The following theorem is about the eigenvectors and eigenvalues of a self adjoint operator. Such
operators are also called Hermitian as in the case of matrices. The proof given generalizes to the
situation of a compact self adjoint operator on a Hilbert space and leads to many very useful
results. It is also a very elementary proof because it does not use the fundamental theorem of
algebra and it contains a way, very important in applications, of finding the eigenvalues. This
proof depends more directly on the methods of analysis than the preceding material. Recall the
following notation.
Definition 12.3.1Let X be an inner product space and let S ⊆ X. Then
⊥
S ≡ {x ∈ X : (x,s) = 0 for all s ∈ S }.
Note that even if S is not a subspace, S^{⊥} is.
Theorem 12.3.2Let A ∈ ℒ
(X,X )
be self adjoint (Hermitian)where X is a finitedimensional inner product space of dimension n. Thus A = A^{∗}. Then there exists anorthonormal basis of eigenvectors,
{vj}
_{j=1}^{n}.
Proof: Consider
(Ax,x)
. This quantity is always a real number because
------ ∗
(Ax,x ) = (x,Ax) = (x,A x) = (Ax, x)
thanks to the assumption that A is self adjoint. Now define
λ1 ≡ inf{(Ax, x) : |x| = 1,x ∈ X1 ≡ X }.
Claim:λ_{1} is finite and there exists v_{1}∈ X with
|v1|
= 1 such that
(Av1,v1)
= λ_{1}.
Proof of claim:Let
{uj}
_{j=1}^{n} be an orthonormal basis for X and for x ∈ X, let (x_{1},
⋅⋅⋅
, x_{n})
be defined as the components of the vector x. Thus,
∑n
x = xjuj.
j=1
Since this is an orthonormal basis, it follows from the axioms of the inner product
that
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
= 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
|(Av1 − λ1v1,w )| = θ (Av1 − λ1v1,w).
Then
( - )
|(Av1 − λ1v1,w )| = Re Av1 − λ1v1,θw = 0.
Since this holds for all w, Av_{1} = λ_{1}v_{1}.
Continuing with the proof of the theorem, let X_{2}≡
{v1}
^{⊥}. This is a closed subspace of X and
A : X_{2}→ X_{2} because for x ∈ X_{2},
(Ax, v1) = (x,Av1) = λ1(x,v1) = 0.
Let
λ2 ≡ inf{(Ax,x) : |x| = 1,x ∈ X2}
As before, there exists v_{2}∈ X_{2} such that Av_{2} = λ_{2}v_{2}, λ_{1}≤ λ_{2}. Now let X_{3}≡
{v1,v2}
^{⊥} and
continue in this way. As long as k < n, it will be the case that
{v1,⋅⋅⋅,vk}
^{⊥}≠
{0}
. This is
because for k < n these vectors cannot be a spanning set and so there exists some
w
∕∈
span
(v1,⋅⋅⋅,vk)
. Then letting z be the closest point to w from span
(v1,⋅⋅⋅,vk)
, it follows
that w −z ∈
{v1,⋅⋅⋅,vk}
^{⊥}. Thus there is an decreasing sequence of eigenvalues
{λk}
_{k=1}^{n} and a
corresponding sequence of eigenvectors, {v_{1},
⋅⋅⋅
, v_{n}} with this being an orthonormal set.
■
Contained in the proof of this theorem is the following important corollary.
Corollary 12.3.3Let A ∈ℒ
(X, X )
be self adjoint where X is a finite dimensional inner productspace. Then all the eigenvalues are real and for λ_{1}≤ λ_{2}≤
⋅⋅⋅
≤ λ_{n}the eigenvalues of A, thereexists an orthonormal set of vectors
{u1,⋅⋅⋅,un}
for which
Auk = λkuk.
Furthermore,
λk ≡ inf{(Ax,x) : |x| = 1,x ∈ Xk}
where
Xk ≡ {u1,⋅⋅⋅,uk− 1}⊥ ,X1 ≡ X.
Corollary 12.3.4Let A ∈ℒ
(X, X )
be self adjoint (Hermitian)where X is a finite dimensionalinner product space. Then the largest eigenvalue of A is given by
max {(Ax,x) : |x| = 1} (12.5)
(12.5)
and the minimum eigenvalue of A is given by
min{(Ax, x) : |x| = 1}. (12.6)
(12.6)
Proof:The proof of this is just like the proof of Theorem 12.3.2. Simply replace inf with sup
and obtain a decreasing list of eigenvalues. This establishes 12.5. The claim 12.6 follows from
Theorem 12.3.2. ■
Another important observation is found in the following corollary.
Corollary 12.3.5Let A ∈ℒ
(X, X)
where A is self adjoint. Then A = ∑_{i}λ_{i}v_{i}⊗v_{i}whereAv_{i} = λ_{i}v_{i}and
{vi}
_{i=1}^{n}is an orthonormal basis.
Proof :If v_{k} is one of the orthonormal basis vectors, Av_{k} = λ_{k}v_{k}. Also,
∑ ∑ ∑
i λivi ⊗ vi(vk) = i λivi(vk,vi) = i λiδikvi = λkvk.
Since the two linear transformations agree on a basis, it follows they must coincide.
■
By Theorem 11.4.5 this says the matrix of A with respect to this basis
{v }
i
_{i=1}^{n} is the
diagonal matrix having the eigenvalues λ_{1},
⋅⋅⋅
,λ_{n} down the main diagonal.
The result of Courant and Fischer which follows resembles Corollary 12.3.3 but is more useful
because it does not depend on a knowledge of the eigenvectors.
Theorem 12.3.6Let A ∈ℒ
(X,X )
be self adjoint where X is a finite dimensional inner productspace. Then for λ_{1}≤ λ_{2}≤
⋅⋅⋅
≤ λ_{n}the eigenvalues of A, there exist orthonormal vectors