Page 203 - Advanced Linear Algebra

Eigenvalues and Eigenvectors 187

these vectors that equal , the equation

# bÄb # ~ (8.2 )

has the fewest number of terms. Applying gives

# ~ (8.3 )
# bÄb
Multiplying (8.2) by and subtracting from (8.3) gives

² c ³ # b Ä b ² c ³ # ~
But this equation has fewer terms than (8.2) and so all of its coefficients must

equal . Since the 's are distinct, ~ for ‚ and so ~ as well. This

contradiction implies that the 's are linearly independent.
#

The next theorem describes the spectrum of a polynomial ² ³ in .
(

Theorem 8.3 The spectral mapping theorem ) Let be a vector space over
=
-
an algebraically closed field . Let  ²= ³ and let ²%³  -´%µ . Then
B

Spec² ²³³ ~ ² Spec²³³ ~ ¸ ² ³ “  Spec²³¹
Proof. We leave it as an exercise to show that if is an eigenvalue of , then

² ³ is an eigenvalue of ²³. Hence, ²Spec ²³³ ‹ Spec ² ²³³. For the reverse

inclusion, let  ² Spec ² ³ ³ , that is,
² ² ³ c ³# ~

for #£ . If

²%³ c ~ ²% c ³ IJ% c ³

where  - , then writing this as a product of (not necessarily distinct) linear

factors, we have

²c ³Ä²c ³Ä²c ³Ä²c ³# ~

(The operator is written for convenience.) We can remove factors from

the left end of this equation one by one until we arrive at an operator (perhaps

the identity) for which but #£ ² c ³ . Then # is an eigenvector
#~

for with eigenvalue . But since ² ³ c ~ , it follows that
² ² ³³ ‹ ²Spec
~ ² ³  ²Spec Spec ² ³³. Hence, ² ³³.
The Trace and the Determinant
Let - be algebraically closed and let (  C ² - ³ have characteristic
polynomial
c
(

²%³ ~ % b c % b Äb %b
~ ²% c ³Ä²% c ³