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Unit 14: Invariant Subspaces
In other words, A = 0 if j r and i r. Notes
ij
Schematically, A has the block form
B C
A = ...(3)
0 D
where B is an r × r matrix, C is an r × (n – r) matrix, and D is an (n – r) × (n – r) matrix. The reader
should note that according to (2) the matrix B is precisely the matrix of the induced operator Tw
in the ordered basis ’.
Most often, we shall carry out arguments about T and Tw without making use of the block form
of the matrix A in (3). But we should note how certain relations between Tw and T are apparent
from that block form.
Lemma: Let W be an invariant subspace for T, the characteristic polynomial for the restriction
operator Tw divides the characteristic polynomial for T. The minimal polynomial for Tw divides
the minimal polynomial for T.
Proof: We have
B C
A
0 D
where A = [T] and B = [T ] . Because of the block form of the matrix
w ’
det (xI – A) = det (xI – B) det (xI – D)
That proves the statement about characteristic polynomials. Notice that we used I to represent
identity matrices of three different sizes.
th
The k power of the matrix A has the block form
B k C
k k
A k
0 D
where C is some r × (n – r) matrix. Therefore, any polynomial which annihilates A also annihilates
k
B (and D too). So, the minimal polynomial for B divides the minimal polynomial for A.
Example 5: Let T be any linear operator on a finite-dimensional space V. Let W be the
subspace spanned by all of the characteristic vectors of T. Let c ,...,c be the distinct characteristic
1 k
values of T. For each i, let W be the space of characteristic vectors associated with the characteristic
i
value c , and let be an ordered basis for W . The lemma before Theorem 2 of unit 12 tells us that
i i i
’ = ( ,..., ) is an ordered basis for W. In particular,
1 k
dim W = dim W + ... + dim W .
1 k
Let ’ = { ,..., } so that the first few ’s form the basis , the next few , and so on. Then
1 r 1 2
T = t , i = 1,...,r
i i i
where (t ,..., t ) = (c , c ,..., c ,..., c , c ,..., c ) with c repeated dim W times.
1 r 1 1 1 k k k i i
Now W is invariant under T, since for each in W we have
= x + ... + x
1 1 r r
T = t x + ... + t x
1 1 1 r r r
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