Page 245 - Linear Algebra Done Right

Chapter 10. Trace and Determinant
236
Proof: Suppose S, T ∈L(V). Choose any basis of V. Then
det(ST) = det M(ST)

= det M(S)M(T)

= det M(S) det M(T)
= (det S)(det T),
where the first and last equalities come from 10.33 and the third equal-
ity comes from 10.31.
In the paragraph above, we proved that det(ST) = (det S)(det T). In-
terchanging the roles of S and T, we have det(TS) = (det T)(det S). Be-
cause multiplication of elements of F is commutative, the last equation
can be rewritten as det(TS) = (det S)(det T), completing the proof.

Volume

We proved the basic results of linear algebra before introducing de-
terminants in this final chapter. Though determinants have value as a
research tool in more advanced subjects, they play little role in basic
Most applied linear algebra (when the subject is done right). Determinants do have
mathematicians agree one important application in undergraduate mathematics, namely, in
that determinants computing certain volumes and integrals. In this final section we will
should rarely be used use the linear algebra we have learned to make clear the connection
in serious numeric between determinants and these applications. Thus we will be dealing
calculations. with a part of analysis that uses linear algebra.
We begin with some purely linear algebra results that will be use-
ful when investigating volumes. Recall that an isometry on an inner-
product space is an operator that preserves norms. The next result
shows that every isometry has determinant with absolute value 1.

10.35 Proposition: Suppose that V is an inner-product space. If
S ∈L(V) is an isometry, then |det S|= 1.

Proof: Suppose S ∈L(V) is an isometry. First consider the case
where V is a complex inner-product space. Then all the eigenvalues of S
have absolute value 1 (by 7.37). Thus the product of the eigenvalues
of S, counting multiplicity, has absolute value one. In other words,
|det S|= 1, as desired.