Page 94 - Linear Algebra Done Right

Upper-Triangular Matrices
(pq)(T) = p(T)q(T)
for all polynomials p and q with coefficients in F. Note that any two
polynomials in T commute, meaning that p(T)q(T) = q(T)p(T), be- 81
cause
p(T)q(T) = (pq)(T) = (qp)(T) = q(T)p(T).
Upper-Triangular Matrices

Now we come to one of the central results about operators on com-
plex vector spaces.

5.10 Theorem: Every operator on a finite-dimensional, nonzero, Compare the simple
complex vector space has an eigenvalue. proof of this theorem
given here with the
Proof: Suppose V is a complex vector space with dimension n> 0 standard proof using
and T ∈L(V). Choose v ∈ V with v = 0. Then determinants. With the
standard proof, first
n
2
(v, Tv, T v,...,T v)
the difficult concept of
determinants must be
cannot be linearly independent because V has dimension n and we have
defined, then an
n + 1 vectors. Thus there exist complex numbers a 0 ,...,a n , not all 0,
operator with 0
such that determinant must be
n
0 = a 0 v + a 1 Tv +· · ·+ a n T v.
shown to be not
Let m be the largest index such that a m = 0. Because v = 0, the invertible, then the
coefficients a 1 ,...,a m cannot all be 0, so 0 <m ≤ n. Make the a’s characteristic
the coefficients of a polynomial, which can be written in factored form polynomial needs to be
(see 4.8) as defined, and by the
time the proof of this
n
a 0 + a 1 z +· · ·+ a n z = c(z − λ 1 )...(z − λ m ), theorem is reached, no
insight remains about
where c is a nonzero complex number, each λ j ∈ C, and the equation
why it is true.
holds for all z ∈ C. We then have
n
0 = a 0 v + a 1 Tv + ··· + a n T v
n
= (a 0 I + a 1 T +· · ·+ a n T )v
= c(T − λ 1 I)...(T − λ m I)v,

which means that T − λ j I is not injective for at least one j. In other
words, T has an eigenvalue.