Diagonalising an intensity matrix of a reversible Markov chain

Recently I needed to build a Monte Carlo simulator of a continuous-time Markov chain. This is a pretty straightforward exercise; the only catch was that I wanted it to perform well, so I had to use a fast algorithm for matrix exponentiation.

One way to calculate a matrix exponent efficiently is to calculate the eigendecomposition of the matrix: A=Q\Lambda Q^{-1}, where \Lambda is a diagonal matrix. Then \exp(A) = Q\exp(\Lambda)Q^{-1} (details here). The process of finding such eigendecomposition is also known as diagonalisation.
Not every matrix can be diagonalised, though. One particularly nice case is when A is Hermitian: then there exists a decomposition A=Q\Lambda Q^{-1} where \Lambda is real, and Q is unitary. Numerical diagonalisation of a Hermitian matrix is implemented in many software packages.
Even if the matrix is question is not Hermitian, not all is lost: there is a big class of non-Hermitian matrices that can be easily diagonalised. This is what this post is about.

If we can find matrix M such that

then there exists a decomposition A=Q\Lambda Q^{-1} where \Lambda is real. Note: in this post, I discuss a general case where elements of A are complex numbers. All of that is also applicable, as a special case, to real-valued matrices. For them, “Hermitian” means “symmetrical”, and “unitary” means “orthogonal”. The elements of an intensity matrix of a Markov chain are, of course, real.

To find the eigendecomposition of a non-Hermitian matrix A, we start with the eigendecomposition of the matrix V=L^*AL^{-*}. We use abbreviation L^{-*}=\left(L^*\right)^{-1}=\left(L^{-1}\right)^*. Matrix V is Hermitian, which is easy to see by checking that V^*=V using the fact that LL^*A=A^*LL^*.

Suppose we have found the eigendecomposition

V=P\Lambda P^{-1}

(using our favourite matrix manipulation package). Then, by definition of V, we have

L^*AL^{-*}=P\Lambda P^{-1}.

Multiplying both sides of this equation by L^{-*} on the left and L^* on the right, we get

A=L^{-*}P\Lambda P^{-1}L^*.

Let Q=L^{-*}P, then Q^{-1}=P^{-1}L^*. Thus,

A=Q\Lambda Q^{-1},

which is the eigendecomposition we were looking for.

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