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M
T
M
e
= λ
e
Multiply both sides of this equation by
M
on the left. Then
MM
T
(
M
e
) =
Mλ
e
= λ(
M
e
)
Thus, as long as
M
e
is not the zero vector
0
, it will be an eigenvector of
MM
T
and λ will be
an eigenvalue of
MM
T
as well as of
M
T
M
.
The converse holds as well. That is, if
e
is an eigenvector of
MM
T
with corresponding
eigenvalue λ, then start with
MM
T
e
= λ
e
and multiply on the left by
M
T
to conclude that
M
T
M
(
M
T
e
) = λ(
M
T
e
). Thus, if
M
T
e
is not
0
, then λ is also an eigenvalue of
M
T
M
.
We might wonder what happens when
M
T
e
=
0
. In that case,
MM
T
e
is also
0
, but
e
is not
0
because
0
cannot be an eigenvector. However, since
0
= λ
e
, we conclude that λ = 0.
We conclude that the eigenvalues of
MM
T
are the eigenvalues of
M
T
M
plus additional
0's. If the dimension of
MM
T
were less than the dimension of
M
T
M
, then the opposite would
be true; the eigenvalues of
M
T
M
would be those of
MM
T
plus additional 0's.
EXAMPLE
11.7 The eigenvalues of
MM
T
for our running example must include 58 and 2,
a 4 × 4 matrix, it has two other eigenvalues, which must both be 0. The matrix of eigenvec-
□
Figure 11.4
Eigenvector matrix for
MM
T
11.2.4
Exercises for Section 11.2
EXERCISE
11.2.1 Let
M
be the matrix of data points
(a) What are
M
T
M
and
MM
T
?
(b) Compute the eigenpairs for
M
T
M
.
!
(c) What do you expect to be the eigenvalues of
MM
T
?
!
(d) Find the eigenvectors of
MM
T
using your eigenvalues from part (c).