Page 477 - Advanced Linear Algebra

An Introduction to Algebras 461

²%³ ~ % ²%³ for some nonzero polynomial ²%³ and so ~ ² ³, which

implies that ² ³ ~ , a contradiction to the minimality of ²%³ .

We have seen that the eigenvalues of a linear operator on a finite-dimensional

vector space are the roots of the minimal polynomial of , or equivalently, the

scalars for which c is not invertible. By analogy, we can define the
eigenvalues of an element of an algebra .

(
Theorem 18.7 Let be an algebra and let  ( be algebraic. An element
(
 - is a root of the minimal polynomial ²%³ if and only if c is not

invertible in .
(
Proof. If c is not invertible, then
c ²%³ ~ % ²%³
and since ²% b ³ is satisfied by c , it follows that

% ²%³ ~ c ²%³ “ ²% b ³

Hence, ²% c ³ “ ²%³ . Alternatively, if c is not invertible, then there is

a nonzero  ( such that ² c ³ ~ , that is, ~ . Hence, for any
polynomial ²%³ we have ² ³ ~ ² ³ . Setting ²%³ ~ ²%³ gives

² ³ ~ .

Conversely, if ² ³ ~ , then ²%³ ~ ²% c ³ ²%³ and so

~² c ³ ² ³, which shows that c is a zero divisor and therefore not
invertible.
-
Definition Let A be an -algebra and let  ( be algebraic. The roots of the
minimal polynomial of are called the eigenvalues of . The set of all

eigenvalues of
Spec² ³ ~¸  - “ ² ³ ~ ¹

is called the spectrum of .

Note that  ( is invertible if and only if ¤ Spec ² ³ .

(
Theorem 18.8 The spectral mapping theorem ) Let be an algebra over an
(
algebraically closed field . Let ( and let ²%³ -´%µ . Then
-
Spec² ² ³³ ~ ² Spec² ³³ ~ ¸ ² ³ “  Spec² ³¹
Proof. We leave it as an exercise to show that ²Spec ² ³³ ‹ Spec ² ² ³³ . For
the reverse inclusion, let  Spec ² ² ³³ and suppose that

²%³ c ~ ²%c ³ IJ%c ³

Then