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Linear Algebra
Notes Choose any other vectors ,..., in V such that = { ,..., } is a basis for V. The matrix of T
r+1 n 1 n
relative to has the block form (3), and the matrix of the restriction operator Tw relative to the
basis ’ is
t 1 0 0
0 t 0
B 2
0
0 t r
The characteristic polynomial of B (i.e., of Tw) is
e
g = (x – c ) 1 ... (x – c ) k e
1 k
where e = dim W . Furthermore, g divides f, the characteristic polynomial for T. Therefore, the
i i
multiplicity of c as a root of f is at least dim W .
i i
All of this should make Theorem 2 of unit 12 transparent. It merely says that T is diagonalizable
if and only if r = n, if and only if e + ... + e = n. It does not help us too much with the
1 k
non-diagonalizable case, since we don’t know the matrices C and D of (3).
Lemma: If W is an invariant subspace for T, then W is invariant under every polynomial in T.
Thus, for each in V, the conductor S (; W) is an ideal in the polynomial algebra F[x].
T
2
k
Proof: If is in W, then T is in W. Consequently, T(T) = T is in W. By induction, T is in W for
each k. Take linear combinations to see that f(T) is in W for every polynomial f.
The definition of S (; W) makes sense if W is any subset of V. If W is a subspace, then S (; W) is
T T
a subspace of F[x], because
(cf + g)(T) = cf(T) + g(T)
If W is also invariant under T, let g be a polynomial in S (; W), i.e., let g(T) be in W. If f is any
T
polynomial, then f(T)[g(T)] will be in W. Since
(fg)(T) = f(T)g(T)
fg is in S (; W). Thus the conductor absorbs multiplication by any polynomial.
T
The unique monic generator of the ideal S (; W) is also called the T-conductor of into W
T
(the T-annihilator in case W = {0}). The T-conductor of into W is the monic polynomial g of least
degree such that g(T) is in W. A polynomial f is in S (; W) if and only if g divides f. Note that
T
the conductor S (; W) always contains the minimal polynomial for T; hence, every T-conductor
T
divides the minimal polynomial for T.
As the first illustration of how to use the conductor S (; W), we shall characterize triangulable
T
operators. The linear operator T is called triangulable if there is an ordered basis in which T is
represented by a triangular matrix.
Lemma. Let V be a finite-dimensional vector space over the field F. Let T be a linear operator on
V such that the minimal polynomial for T is a product of linear factors
r
r
p = (x – c ) 1 ... (x – c ) k , c in F
1 k i
Let W be a proper (W V) subspace of V which is invariant under T. There exists a vector in V
such that
(a) is not in W;
(b) (T – cI) is in W, for some characteristic value c of the operator T.
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