Page 374 - A Course in Linear Algebra with Applications

358 Chapter Nine: Advanced Topics

n
We now assert that these vectors form a basis of C which
consists of Jordan strings of A. Certainly

AY k = (A' + d)Y k = A'Y k + cY k = cY k + X klk.

Thus the Jordan string X ki,..., X ki k has been extended by
adjoining Y k. Also AZ m = cZ m since Z m belongs to the null
space of A'; thus Z m is a Jordan string of A with length 1.
Hence the vectors in question constitute a set of Jordan strings
of A
What remains to be done is to prove that the vectors
n
Xij,Y k, Z m form a basis of C , and by 5.1.9 it is enough to
show that they are linearly independent. To accomplish this,
f
we assume that e^, k,g m are scalars such that
^ ^ eijXij + ^ f kY k + ^ 9mZm = 0.

Multiplying both sides of this equation on the left by A', we
get
53 53 e (0 or X^) + J2 fk*ki k = 0.
y
Now X ki k does not appear among the terms of the first sum
in the above equation since j — 1 < l k. Hence f k = 0 for all
k. Thus
= e
/ J QrnZm ~ / v / _, ijX{j,
which therefore belongs to D. Hence e^ = 0 if j > 1, and
e
J2 9mZ m = — Yl aXn. This can only mean that g m = 0
and eji = 0 since the Xn and Z m are linearly independent.
Hence the theorem is established.

Corollary 9.4.6
Every complex n x n matrix is similar to an upper triangular
matrix with zeros above the superdiagonal.

This follows at once from the theorem since every Jordan
block is an upper triangular matrix of the specified type.