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Chapter 5. Eigenvalues and Eigenvectors
76
Invariant Subspaces
In this chapter we develop the tools that will help us understand the
structure of operators. Recall that an operator is a linear map from a
vector space to itself. Recall also that we denote the set of operators
on V by L(V); in other words, L(V) =L(V, V).
Let’s see how we might better understand what an operator looks
like. Suppose T ∈L(V). If we have a direct sum decomposition
5.1 V = U 1 ⊕· · · U m ,

where each U j is a proper subspace of V, then to understand the be-
; here
havior of T, we need only understand the behavior of each T| U j
denotes the restriction of T to the smaller domain U j . Dealing
T| U j
should be easier than dealing with T because U j is a smaller
with T| U j
vector space than V. However, if we intend to apply tools useful in the
study of operators (such as taking powers), then we have a problem:
may not be an
T| U j may not map U j into itself; in other words, T| U j
operator on U j . Thus we are led to consider only decompositions of
the form 5.1 where T maps each U j into itself.
The notion of a subspace that gets mapped into itself is sufficiently
important to deserve a name. Thus, for T ∈L(V) and U a subspace
of V, we say that U is invariant under T if u ∈ U implies Tu ∈ U.
In other words, U is invariant under T if T| U is an operator on U. For
example, if T is the operator of differentiation on P 7 (R), then P 4 (R)
(which is a subspace of P 7 (R)) is invariant under T because the deriva-
tive of any polynomial of degree at most 4 is also a polynomial with
degree at most 4.
The most famous Let’s look at some easy examples of invariant subspaces. Suppose
unsolved problem in T ∈L(V). Clearly {0} is invariant under T. Also, the whole space V is
functional analysis is obviously invariant under T. Must T have any invariant subspaces other
called the invariant than {0} and V? Later we will see that this question has an affirmative
subspace problem. It answer for operators on complex vector spaces with dimension greater
deals with invariant than 1 and also for operators on real vector spaces with dimension
subspaces of operators greater than 2.
on infinite-dimensional If T ∈L(V), then null T is invariant under T (proof: if u ∈ null T,
vector spaces. then Tu = 0, and hence Tu ∈ null T). Also, range T is invariant under T
(proof: if u ∈ range T, then Tu is also in range T, by the definition of
range). Although null T and range T are invariant under T, they do not
necessarily provide easy answers to the question about the existence