Page 176 - Linear Algebra Done Right

Generalized Eigenvectors
T(z 1 ,z 2 ,z 3 ) = (z 2 , 0,z 3 ),
2
then T (z 1 ,z 2 , 0) = 0 for all z 1 ,z 2 ∈ C. Hence every element of C
whose last coordinate equals 0 is a generalized eigenvector of T.As 3 165
you should verify,
3
C ={(z 1 ,z 2 , 0) : z 1 ,z 2 ∈ C}⊕{(0, 0,z 3 ) : z 3 ∈ C},
where the first subspace on the right equals the set of generalized eigen-
vectors for this operator corresponding to the eigenvalue 0 and the sec-
ond subspace on the right equals the set of generalized eigenvectors
corresponding to the eigenvalue 1. Later in this chapter we will prove
that a decomposition using generalized eigenvectors exists for every
operator on a complex vector space (see 8.23).
Though j is allowed to be an arbitrary integer in the definition of a Note that we do not
generalized eigenvector, we will soon see that every generalized eigen- define the concept of a
vector satisfies an equation of the form 8.3 with j equal to the dimen- generalized eigenvalue
sion of V. To prove this, we now turn to a study of null spaces of because this would not
powers of an operator. lead to anything new.
j
k
Suppose T ∈L(V) and k is a nonnegative integer. If T v = 0, then Reason: if (T − λI) is
k
k
T k+1 v = T(T v) = T(0) = 0. Thus null T ⊂ null T k+1 . In other words, not injective for some
we have positive integer j, then
T − λI is not injective,
0
1
k
8.4 {0}= null T ⊂ null T ⊂· · ·8 null T ⊂ null T k+1 ⊂· · · . and hence λ is an
eigenvalue of T.
The next proposition says that once two consecutive terms in this se-
quence of subspaces are equal, then all later terms in the sequence are
equal.
8.5 Proposition: If T ∈L(V) and m is a nonnegative integer such
that null T m = null T m+1 , then
0
1
null T ⊂ null T ⊂ ··· ⊂ null T m = null T m+1 = null T m+2 =· · · .
Proof: Suppose T ∈L(V) and m is a nonnegative integer such
that null T m = null T m+1 . Let k be a positive integer. We want to prove
that
null T m+k = null T m+k+1 .

We already know that null T m+k ⊂ null T m+k+1 . To prove the inclusion
in the other direction, suppose that v ∈ null T m+k+1 . Then