Self Adjoint Linear Maps
If V is a vector space of 2 dimensions and an inner product given by
< , > we sat that a linear map A : V --> V is
self-adjoint if <Av,w>=<v,Aw> for all v and w in V.
Note that if
is an orthonormal basis for V and (aij), i,j=1,2, is the
matrix of A relative to that basis, then
<Aei,ej> = aij =
<Aej,ei> = aji
and (aij) is symmetric.
To each self-adjoint map A we associate a map B : V x V --> R defined
by B(v,w) = <Av,w>. Because of the definition of the inner product
and the properties of matrixes, B is linear in both v and w. Further,
since A is self-adjoint, B(v,w) = B(w,v). Therefore, B is a bilinear
symmetric form in V.
On the other hand, if B is a bilinear symmetric form in V, we can define
a linear map A : V --> V by <Av,w> = B(v,w) and the symmetry of
B implies that A is self-adjoint.
For each symmetric bilinear form B in V, there corresponds a quadratic
form Q in V given by Q(v) = B(v,v) for v in V. This relationship is
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)
Now, using the lemma, we can construct the following proposition:
Given a quadratic form Q in V, there exists an orthonormal basis
of V such that if v in V is given by v = x
e1 + y
Q(v) = l1
where l1 and
l2 are the
maximum and minimum,
respectively, of Q on the unit circle |v| = 1.
We say that a vector v != 0 is an eignevector of a linear map A : V
--> V if Av = lv for some real number
l; l is then called an
eigenvalue of A.
The previous definitions allow us to prove the following theorem:
Let A : V --> V be a self-adjoint linear map. Then there exists an
of V such that A(e1) =
(that is, e1 and
e2 are eigenvectors, and
are eigenvalues of A). In the basis
the matrix of A is clearly diagonal adn the elements
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)