Page 177 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
166
k
m+k+1
0 = T
v = T
Hence
m
k
T v ∈ null T m+1 m+1 (T v).
= null T .
m
k
Thus 0 = T (T v) = T m+k v, which means that v ∈ null T m+k . This
implies that null T m+k+1 ⊂ null T m+k , completing the proof.
The proposition above raises the question of whether there must ex-
ist a nonnegative integer m such that null T m = null T m+1 . The propo-
sition below shows that this equality holds at least when m equals the
dimension of the vector space on which T operates.
8.6 Proposition: If T ∈L(V), then
null T dim V = null T dim V+1 = null T dim V+2 =· · · .
Proof: Suppose T ∈L(V). To get our desired conclusion, we need
only prove that null T dim V = null T dim V+1 (by 8.5). Suppose this is not
true. Then, by 8.5, we have
1
0
{0}= null T ⊊ null T ⊊ ··· ⊊ null T dim V ⊊ null T dim V+1 ,
where the symbol ⊊ means “contained in but not equal to”. At each of
the strict inclusions in the chain above, the dimension must increase by
at least 1. Thus dim null T dim V+1 ≥ dim V + 1, a contradiction because
a subspace of V cannot have a larger dimension than dim V.

Now we have the promised description of generalized eigenvectors.

This corollary implies 8.7 Corollary: Suppose T ∈L(V) and λ is an eigenvalue of T. Then
that the set of the set of generalized eigenvectors of T corresponding to λ equals
generalized null(T − λI) dim V .
eigenvectors of
T ∈L(V) Proof: If v ∈ null(T − λI) dim V , then clearly v is a generalized
corresponding to an eigenvector of T corresponding to λ (by the definition of generalized
eigenvalue λ is a eigenvector).
subspace of V. Conversely, suppose that v ∈ V is a generalized eigenvector of T
corresponding to λ. Thus there is a positive integer j such that

j
v ∈ null(T − λI) .
From 8.5 and 8.6 (with T −λI replacing T), we get v ∈ null(T −λI) dim V ,
as desired.