Lemma: If the function Q(x,y) = ax2+2bxy+cy2, restricted to the unit circle, has a maximum at the point (1,0), then b = 0.
Proof: Parametrize the circle by x = cos t, y = sin t, t in (0-e, 2p+e). Then Q, restricted to the circle, becomes a function of t:
Q(t) = a cos2t + 2b cos t sin t + c sin2t
Since Q has a maximum at the point (1,0), we have:
| ( |
dQ
dt |
)t=0 = 2b = 0. |
| Therefore, b = 0. |
Proposition: Given a quadratic form Q in V, there exists an orthonormal basis {e1,e2} of V such that if v in V is given by v = xe1 + ye2, then
Q(v) = l1x2 + l2y2,
where l1 and l2 are the maximum and minimum, respectively, of Q on the unit circle |v| = 1.
Proof: Let l1 be the maximum of Q on the unit circle |v| = 1, and let e1 be a unit vector with Q(e1) = l1. Such an e1 exists by continuity of Q on the compact set |v| = 1. Let e2 be a unit vector that is orthogonal to e1, and set l2 = Q(e2). We shall show that the basis {e1,e2} satisfies the conditions of the proposition.
Let B be the symmetric bilinear form that is associated to Q and set v = xe1 + ye2. Then
Q(v) = B(v,v) = B(xe1 + ye2, xe1 + ye2) = l1x2 + 2bxy + l2y2,
where b = B(e1,e2). By the lemma, b = 0, and it only remains to prove that k2 is the minimum of Q in the circle |v| = 1. This is immediate because, for any v = xe1 + ye2, with x2 + y2 = 1, we have that
Q(v) = l1x2 + l2y2 > l2(x2 + y2) = l2,
since l2 < l1.
Theorem: Let A : V --> V be a self-adjoint linear map. Then there exists an orthonormal basis {e1,e2} of V such that A(e1) = l1e1, A(e2) = l2e2 (that is, e1 and e2 are eigenvectors, and l1 and l2 are eigenvalues of A). In the basis {e1,e2}, the matrix of A is clearly diagonal and the elements l1, l2, l1 > l2, on the diagonal are the maximum and the minimum, respectively, of the quadratic form Q(v) = <Av,v> on the unit circle of V.
Proof: Consider the quadratic form Q(v) = <Av,v>. By the proposition above, there exists an orthonormal basis {e1,e2} of V, with Q(e1) = l1, Q(e2) = l2 < l1, where l1 and l2 are the maximum and minimum, respectively, of Q in the unit circle. It remains, therefore, to prove that
A(e1) = l1e1 and A(e2) = l2e2.
Since
B(e1,e2) =
<Ae1,e2>
= 0 (by the lemma) and
e2!=0, we have that either
Ae1 is parallel to
e1 or
Ae1=0. If
Ae1 is parallel to
e1, then
Ae1 = ae1,
and since
<Ae1,e1>
= l1 =
<ae1,e1>
= a, we conclude that Ae1 =
l1e1;
if Ae1 = 0, then
l1 =
<Ae1,e1>
= 0, and Ae1 = 0 =
l1e1.
Thus, we have in any case that Ae1 =
l1e1.
Now using the fact that
B(e1,e2) = <Ae2,e1> = 0
and that
<Ae2,e2> = l2,
we can prove in the same way that Ae2 = l2e2.