Page 233 - Linear Algebra Done Right

Chapter 10. Trace and Determinant
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The next lemma gives the analogous result for eigenpairs. We will use
this lemma to show that the characteristic polynomial can be expressed
as a certain determinant.
Real vector spaces are 10.16 Lemma: Suppose V is a real vector space, T ∈L(V), and
2
harder to deal with α, β, x ∈ R with α < 4β. Then (α, β) is an eigenpair of T if and only
2
than complex vector if (−2x −α, x +αx +β) is an eigenpair of xI −T. Furthermore, these
spaces. The first time eigenpairs have the same multiplicities.
you read this chapter,
2
you may want to Proof: First we need to check that (−2x −α, x +αx +β) satisfies
concentrate on the the inequality required of an eigenpair. We have
basic ideas by 2 2 2
considering only (−2x − α) = 4x + 4αx + α
2
complex vector spaces < 4x + 4αx + 4β
and ignoring the = 4(x + αx + β).
2
special procedures
2
needed to deal with Thus (−2x − α, x + αx + β) satisfies the required inequality.
real vector spaces. Now
2
2
2
T + αT + βI = (xI − T) − (2x + α)(xI − T) + (x + αx + β)I,
as you should verify. Thus (α, β) is an eigenpair of T if and only if
2
(−2x − α, x + αx + β) is an eigenpair of xI − T. Furthermore, raising
both sides of the equation above to the dim V power and then taking
null spaces of both sides shows that the multiplicities are equal.

Most textbooks take the theorem below as the definition of the char-
acteristic polynomial. Texts using that approach must spend consider-
ably more time developing the theory of determinants before they get
to interesting linear algebra.

10.17 Theorem: Suppose T ∈L(V). Then the characteristic poly-
nomial of T equals det(zI − T).

Proof: First suppose V is a complex vector space. Let λ 1 ,...,λ n
denote the eigenvalues of T, repeated according to multiplicity. Thus
for z ∈ C, the eigenvalues of zI − T are z − λ 1 ,...,z − λ n , repeated
according to multiplicity. The determinant of zI − T is the product of
these eigenvalues. In other words,

det(zI − T) = (z − λ 1 )...(z − λ n ).