Page 187 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
176
Now let’s think about the matrix of N with respect to this basis. The
first column, and perhaps additional columns at the beginning, consists
of all 0’s because the corresponding basis vectors are in null N. The
2
next set of columns comes from basis vectors in null N . Applying N
to any such vector, we get a vector in null N; in other words, we get a
vector that is a linear combination of the previous basis vectors. Thus
all nonzero entries in these columns must lie above the diagonal. The
3
next set of columns come from basis vectors in null N . Applying N
2
to any such vector, we get a vector in null N ; in other words, we get a
vector that is a linear combination of the previous basis vectors. Thus,
once again, all nonzero entries in these columns must lie above the
diagonal. Continue in this fashion to complete the proof.
Note that in the next theorem we get many more zeros in the matrix
of T than are needed to make it upper triangular.

8.28 Theorem: Suppose V is a complex vector space and T ∈L(V).
Let λ 1 ,...,λ m be the distinct eigenvalues of T. Then there is a basis
of V with respect to which T has a block diagonal matrix of the form

 
A 1 0
 . 
 . .  ,
 
0 A m

where each A j is an upper-triangular matrix of the form

 
λ j ∗
 . 
8.29 A j =  . .  .
 
0 λ j

Proof: For j = 1,...,m, let U j denote the subspace of generalized
is nilpotent
eigenvectors of T corresponding to λ j . Thus (T − λ j I)| U j
(see 8.23(c)). For each j, choose a basis of U j such that the matrix of
with respect to this basis is as in 8.26. Thus the matrix of
(T − λ j I)| U j
with respect to this basis will look like 8.29. Putting the bases for
T| U j
the U j ’s together gives a basis for V (by 8.23(a)). The matrix of T with
respect to this basis has the desired form.