Page 217 - Linear Algebra Done Right

Now we can prove a result that was promised in the last chapter,
where we proved the analogous theorem (8.20) for operators on com-
plex vector spaces. The Characteristic Polynomial 207
9.20 Cayley-Hamilton Theorem: Suppose V is a real vector space
and T ∈L(V). Let q denote the characteristic polynomial of T. Then
q(T) = 0.

Proof: Choose a basis of V with respect to which T has a block This proof uses the
upper-triangular matrix of the form 9.18, where each A j is a 1-by-1 same ideas as the proof
matrix or a 2-by-2 matrix with no eigenvalues. Suppose U j is the one- or of the analogous result
two-dimensional subspace spanned by the basis vectors corresponding on complex vector
to A j . Define q j as in 9.19. To prove that q(T) = 0, we need only show spaces (8.20).
= 0 for j = 1,...,m. To do this, it suffices to show that
that q(T)| U j
9.21 q 1 (T)...q j (T)| U j = 0

for j = 1,...,m.
We will prove 9.21 by induction on j. To get started, suppose that
= 0 (obvious if
j = 1. Because M(T) is given by 9.18, we have q 1 (T)| U 1
dim U 1 = 1; from 9.7(a) if dim U 1 = 2), giving 9.21 when j = 1.
Now suppose that 1 <j ≤ n and that

0 = q 1 (T)| U 1
0 = q 1 (T)q 2 (T)| U 2
. . .

.
0 = q 1 (T)...q j−1 (T)| U j−1
If v ∈ U j , then from 9.18 we see that
q j (T)v = u + q j (S)v,

where u ∈ U 1 + ··· + U j−1 and S ∈L(U j ) has characteristic poly-
nomial q j . Because q j (S) = 0 (obvious if dim U j = 1; from 9.7(a) if
dim U j = 2), the equation above shows that

q j (T)v ∈ U 1 + ··· + U j−1
whenever v ∈ U j . Thus, by our induction hypothesis, q 1 (T)...q j−1 (T)
applied to q j (T)v gives 0 whenever v ∈ U j . In other words, 9.21 holds,
completing the proof.