Page 218 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes 20.1 Overview

In this unit we are interested in dealing with a linear operator T on a finite-dimensional space V,
and dealing with the cyclic subspaces Z ( ; T) where the vectors , , ... in V. In this case the
1 1 2 r
finite dimensional space V can be decomposed as the direct sum, i.e.,
V = Z ( ; T) .... Z ( ; T)
1 r
This will also show that T is the direct sum of a finite number of linear operators each of which
has a cyclic vector. The effect of this will be to reduce many questions about the general linear
operator.
Let us consider T-invariant subspaces W and W such that V = W W . Here for any invariant
subspace W, W is complementary to W. We are interested in those W which are also T-invariant.

T-admissible invariant subspace W
Let T be a linear operator on a vector space V and let W be a subspace of V. We say that W is
T-admissible if W is invariant under T and if f(T) is in W where f is a polynomial, and there exists
a vector Y in W such that
f(T) = f(T)
our method for arriving at a decomposition
V = Z ( ; T) ... Z ( ; T)
i r
will be to inductively select the vectors , , ... . Suppose that by some process or another
1 2 r
we have selected , , .... and the sub-space
1 2 j
W = Z ( ; T) Z ( ; T) + ... + Z ( ; T)
j i 2 j
is proper. We would like to find a non-zero vector such that
j + 1
W Z ( ; T) = {0}.
j j + 1
Thus W will be a proper T-invariant subspace if there is a non-zero vector such that
W Z ( ; T) = {0} ...(1)
Thus the subspace Z ( ; T) and W are independent if (1) is satisfied and the polynomial f is the
T-annihilator of i.e. f(T) = 0.

20.2 Cyclic Decomposition

With the above definition we arrive at the following theorem for the cyclic decomposition of
the finite vector space.
Theorem 1 (Cyclic Decomposition Theorem). Let T be a linear operator on a finite-dimensional
vector space V and let W be a proper T-admissible subspace of V. There exist non-zero vectors
0
, ...., in V with respective T- annihilators p ,...., p such that
1 r 1 r
(i) V = W Z ( ; T) ... Z ( ; T);
0 1 r
(ii) p divides p , k = 2, ..., r.
k k–1
Furthermore, the integer r and the annihilators p , ...., p are uniquely determined by (i), (ii), and
1 r
the fact that no is 0.
k
Proof: The proof is rather long; hence, we shall divide it into four steps. For the first reading it
may seem easier to take W = {0}, although it does not produce any substantial simplification.
0
Throughout the proof, we shall abbreviate f(T) to f .

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