Page 231 - DMTH502_LINEAR_ALGEBRA

Unit 21: The Jordan Form

Now we wish to make some further observations about the operator T and the Jordan matrix A Notes
which represents T in some ordered basis. We shall list a string of observations:
(1) Every entry of A not on or immediately below the main diagonal is 0. On the diagonal of
A occur the k distinct characteristic values c ,..., c of T. Also, c is repeated d times, where d
1 k i i i
is the multiplicity of c as a root of the characteristic polynomial, i.e., d = dim W .
i i i
j with
(2) For each i, the matrix A is the direct sum of n elementary Jordan matrices J ( ) i
i i
characteristic values c . The number n is precisely the dimension of the space of characteristic
i i
vectors associated with the characteristic value c . For, n is the number of elementary
i i
nilpotent blocks in the rational form for (T – c I), and is thus equal to the dimension of the
i i
null space of (T – c I). In particular notice that T is diagonalizable if and only if n = d for
i i i
each i.
t
( )
(3) For each i, the first block J in the matrix A, is an r × r matrix, where r is the multiplicity
1 i i i
of c as a root of the minimal polynomial for T. This follows from the fact that the minimal
i
polynomial for the nilpotent operator (T – c I) is x i .
r
i i
Of course we have as usual the straight matrix result. If B is an n × n matrix over the field F and
if the characteristic polynomial for B factors completely over F, then B is similar over F to an
n × n matrix A in Jordan form, and A is unique up to a rearrangement of the order of its
characteristic values. We call A the Jordan form of B.
Also, note that if F is an algebraically closed field, then the above remarks apply to every linear
operator on a finite-dimensional space over F, or to every n × n matrix over F. Thus, for example,
every n × n matrix over the field of complex numbers is similar to an essentially unique matrix
in Jordan form.
If the linear transformation T is nilpotent then T n 1 0 where n is the index of nilpotency. If
1
T n 1 1 0 we can find a vector v in the space V such that T n 1 1 0 . Then we can form the vectors
v = v, v = T v, v = T v, … v T n 1 1 v vectors which are claimed to be linearly independent over
2
1 2 3 n 1
the field F.
Let V be the subspace of V spanned by v = v, v = Tv, … v T n 1 1 v , V is invariant under T, and
1 1 2 n 1 1
in the basis above, the linear transformation induced by T on V has a matrix A of the form (1).
n
1 1
Let the vector space V is of the form V = V W where W is invariant under T. Using the basis
1
v , v , … v of V and any basis of W as a basis of V, the matrix of T in this basis has the form
n
1 2 1 1
A 0
n 1
0 A
n 2
where A is the matrix of T , the linear transformation induced on W by T. Since T n 1 0, T 1/2 0
2 2 2
for some n n .
2 1
Let T is a linear operator on C . The characteristic polynomial for T is either (x – C ) (x – C ) where
2
1 2
C and C are distinct or is (x – C) . In the former case T is diagonalizable and is represented in
2
1 2
some ordered basis by the matrix
C 1 0
0 C .
2
In the later case, the minimal polynomial for T may be (x – C), in which case T = C I, or may be
2
(x – C) , in which case T is represented in some order basis by the matrix

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