Page 225 - DMTH502_LINEAR_ALGEBRA

Unit 20: Cyclic Decomposition and the Rational Form

that is, that T is represented by B in some ordered basis. How can we find suitable vectors and Notes
1
? Well, we know that any vector which generates a T-cyclic subspace of dimension 2 is a
2
suitable . So let’s just try . We have
1 1
T = (5, –1, 3)
1
which is not a scalar multiple of ; hence Z( ; T) has dimension 2. This space consists of all
1 1
vectors a + b (T ):
1 1
b
b
a (1,0,0) b (5, 1,3) (a 5 , b ,3 )
or, all vectors (x , x , x ) satisfying x = –3x . Now what we want is a vector such that T = 2
1 2 3 3 2 2 2 2
and Z( ; T) is disjoint from Z( ; T). Since is to be a characteristic vector for T, the space Z( ;
2 1 2 2
T) will simply by the one-dimensional space spanned by , and so what we require is that not
2 2
be in Z( ; T). If = (x , x , x ), one can easily compute that T = 2 if and only if x = 2x + 2x . Thus
1 1 2 3 1 2 3
= (2, 1, 0) satisfies T = 2 and generates a T-cyclic subspace disjoint from Z( ; T). The reader
2 2 2 1
should verify directly that the matrix of T is the ordered basis.
{(1, 0, 0) , (5, –1, 3), (2, 1, 0)}
is the matrix B above.
Example 3: Suppose that T is a diagonalizable linear operator on V. It is interesting to
relate a cyclic decomposition for T to a basis which diagonalizes the matrix of T. Let c , ... c be the
1 k
distinct characteristic values of T and let V be the space of characteristic vectors associated with
i
the characteristic value c . Then
i
V V ... V
i k
and if d = dim V then
i i
1 d
f (x c ) ....(x c ) k d
1 k
is the characteristic polynomial for T. If is a vector in V, it is easy to relate the cyclic subspace

Z( ; T) to the subspaces V , ..., V . There are unique vectors , ..., such that is in V and
1 k 1 k i i
.... .
1 k
Since T = c , we have
1 i i
c
c
f ( ) f ( ) ... f ( ) ...(18)
T
1 1 k k
for every polynomial f. Given any scalars t , ...., t there exists a polynomial f such that f(c ) = t ,
1 k i i
1 i k. Therefore Z( ; T) is just the subspace spanned by the vectors , ...., . What is the
1 k
annihilator of ? According to (18), we have f(T) = 0 if and only if f(c ) = 0 for each i. In other
i i
words, f(T) = 0 provided f(c ) = 0 for each i such that 0. Accordingly, the annihilator of is the
i i
product
(x c i ). ...(19)
i 0
Now, let i { t 1 ,....., t i d } be an ordered basis for V . Let
i
r max d i .
u
We define vectors , ..., by
1 r
t
j j , 1 j . r ...(20)
i d j

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