Page 357 - A Course in Linear Algebra with Applications

9.3: Bilinear Forms 341

triangular matrix and its diagonal entries are t + (1 — t)ru\
these cannot be zero since ra > 0 and 0 ^ t ^ 1. Hence U is
invertible, while Q is certainly invertible since it is orthogonal.
It follows that S(t) = QU is invertible; thus det(S(t)) ^ 0.
T
Now consider A(t) = S(t) AS{t); since

2
det(A{t)) = det(A) det(S(t)) ^ 0,
it follows that A(t) cannot have zero eigenvalues. Now as t
T
goes from 0 to 1, the eigenvalues of A(0) = S AS gradually
T
change to those of A(l) = Q AQ, that is, to those of A.
But in the process no eigenvalue can change sign because the
eigenvalues that appear are continuous functions of t and they
are never zero. Consequently the numbers of positive and
T
negative eigenvalues of S AS are equal to those of A.
Finally, what if A is singular? In this situation the trick
is to consider the matrix A + el, which may be thought of as
a "perturbation" of A. Now A + el will be invertible provided
that € is sufficiently small and positive: for det(A + xl) is a
polynomial of degree n in x, so it vanishes for at most n values
of x. The previous argument shows that the result is true for
A + el if e is small and positive; then by taking the limit as
e —•> 0, we can deduce the result for A.

It follows from this theorem that the numbers of posi-
tive and negative signs that appear in the canonical form of
9.3.4 are uniquely determined by the bilinear form and do not
depend on the particular basis chosen.

Example 9.3.3

Show that the matrices ( j and I J are not con-
gruent.
All one need do here is note that the first matrix has
3
eigenvalues 1,3, while the second has eigenvalues , - 1 . Hence
by 9.3.5 they cannot be congruent.