Page 249 - Linear Algebra Done Right

240
We need one more ingredient before getting to the main result in
n
this section. Suppose S ∈L(R ) is an isometry. For x, y ∈ R ,we
have Chapter 10. Trace and Determinant n
Sx − Sy = S(x − y)
= x − y .
In other words, S does not change the distance between points. As you
can imagine, this means that S does not change volumes. Specifically,
n
if Ω ⊂ R , then volume S(Ω) = volume Ω.
n
Now we can give our pseudoproof that an operator T ∈L(R )
changes volumes by a factor of |det T|.

n
10.38 Theorem: If T ∈L(R ), then
volume T(Ω) =|det T|(volume Ω)
n
for Ω ⊂ R .
n
Proof: First consider the case where T ∈L(R ) is a positive
operator. Let λ 1 ,...,λ n be the eigenvalues of T, repeated according
to multiplicity. Each of these eigenvalues is a nonnegative number
(see 7.27). By the real spectral theorem (7.13), there is an orthonormal
basis (e 1 ,...,e n ) of V such that Te j = λ j e j for each j. As discussed
above, this implies that T changes volumes by a factor of det T.
n
Now suppose T ∈L(R ) is an arbitrary operator. By the polar de-
composition (7.41), there is an isometry S ∈L(V) such that
√
T = S T T.
∗
√
n
If Ω ⊂ R , then T(Ω) = S T T(Ω) . Thus
∗
√
volume T(Ω) = volume S T T(Ω)
∗
√
= volume T T(Ω)
∗
√
= (det T T)(volume Ω)
∗
=|det T|(volume Ω),
where the second equality holds because volumes are not changed by
the isometry S (as discussed above), the third equality holds by the
√
previous paragraph (applied to the positive operator T T), and the
∗
fourth equality holds by 10.37.