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Unit 21: The Jordan Form
Ki
The companion matrix of x is the K × K matrix Notes
i i
0 0 0 0
1 0 0 0
A 0 1 0 0
i
0 0 0 1 0
...(1)
Thus Theorem 1 of unit 20 gives us an ordered basis for V in which the matrix of N is the direct
sum of the elementary nilpotent matrices (1). Thus with a nilpotent n × n matrix we associate an
integer r such that k + k + ... + k = n and k k and which determines the rational form of matrix.
1 2 r i i+1
The positive integer is precisely the nullity of N, as the null space has a basis the r vectors
k –1
N i ...(2)
i
For, let be in the null space of N, we write as
= f + ... + f
1 1 r r
where f is a polynomial, the degree of f is assumed to be less than k . Since N = 0 for each i we
i i i
have
0 = N(f )
i i
= Nf (N),
i i
= (xf )
i i
k
Thus x f is divisible by x and since deg (f ) < k , this means that
i i i
f = c x i k –1
i i
where c is some scalar. But then
i
k
= c (x 1 ) + ... + c (x r –1 )
k –1
1 1 r r
which shows that the vectors (2) form a basis for the null space of N.
21.2 Jordan Form
Now we combine our findings about nilpotent operators or matrices with the primary
decomposition theorem of unit 18. Suppose that T is a linear operator on V and that the
characteristic polynomials for T factors over F as follows:
d ... d
f = (x – c ) 1 (x – c ) k
1 k
where c ,..., c are distinct elements of F and d 1. Then the minimal polynomial for T will be
1 k i
r ... r
p = (x – c ) 1 (x – c ) k
1 k
r
where 1 r d . If W is the null space of (T – c I) i , then the primary decomposition theorem tells
i i i i
us that
...
V = W W
1 k
r
and that the operator T induced on W by T has minimal polynomial (x – c ) i . Let N be the linear
i i i i
r
operator on W defined by N = T – c I. Then N is nilpotent and has minimal polynomial x i . On W ,
i i i i i
T acts like N plus the scalar c times the identity operator. Suppose we choose a basis for the
i i
subspace W corresponding to the cyclic decomposition for the nilpotent operator N . Then the
i i
matrix of T in this ordered basis will be the direct sum of matrices
i
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