16)n 
13  2 3 
sn
1
1  - 3(0.9)n + 16(1.16)n 
(e)  g  = An 2  =
13  2(0.9)n + 24(1.16)n 
n
Answers: (a) u1 =
One consequence of proposition 1 is that the product of two orthogonal matrices is
orthogonal and the inverse of an orthogonal matrix is orthogonal.
Proposition 3. If S and T are orthogonal then so is ST and S-1.
Proof. S and T are orthogonal  S-1 = ST and T-1 = TT  (ST)-1 = T-1S-1 = ST = (ST)T 
ST is orthogonal. S is orthogonal  S-1 = ST  (S-1)-1 = (ST)-1 = (S-1)T  S-1 is
orthogonal. //
Proposition 1 has one characterization of orthogonal matrices. Another is the following
that says the orthogonal matrices are the ones that preserve inner products and lengths.
Proposition 4. (a) S is orthogonal if and only if (Sx) . (Sy) = x . y for all x and y. (b) S
is orthogonal if and only if | Sx | = | x | for all x.
Proof. (a)
8.2 - 3