Page 185 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
174
If T ∈L(V) and p ∈P(F), then null p(T) is
Proposition:
8.22
invariant under T.
Proof: Suppose T ∈L(V) and p ∈P(F). Let v ∈ null p(T). Then
p(T)v = 0. Thus
(p(T))(Tv) = T(p(T)v) = T(0) = 0,
and hence Tv ∈ null p(T). Thus null p(T) is invariant under T,as
desired.
The following major structure theorem shows that every operator on
a complex vector space can be thought of as composed of pieces, each
of which is a nilpotent operator plus a scalar multiple of the identity.
Actually we have already done all the hard work, so at this point the
proof is easy.

8.23 Theorem: Suppose V is a complex vector space and T ∈L(V).
Let λ 1 ,...,λ m be the distinct eigenvalues of T, and let U 1 ,...,U m be
the corresponding subspaces of generalized eigenvectors. Then

(a) V = U 1 ⊕· · · U m ;
(b) each U j is invariant under T;
(c) each (T − λ j I)| U j is nilpotent.

Proof: Note that U j = null(T − λ j I) dim V for each j (by 8.7). From
8.22 (with p(z) = (z − λ j ) dim V ), we get (b). Obviously (c) follows from
the definitions.
To prove (a), recall that the multiplicity of λ j as an eigenvalue of T
is defined to be dim U j . The sum of these multiplicities equals dim V
(see 8.18); thus

8.24 dim V = dim U 1 +· · ·+ dim U m .

Let U = U 1 +· · ·+ U m . Clearly U is invariant under T. Thus we can
define S ∈L(U) by
S = T| U .

Note that S has the same eigenvalues, with the same multiplicities, as T
because all the generalized eigenvectors of T are in U, the domain of S.
Thus applying 8.18 to S, we get