Linear Algebra, 2020a

416 Chapter Five. Similarity
3.8 Example If
( )
π 1
S =
0 3
(here π is not a projection map, it is the number 3.14 . . .) then
π − x 1
=(x − π)(x − 3)
∣ 0 3− x∣ so S has eigenvalues of λ 1 = π and λ 2 = 3. To find associated eigenvectors, first
plug in λ 1 for x
(
)( ) ( )
π − π 1 z 1 0
=
0 3− π z 2 0
=⇒
( ) ( )
z 1 a
=
z 2 0
for a scalar a ≠ 0. Then plug in λ 2
(
)( ) ( )
π − 3 1 z 1 0
=
0 3− 3 z 2 0
where b ≠ 0.
=⇒
( ) ( )
z 1 −b/(π − 3)
=
z 2 b
3.9 Definition The characteristic polynomial of a square matrix T is the
determinant |T − xI| where x is a variable. The characteristic equation is
|T − xI| = 0. The characteristic polynomial of a transformation t is the
characteristic polynomial of any matrix representation Rep B,B (t).
The characteristic polynomial of an n × n matrix, or of a transformation
t: C n → C n , is of degree n. Exercise 35 checks that the characteristic polynomial
of a transformation is well-defined, that is, that the characteristic polynomial is
the same no matter which basis we use for the representation.
3.10 Lemma A linear transformation on a nontrivial vector space has at least
one eigenvalue.
Proof Any root of the characteristic polynomial is an eigenvalue. Over the
complex numbers, any polynomial of degree one or greater has a root. QED
3.11 Remark That result is the reason that in this chapter we use scalars that
are complex numbers. Had we stuck to real number scalars then there would be
characteristic polynomials, such as x 2 + 1, that do not factor.