Page 213 - Advanced Linear Algebra

Eigenvalues and Eigenvectors 197

#~ #
for all ~ Á Ã Á .

The previous definition leads immediately to the following simple
characterization of diagonalizable operators.

B
Theorem 8.10 Let ²= ³ . The following are equivalent:

)
1 is diagonalizable.
)
2 = has a basis consisting entirely of eigenvectors of .

)
3 = has the form
=~ ; l Ä l ;

where ÁÃÁ are the distinct eigenvalues of .
Diagonalizable operators can also be characterized in a simple way via their
minimal polynomials.
B

Theorem 8.11 A linear operator ²= ³ on a finite-dimensional vector space
is diagonalizable if and only if its minimal polynomial is the product of distinct
linear factors.
Proof. If is diagonalizable, then

=~ ; l Ä l ;
and Theorem 7.7 implies that ²%³ is the least common multiple of the

minimal polynomials %c of restricted to . Hence, ²%³ is a product of
;

distinct linear factors. Conversely, if ²%³ is a product of distinct linear

factors, then the primary decomposition of has the form
=
=~ = l Ä l =
where
= ~ ¸#  = “ ² c ³# ~ ¹ ~ ;

and so is diagonalizable.
Spectral Resolutions
We have seen (Theorem 2.25) that resolutions of the identity on a vector space
= = correspond to direct sum decompositions of . We can do something similar
for any diagonalizable linear operator on (not just the identity operator).
=

Suppose that has the form

~ b Ä
b
where bÄb ~ is a resolution of the identity and the  - are
distinct. This is referred to as a spectral resolution of .