Page 213 - Linear Algebra Done Right

The Characteristic Polynomial
Thus, by our induction hypothesis,
n
9.12 precisely (1/d) dim null p(T| U ) of the matrices 203
A 1 ,...,A m−1 have characteristic polynomial p.
Actually the induction hypothesis gives 9.12 with exponent dim U in-
stead of n, but then we can replace dim U with n (by 8.6) to get the
statement above.
Suppose u m ∈ U m . Let S ∈L(U m ) be the operator whose matrix
(with respect to the basis corresponding to U m ) equals A m . In particu-
lar, Su m = P U m ,U Tu m . Now

Tu m + P U m ,U Tu m
Tu m = P U,U m
=∗ U + Su m ,
where ∗ U denotes a vector in U. Note that Su m ∈ U m ; thus applying
T to both sides of the equation above gives
2
2
T u m =∗ U + S u m ,
where again ∗ U denotes a vector in U, though perhaps a different vector
than the previous usage of ∗ U (the notation ∗ U is used when we want
to emphasize that we have a vector in U but we do not care which
particular vector—each time the notation ∗ U is used, it may denote a
different vector in U). The last two equations show that

9.13 p(T)u m =∗ U + p(S)u m
for some ∗ U ∈ U. Note that p(S)u m ∈ U m ; thus iterating the last
equation gives

n
n
9.14 p(T) u m =∗ U + p(S) u m
for some ∗ U ∈ U.
The proof now breaks into two cases. First consider the case where
the characteristic polynomial of A m does not equal p. We will show
that in this case
n
9.15 null p(T) ⊂ U.
Once this has been verified, we will know that

n
n
null p(T) = null p(T| U ) ,