Page 186 - Linear Algebra Done Right

Decomposition of an Operator
dim U = dim U 1 +· · ·+ dim U m .
This equation, along with 8.24, shows that dim V = dim U. Because U
is a subspace of V, this implies that V = U. In other words, 175
V = U 1 +· · ·+ U m .
This equation, along with 8.24, allows us to use 2.19 to conclude that
(a) holds, completing the proof.

As we know, an operator on a complex vector space may not have
enough eigenvectors to form a basis for the domain. The next result
shows that on a complex vector space there are enough generalized
eigenvectors to do this.

8.25 Corollary: Suppose V is a complex vector space and T ∈L(V).
Then there is a basis of V consisting of generalized eigenvectors of T.

Proof: Choose a basis for each U j in 8.23. Put all these bases
together to form a basis of V consisting of generalized eigenvectors
of T.

Given an operator T on V, we want to find a basis of V so that the
matrix of T with respect to this basis is as simple as possible, meaning
that the matrix contains many 0’s. We begin by showing that if N is
nilpotent, we can choose a basis of V such that the matrix of N with
respect to this basis has more than half of its entries equal to 0.

8.26 Lemma: Suppose N is a nilpotent operator on V. Then there is If V is complex vector
a basis of V with respect to which the matrix of N has the form space, a proof of this
lemma follows easily
 
0 ∗ from Exercise 6 in this
 . 
8.27  . .  ; chapter, 5.13, and 5.18.
 
0 0 But the proof given
here uses simpler ideas
here all entries on and below the diagonal are 0’s. than needed to prove
5.13, and it works for
Proof: First choose a basis of null N. Then extend this to a basis both real and complex
3
2
of null N . Then extend to a basis of null N . Continue in this fashion, vector spaces.
eventually getting a basis of V (because null N m = V for m sufficiently
large).