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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 ) = 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,
because those are the eigenvalues of M T M as we observed in Section 11.2.1 . Since MM T is
a 4 × 4 matrix, it has two other eigenvalues, which must both be 0. The matrix of eigenvec-
tors corresponding to 58, 2, 0, and 0 is shown in Fig. 11.4 .
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).
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