Page 184 - Linear Algebra Done Right

Cayley-Hamilton Theorem: Suppose that V is a complex vector
8.20
mathematician Arthur
space and T ∈L(V). Let q denote the characteristic polynomial of T.
Then q(T) = 0. Decomposition of an Operator The English 173
Cayley published three
mathematics papers
Proof: Suppose (v 1 ,...,v n ) is a basis of V with respect to which before he completed
the matrix of T has the upper-triangular form 8.19. To prove that his undergraduate
q(T) = 0, we need only show that q(T)v j = 0 for j = 1,...,n.To degree in 1842. The
do this, it suffices to show that Irish mathematician
William Hamilton was
8.21 (T − λ 1 I)...(T − λ j I)v j = 0 made a professor in
1827 when he was 22
for j = 1,...,n. years old and still an
We will prove 8.21 by induction on j. To get started, suppose j = 1.
undergraduate!
Because M T, (v 1 ,...,v n ) is given by 8.19, we have Tv 1 = λ 1 v 1 , giving
8.21 when j = 1.
Now suppose that 1 <j ≤ n and that

0 = (T − λ 1 I)v 1
= (T − λ 1 I)(T − λ 2 I)v 2
. . .

= (T − λ 1 I)...(T − λ j−1 I)v j−1 .

Because M T, (v 1 ,...,v n ) is given by 8.19, we see that

(T − λ j I)v j ∈ span(v 1 ,...,v j−1 ).
Thus, by our induction hypothesis, (T − λ 1 I)...(T − λ j−1 I) applied to
(T −λ j I)v j gives 0. In other words, 8.21 holds, completing the proof.

Decomposition of an Operator

We saw earlier that the domain of an operator might not decompose
into invariant subspaces consisting of eigenvectors of the operator,
even on a complex vector space. In this section we will see that every
operator on a complex vector space has enough generalized eigenvec-
tors to provide a decomposition.
We observed earlier that if T ∈L(V), then null T is invariant un-
der T. Now we show that the null space of any polynomial of T is also
invariant under T.