Page 248 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Then the cyclic decomposition theorem help us in expressing any linear operator as a direct sum
of operators with a cyclic vector. If U is a linear operator with a cyclic vector, there is a basis
( , , ..., ) with
1 2 n
U = i = 1, .... n – 1,
i i + 1
U = –c – c , .... –c .
n 0 1 1 1 n–1 n
This means that action of U on this basis is to shift each to the next vector , except that U
i j + 1 n
is some prescribed linear combination of the vectors in the basis. The general operator T is the
direct sum of a finite number of such operators U and got reasonably elementary description of
the action of T. Then cyclic decomposition theorem to nilpotent operators is applied and with
the help of the primary decomposition theorem Jordan form is obtained.

The importance of the rational form or the Jordan form is obtained from the fact that these forms
can be computed in specific cases. Of course, if one is given a specific linear operator T and if its
cyclic or Jordan form can be computed, one can obtain vast amounts of information about T.
However there are some difficulties in this method. At first the computation may be lengthy.
The other difficulty is there may not be any method for doing computations. In the case of
rational form the difficulty may be due to lengthy calculation. It is also worthwhile to mention
a theorem which states that if T is a linear operator on a finite-dimensional vector space over an
algebraically closed field then T is uniquely expressible as the sum of a diagonalizable operator
and a nilpotent operate which commute.
In this unit we shall prose analogous theorem without assuming that the scalar field is
algebraically closed. We begin by defining the operators which will play the role of the
diagonalizable operators.

23.2 Semi-simple Operators

We say that T a linear operator on a finite dimensional space V over the field F, is semi-simple
if every T-invariant subspace has a complementary T-invariant subspace.
We are going to characterize semi-simple operators by means of their minimal polynomials,
and this characterization will show us, that, when F is algebraically closed, an operator is semi-
simple if and only it is diagonalizable.
Lemma: Let T be a linear operator on the finite dimensional vector space V and let
V = W ... + W
1 k
be the primary decomposition for T. In other words, if p is the minimal polynomial for T and
1 r k r r j
T
p p ... p is the prime factorization of p, then W is the null space of p ( ) . Let W be any
1 k j j
subspace of V which is invariant under T. Then
W = (W W ) ... (W W )
1 k
Proof: If E , E , ..... E the projections associated with the decomposition V = W ... W , then
1 2 k I k
each E is a polynomial in T. That is, there are polynomials h , ..., h such that E = h (T).
j 1 k j j
Now let W be a subspace which is invariant under T. If is any vector in W, then = + ... + ,
l k
where is in W . Now = E = h (T) , and since W is invariant under T, each is also in W. Thus
j j j j j j
each vector in W is of the form = + ... + , where is in the intersection W· W . This
1 k j j
expression is unique, since V = W ... W . Therefore
1 k
W = (W W ) ... (W W )
1 k
Lemma: Let T be a linear operator on V, and suppose that the minimal polynomial for T is
irreducible over the scalar field F. Then T is semi-simple.

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