432 CHAPTER 18. LINEAR TRANSFORMATIONS

Corollary 18.3.8 In the above theorem, each of the scalars λ k has the property that thereexists a nonzero v such that (A−λ iI)v = 0. Furthermore the λ i are the only scalars withthis property.

Proof: For the first claim, just factor out (A−λ iI) instead of (A−λ mI) . Next suppose

(A−µI)v = 0

for some µ and v ̸= 0. Then

0 =m

∏k=1

(A−λ kI)v =m−1

∏k=1

(A−λ kI)

 =µv︷︸︸︷Av −λ mv

= (µ−λ m)

(m−1

∏k=1

(A−λ kI)

)v

= (µ−λ m)

(m−2

∏k=1

(A−λ kI)

)(Av−λ m−1v)

= (µ−λ m)(µ−λ m−1)

(m−2

∏k=1

(A−λ kI)

)v

continuing this way yields = ∏mk=1 (µ−λ k)v, a contradiction unless µ = λ k for some k. ■

Therefore, these are eigenvectors and eigenvalues with the usual meaning. This leadsto the following definition.

Definition 18.3.9 For A ∈ L (V,V ) where dim(V ) = n, the scalars, λ k in the minimalpolynomial,

p(λ ) =m

∏k=1

(λ −λ k)≡p

∏k=1

(λ −λ k)rk

are called the eigenvalues of A. In the last expression, λ k is a repeated root which occurs rktimes. The collection of eigenvalues of A is denoted by σ (A). The generalized eigenspacesare

ker(A−λ kI)rk ≡Vk.

Theorem 18.3.10 In the situation of the above definition,

V =V1⊕·· ·⊕Vp

That is, the vector space equals the direct sum of its generalized eigenspaces.

Proof: Since V = ker(∏

pk=1 (A−λ kI)rk

), the conclusion follows from Theorem 18.3.5.

■