Page 246 - Linear Algebra Done Right

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Now suppose V is a real inner-product space. Then there is an ortho-
normal basis of V with respect to which S has a block diagonal matrix,
where each block on the diagonal is a 1-by-1 matrix containing 1 or −1 237
or a 2-by-2 matrix of the form

cos θ − sin θ
10.36 ,
sin θ cos θ
with θ ∈ (0,π) (see 7.38). Note that the constant term of the charac-
teristic polynomial of each matrix of the form 10.36 equals 1 (because
2
2
cos θ + sin θ = 1). Hence the second coordinate of every eigenpair
of S equals 1. Thus the determinant of S is the product of 1’s and −1’s.
In particular, |det S|= 1, as desired.
Suppose V is a real inner-product space and S ∈L(V) is an isometry.
By the proposition above, the determinant of S equals 1 or −1. Note
that
{v ∈ V : Sv =−v}
is the subspace of V consisting of all eigenvectors of S corresponding
to the eigenvalue −1 (or is the subspace {0} if −1 is not an eigenvalue
of S). Thinking geometrically, we could say that this is the subspace
on which S reverses direction. A careful examination of the proof of
the last proposition shows that det S = 1 if this subspace has even
dimension and det S =−1 if this subspace has odd dimension.
A self-adjoint operator on a real inner-product space has no eigen-
pairs (by 7.11). Thus the determinant of a self-adjoint operator on a
real inner-product space equals the product of its eigenvalues, count-
ing multiplicity (of course, this holds for any operator, self-adjoint or
not, on a complex vector space).
Recall that if V is an inner-product space and T ∈L(V), then T T
∗
is a positive operator and hence has a unique positive square root, de-
√ √
noted T T (see 7.27 and 7.28). Because T T is positive, all its eigen-
∗
∗
values are nonnegative (again, see 7.27), and hence its determinant is
nonnegative. Thus in the corollary below, taking the absolute value of
√
det T T would be superfluous.
∗
10.37 Corollary: Suppose V is an inner-product space. If T ∈L(V),
then
√
|det T|= det T T.
∗