Linear Algebra, Theory And Applications, 2012a

314 SELF ADJOINT OPERATORS
As before, there exists v 2 ∈ X 2 such that (Av 2 ,v 2 )=λ 2 ,λ 1 ≤ λ 2 .NowletX 3 ≡{v 1 ,v 2 } ⊥
and continue in this way. This leads to an increasing sequence of real numbers, {λ k } n k=1 and
an orthonormal set of vectors, {v 1 , ···, v n }. It only remains to show these are eigenvectors
and that the λ j are eigenvalues.
Consider the first of these vectors. Letting w ∈ X 1 ≡ X, the function of the real variable,
t, given by
f (t) ≡ (A (v 1 + tw) ,v 1 + tw)
|v 1 + tw| 2
= (Av 1,v 1 )+2t Re (Av 1 ,w)+t 2 (Aw, w)
|v 1 | 2 +2t Re (v 1 ,w)+t 2 |w| 2
achieves its minimum when t = 0. Therefore, the derivative of this function evaluated at
t = 0 must equal zero. Using the quotient rule, this implies, since |v 1 | =1that
2Re(Av 1 ,w) |v 1 | 2 − 2Re(v 1 ,w)(Av 1 ,v 1 )
=2(Re(Av 1 ,w) − Re (v 1 ,w) λ 1 )=0.
Thus Re (Av 1 − λ 1 v 1 ,w) = 0 for all w ∈ X. This implies Av 1 = λ 1 v 1 . To see this, let w ∈ X
be arbitrary and let θ be a complex number with |θ| = 1 and
|(Av 1 − λ 1 v 1 ,w)| = θ (Av 1 − λ 1 v 1 ,w) .
Then
|(Av 1 − λ 1 v 1 ,w)| =Re ( Av 1 − λ 1 v 1 , θw ) =0.
Since this holds for all w, Av 1 = λ 1 v 1 .
Now suppose Av k = λ k v k for all k