Page 222 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes which shows that s 2. Now it makes sense to ask whether or not p = g . From the two
2 2
decompositions of V, we obtain two decompositions of the subspace p V:
2
p V p W 0 ( Z p 2 1 ; )
T
2
2
...(12)
T
p V p W ( Z p ; ) ... Z (p ; ).
T
2 2 0 2 1 2 s
We have made use of facts (1) and (2) above and we have used the fact that p 2 i 0, i 2. Since
T
T
Z
we know that p = g , fact (3) above tells us that (Z p 2 1 ; ) and (p 2 1 ; ) have the same
1 1
dimension. Hence, it is apparent from (12) that
T
Z
dim (p ; ) 0, i 2.
2 i
We conclude that p = 0 and g divides p . The argument can be reversed to show that p divides
2 2 2 2 2
g . Therefore p = g .
2 2 2
Corollary: If T is a linear operator on a finite-dimensional vector space, then every T-admissible
subspace has a complementary subspace which is also invariant under T.
Proof: Let W be an admissible subspace of V. If W = V, the complement we seek is {0}. If W is
0 0 0
proper, apply Theorem 1 and let
T
T
W 0 ( Z 1 ; ) ... ( Z r ; ).
Then W is invariant under T and V = W W .
0 0 0
Corollary: Let T be a linear operator on a finite-dimensional vector space V.
(a) There exists a vector in V such that the T-annihilator of is the minimal polynomial for T.
(b) T has a cyclic vector if and only if the characteristic and minimal polynomials for T are
identical.
Proof: If V = {0}, the results are trivially true. If V {0}, let
T
T
V ( Z 1 ; ) ... Z ( r ; ) ...(13)
where the T-annihilators p 1 ,....,p are such that p k 1 divides p k , 1 k r 1. As we noted in the
r
proof of Theorem 1, it follows easily that p is the minimal polynomial for T, i.e., the T-conductor
1
of V into {0}. We have proved (a).
We saw in unit 19 that, if T has a cyclic vector, the minimal polynomial for T coincides with the
characteristic polynomial. The content of (b) is in the converse. Choose any as in (a). If the
degree of the minimal polynomial is dim V, then V = Z ( ; T).
Theorem 2 (Generalized Cayley-Hamilton Theorem): Let T be a linear operator on a finite-
dimensional vector space V. Let p and f be the minimal and characteristic polynomials for T,
respectively.

1. p divides f.
2. p and f have the same prime factors, except for multiplication.
3. If

p f 1 T .... f k T ...(14)
1 1
is the prime factorization of p, then
f f 1 1 d ....f k k d ...(15)

n
where d is the nullity of f (T) divided by the degree of f .
i i i

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