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9.3 Some Special Types of Matrices 285

THEOREM 9.8

Let A be an n × n matrix of real numbers. Then
n
1. A is orthogonal if and only the row vectors are mutually orthogonal unit vectors in R .
2. A is orthogonal if and only if the column vectors are mutually orthogonal unit vectors
in R .
n
We say that the row vectors of an orthogonal matrix form an orthonormal set of vectors in
n
R . The column vectors also form an orthonormal set.
t
t
Proof The i, j element of AA is the dot product of row i of A with column j of A , and this
is the dot product of row i of A with row j of A.
If i = j, then this dot product is zero, because the i, j− element of I n is zero. And if i =
j, then this dot product is 1 because the i,i− element of I n is 1. This proves that, if A is an
n
orthogonal matrix, then its rows form an orthonormal set of vectors in R .
n
Conversely, suppose the rows are mutually orthogonal unit vectors in R . Then the i, j
t
t
element of AA is 0 if i = j and 1 if i = j,so AA = I n .
t
By applying this argument to A , this transpose is orthogonal if and only if its rows are
orthogonal unit vectors, and these rows are the columns of A.
We now know a lot about orthogonal matrices. We will use this information to determine all
2 × 2 real orthogonal matrices. Suppose

a b
Q =
c d
is orthogonal. What does this tell us about a,b,c and d? Because the row (column) vectors are
mutually orthogonal unit vectors,
ac + bd = 0
ab + cd = 0
2
2
a + b = 1
2
2
c + d = 1.
Furthermore, |Q|=±1, so
ad − bc = 1or ad − bc =−1.
By analyzing these equations in all cases, we find that there must be some θ in [0,2π) such that
a = cos(θ) and b = sin(θ), and Q must have one of the two forms:

cos(θ) sin(θ) cos(θ) sin(θ)
or ,
−sin(θ) cos(θ) sin(θ) −cos(θ)
depending on whether the determinant is 1 or −1. For example, with θ = π/6, we obtain the
orthogonal 2 × 2 matrices
√ √
3/2 1/2 3/2 1/2
√ or √ .
−1/2 3/2 1/2 − 3/2
If we put Theorems 9.4 and 9.8 together, we obtain an interesting conclusion. Suppose S is
a real, symmetric n × n matrix with n distinct eigenvalues. Then the associated eigenvectors are
orthogonal. These may not be unit vectors. However, a scalar multiple of an eigenvector is still
an eigenvector. Divide each eigenvector by its length and use these unit eigenvectors as columns
of an orthogonal matrix Q that diagonalizes S. This proves the following.

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October 14, 2010 14:49 THM/NEIL Page-285 27410_09_ch09_p267-294