Page 214 - Linear Algebra Done Right

Chapter 9. Operators on Real Vector Spaces
204
n
and hence 9.12 will tell us that precisely (1/d) dim null p(T)
of the
matrices A 1 ,...,A m have characteristic polynomial p, completing the
proof in the case where the characteristic polynomial of A m does not
equal p.
To prove 9.15 (still assuming that the characteristic polynomial of
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A m does not equal p), suppose v ∈ null p(T) . We can write v in the
form v = u + u m , where u ∈ U and u m ∈ U m . Using 9.14, we have
n n n n n
0 = p(T) v = p(T) u + p(T) u m = p(T) u +∗ U + p(S) u m
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for some ∗ U ∈ U. Because the vectors p(T) u and ∗ U are in U and
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p(S) u m ∈ U m , this implies that p(S) u m = 0. However, p(S) is in-
vertible (see the discussion preceding this theorem about one- and two-
dimensional subspaces and note that dim U m ≤ 2), so u m = 0. Thus
v = u ∈ U, completing the proof of 9.15.
Now consider the case where the characteristic polynomial of A m
equals p. Note that this implies dim U m = d. We will show that
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9.16 dim null p(T) = dim null p(T| U ) + d,
which along with 9.12 will complete the proof.
Using the formula for the dimension of the sum of two subspaces
(2.18), we have
n
n
n
dim null p(T) = dim(U ∩ null p(T) ) + dim(U + null p(T) ) − dim U
n
n
= dim null p(T| U ) + dim(U + null p(T) ) − (n − d).
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If U +null p(T) = V, then dim(U +null p(T) ) = n, which when com-
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bined with the last formula above for dim null p(T) would give 9.16,
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as desired. Thus we will finish by showing that U + null p(T) = V.
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To prove that U + null p(T) = V, suppose u m ∈ U m . Because the
characteristic polynomial of the matrix of S (namely, A m ) equals p,we
have p(S) = 0. Thus p(T)u m ∈ U (from 9.13). Now
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p(T) u m = p(T) n−1 (p(T)u m ) ∈ range p(T| U ) n−1 = range p(T| U ) ,
where the last equality comes from 8.9. Thus we can choose u ∈ U
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such that p(T) u m = p(T| U ) u. Now
n
n
n
p(T) (u m − u) = p(T) u m − p(T) u
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= p(T) u m − p(T| U ) u
= 0.