Dirichlet-Forms and Rayleigh-Theorem

- Let be an arbitrary finite set and let be an -dimensional transition matrix, which is irreducible and aperiodic (i.e. quasi-positive) as well as reversible.
- Recall that

**Remarks**-
- Instead of ordering the eigenvalues according to their absolute
values (like above) we will now order them with respect to their
*own*size and denote them by such that - Moreover, for the multiplicative reversible version
of the transition matrix
that
was introduced in Section 2.3.4 we have
- For large ,
- the calculation of the second largest absolute value among the eigenvalues can cause difficulties.
- Therefore, in Section 2.3.7 we will derive bounds for and , whose calculation is very simple.

- These bounds are particularly useful if
- the stationary (limit) distribution is at least in principle known,
- but in spite of this the corresponding Markov chain is started with an instationary initial distribution ; for example it could be started in a predetermined state , i.e. and for .

- Instead of ordering the eigenvalues according to their absolute
values (like above) we will now order them with respect to their

In order to derive an upper bound for , we need a representation formula for ,

- that is usually called the
*Rayleigh-theorem*in literature - and that is expressed based on the so-called
*Dirichlet-form*

of the reversible pair , where denotes the inner product of and with respect to ; see (103).

First of all we will show the following lemma.

**Proof**-
From the definition (103) of the inner product and the
reversibility of the pair
we obtain

We will now prove the *Rayleigh-theorem * that yields a
representation formula for the second largest eigenvalue
of the reversible pair
.

- Let denote the set of all vectors in whose components are not all equal.
- For the eigenvalue of the reversible pair
the following holds

where denotes the variance of the components of with respect to defined in .

**Proof**-
- Lemma 2.8 implies for arbitrary
and
- Thus, the assertion (117) is equivalent to

where . - Let now the left eigenvectors of be chosen such that they are an orthonormal basis of with respect to the inner product , i.e. if and if where .
- First of all, the eigenvectors of the symmetric vectors are chosen such that they are orthonormal with respect to the ordinary euclidian inner product. Then we can define for all (see also Section 2.3.3).
- For every
there is now a uniquely determined
vector
such that
- As
we obtain
and hence

- Thus, the assertion (117) is equivalent to
- On the other hand as
and the eigenvectors
are orthonormal with respect to the inner product
we can conclude that
and if
- Thus for every

- This shows that (118) holds as the last
expression for the quotient
implies

- Thus for every

- Lemma 2.8 implies for arbitrary
and