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3.2 Eigenvalue equations

We have all seen simple matrix eigenvalue problems; this is now generalised to linear operators, and we shall first of all consider eigenvalue equations of the form

La = λa.

Theorem 3.1. For an Hermitian operator L,

  1. the eigenvalues are real and
  2. eigenvectors corresponding to different eigenvalues are orthogonal.

Proof. Let’s consider the first property first. Calculate

\eqalignno{ (a,La) & = λ(a,a), & & \cr ({L}^{†}a,a) & = {λ}^{∗}(a,a), & & }

but Hermiticity says the left-hand sides are equal. Subtract both sides of the equations and find

0 = (λ − {λ}^{∗})(a,a).

Positivity of (a,a), ((a,a) ≥ 0 and is only 0 if a = 0) allows us to conclude that λ = {λ}^{∗}.

For the second property we consider two cases. First assume there is a second solution to the eigenvalue problem of the form Lb = μb, with λ\mathrel{≠}μ, λ,μ ∈ ℝ. Then, using Hermiticity we can show that we have two expressions (obtained by having L act on a or {L}^{†} = L on b) for

(b,La) = λ(b,a) = μ(b,a).

Taking the difference between the two right-hand sides, we find (λ − μ)(a,b) = 0, and since μ\mathrel{≠}λ, (a,b) = 0.

This does not apply to the case when we have two different eigenvalues for the same eigenvalue (degeneracy). There is no rule precluding (a,b) to be zero, just no requirement for it to be so. In that case we can construct from the subspace of degenerate eigenvalues a set of vectors that are orthogonal, using the procedures set out in the previous chapter, since any linear combination of the degenerate eigenvectors still correspond to the same eigenvalue. □

Example 3.4: 

Find the eigenvalues and eigenvectors of the matrix (“diagonalise the matrix”)

M = \left (\array{ 4&1&1\cr 1&4 &1 \cr 1&1&4} \right )

Solution: 

The eigenvalues can be found from

Me = λe.

This equation only has interesting (non-zero) solutions if the determinant of coefficients is zero,

\eqalignno{ 0 & =\mathop{ det}(M − λI) & & \cr & = (4 − λ)({(4 − λ)}^{2} − 1) − 1((4 − λ) − 1) + 1(1 − (4 − λ)) & & \cr & = (4 − λ)({(4 − λ)}^{2} − 3) + 2 & & \cr & = −{λ}^{3} + 12λ − 45λ + 54\quad . & & }

A little guesswork shows that this can be factorized as

−{(λ − 3)}^{2}(λ − 6) = 0.

The unique eigenvalue 6 has an eigenvector satisfying

\left (\array{ −2& 1 & 1\cr 1 &−2 & 1 \cr 1 & 1 &−2} \right )e = 0,

which has as normalised solution {e}_{3} = \left (\array{ 1\cr 1 \cr 1} \right )∕\sqrt{3}. The degenerate eigenspace for λ = 3 has eigenvectors satisfying

\left (\array{ 1&1&1\cr 1&1 &1 \cr 1&1&1} \right )e = 0,

which describes a plane through the origin, orthogonal to (1,1,1). We can find non-orthogonal eigenvectors, e.g. (1,1,−2) and (1,0,−1), but we can use the Gramm-Schmidt procedure to find orthonormal eigenvectors of the form {e}_{1} = (1,1,−2)∕\sqrt{6} and {e}_{2} = (1,−1,0)∕\sqrt{2}. The general eigenvector for eigenvalue 3 is then a{e}_{1} + b{e}_{2}.

This example shows the reality of the eigenfunctions, the orthogonality of the eigenvectors, etc.

Weight functions. For function spaces, one often meets the generalised eigenvalue equation

Ly(x) = λρ(x)y(x),

where L is a differential operator, ρ(x) is a real and positive “weight function”.

Theorem 3.2. For an operator L, Hermitian with respect to the ordinary inner product (u,v) =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits u{(x)}^{∗}v(x)\kern 1.70406pt dx, the eigenvalues are real and eigenvectors u(x), v(x) corresponding to different eigenvalues are “orthogonal with a weight function ρ(x)”, i.e.

{(u,v)}_{ρ} =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dx\kern 1.70406pt ρ(x){u}^{∗}(x)v(x) = 0.
(3.5)

3.2.1 Problems

1.

Show that the definition (3.5) satisfies the conditions (2.2–2.4).

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