Page 373 - A Course in Linear Algebra with Applications

9.4: Jordan Normal Form 357

qualifies as a Jordan string of length 1, so we can assume that
n > 1.
Since A is complex, it has an eigenvalue c. Thus the
matrix A' = A — cl is singular, and so its column space C
has dimension r < n. Recall from Example 6.3.2 that C
is the image of the linear operator on C" which sends X to
1
A X. Restriction of this linear operator to C produces a linear
operator which is represented by an r x r matrix. Since r < n,
we may assume by induction hypothesis on n that C has a
basis which is a union of Jordan strings for A. Let the ith
such string be written Xij, j = 1,... ,U\ thus AXn = CjXji
and in addition AXij = QXJJ + Xij-i for 1 < j < h- Then
A'Xn = 0 and A'Xij = Xy-_i if j > 1.
Next let D denote the intersection of C with N, the null
space of A', and set p = dim(D). We need to identify the
elements of D. Now any element of C has the form

Y = > y]aijXij
J
i 3

where a^ is a complex number. Assume that Y is in D, and
thus in N, the null space of A'. Suppose that a^- ^ 0 and let
j be as large as possible with this property for the given i. If
j > 1, then the equations A'Xn = 0 and A'Xik = Xik-i will
prevent A'Y from being zero. Hence j = 1. It follows that
the Xii form a basis of .D, so there are exactly p of these Xn.
Every vector in C is of the form A'Y for some Y, since C
is the image space of the linear operator sending X to A'X.
For each i write the vector Xu t in the form X ^ = A'Yi, for
some Yi , i = 1,..., p. There are p of these Yi. Finally, N has
dimension n — r, so we can adjoin a further set of n — r — p
vectors to the Xn to get a basis for N, say Z\,..., Z n _ r _ p .
Altogether we have a total of r + p + (n — r — p) = n
vectors