Page 193 - Linear Algebra Done Right

Chapter 8. Operators on Complex Vector Spaces
182
For example, consider the operator T on C whose matrix is given
by
  5
0 0 0 0 −3
 
 1 0 0 0 6 
 
8.37  0 1 0 0 0  .
 
 0 0 1 0 0 


0 0 0 1 0
Because of the large number of 0’s in this matrix, Gaussian elimination
is not needed here. Simply compute powers of M(T) and notice that
there is no linear dependence until the fifth power. Do the computa-
tions and you will see that the minimal polynomial of T equals
5
8.38 z − 6z + 3.
Now what about the eigenvalues of this particular operator? From 8.36,
we see that the eigenvalues of T equal the solutions to the equation
5
z − 6z + 3 = 0.
Unfortunately no solution to this equation can be computed using ra-
tional numbers, arbitrary roots of rational numbers, and the usual rules
of arithmetic (a proof of this would take us considerably beyond linear
algebra). Thus we cannot find an exact expression for any eigenvalues
of T in any familiar form, though numeric techniques can give good ap-
proximations for the eigenvalues of T. The numeric techniques, which
we will not discuss here, show that the eigenvalues for this particular
operator are approximately
−1.67, 0.51, 1.40, −0.12 + 1.59i, −0.12 − 1.59i.

Note that the nonreal eigenvalues occur as a pair, with each the complex
conjugate of the other, as expected for the roots of a polynomial with
real coefficients (see 4.10).
Suppose V is a complex vector space and T ∈L(V). The Cayley-
Hamilton theorem (8.20) and 8.34 imply that the minimal polynomial
of T divides the characteristic polynomial of T. Both these polynomials
are monic. Thus if the minimal polynomial of T has degree dim V, then
it must equal the characteristic polynomial of T. For example, if T is
5
the operator on C whose matrix is given by 8.37, then the character-
istic polynomial of T, as well as the minimal polynomial of T, is given
by 8.38.