Page 279 - 35Linear Algebra

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2 1
Example 141 The matrix M = has eigenvalues determined by
1 2

2
det(M − λI) = (2 − λ) − 1 = 0.

So the eigenvalues of M are 3 and 1, and the associated eigenvectors turn out to be

1 1
and . It is easily seen that these eigenvectors are orthogonal;
1 −1

1 1
= 0.
1 −1

In chapter 14 we saw that the matrix P built from any orthonormal basis
n
(v 1 , . . . , v n ) for R as its columns,

P = v 1 · · · v n ,

was an orthogonal matrix. This means that

T
T
T
P −1 = P , or PP = I = P P.

Moreover, given any (unit) vector x 1 , one can always find vectors x 2 , . . . , x n
such that (x 1 , . . . , x n ) is an orthonormal basis. (Such a basis can be obtained
using the Gram-Schmidt procedure.)

Now suppose M is a symmetric n × n matrix and λ 1 is an eigenvalue
with eigenvector x 1 (this is always the case because every matrix has at
least one eigenvalue–see Review Problem 3). Let P be the square matrix of
orthonormal column vectors

P = x 1 x 2 · · · x n ,

While x 1 is an eigenvector for M, the others are not necessarily eigenvectors
for M. Then

MP = λ 1 x 1 Mx 2 · · · Mx n .

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