Lemma: If the function Q(x,y) = ax^{2}+2bxy+cy^{2}, 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 (0e, 2p+e). Then Q, restricted to the circle, becomes a function of t: Q(t) = a cos^{2}t + 2b cos t sin t + c sin^{2}t Since Q has a maximum at the point (1,0), we have:
Proposition: Given a quadratic form Q in V, there exists an orthonormal basis {e_{1},e_{2}} of V such that if v in V is given by v = xe_{1} + ye_{2}, then Q(v) = l_{1}x^{2} + l_{2}y^{2}, where l_{1} and l_{2} are the maximum and minimum, respectively, of Q on the unit circle v = 1. Proof: Let l_{1} be the maximum of Q on the unit circle v = 1, and let e_{1} be a unit vector with Q(e_{1}) = l_{1}. Such an e_{1} exists by continuity of Q on the compact set v = 1. Let e_{2} be a unit vector that is orthogonal to e_{1}, and set l_{2} = Q(e_{2}). We shall show that the basis {e_{1},e_{2}} satisfies the conditions of the proposition. Let B be the symmetric bilinear form that is associated to Q and set v = xe_{1} + ye_{2}. Then Q(v) = B(v,v) = B(xe_{1} + ye_{2}, xe_{1} + ye_{2}) = l_{1}x^{2} + 2bxy + l_{2}y^{2}, where b = B(e_{1},e_{2}). By the lemma, b = 0, and it only remains to prove that k_{2} is the minimum of Q in the circle v = 1. This is immediate because, for any v = xe_{1} + ye_{2}, with x^{2} + y^{2} = 1, we have that Q(v) = l_{1}x^{2} + l_{2}y^{2} > l_{2}(x^{2} + y^{2}) = l_{2}, since l_{2} < l_{1}.
Theorem: Let A : V > V be a selfadjoint linear map. Then there exists an orthonormal basis {e_{1},e_{2}} of V such that A(e_{1}) = l_{1}e_{1}, A(e_{2}) = l_{2}e_{2} (that is, e_{1} and e_{2} are eigenvectors, and l_{1} and l_{2} are eigenvalues of A). In the basis {e_{1},e_{2}}, the matrix of A is clearly diagonal and the elements l_{1}, l_{2}, l_{1} > l_{2}, 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 {e_{1},e_{2}} of V, with Q(e_{1}) = l_{1}, Q(e_{2}) = l_{2} < l_{1}, where l_{1} and l_{2} are the maximum and minimum, respectively, of Q in the unit circle. It remains, therefore, to prove that A(e_{1}) = l_{1}e_{1} and A(e_{2}) = l_{2}e_{2}.
Since
B(e_{1},e_{2}) =
<Ae_{1},e_{2}>
= 0 (by the lemma) and
e_{2}!=0, we have that either
Ae_{1} is parallel to
e_{1} or
Ae_{1}=0. If
Ae_{1} is parallel to
e_{1}, then
Ae_{1} = ae_{1},
and since
<Ae_{1},e_{1}>
= l_{1} =
<ae_{1},e_{1}>
= a, we conclude that Ae_{1} =
l_{1}e_{1};
if Ae_{1} = 0, then
l_{1} =
<Ae_{1},e_{1}>
= 0, and Ae_{1} = 0 =
l_{1}e_{1}.
Thus, we have in any case that Ae_{1} =
l_{1}e_{1}. B(e_{1},e_{2}) = <Ae_{2},e_{1}> = 0 and that <Ae_{2},e_{2}> = l_{2}, we can prove in the same way that Ae_{2} = l_{2}e_{2}. 

