Matrix exponential - general form of the solution

Motivation

This question is related to continuous-time Markov chains and models of DNA evolution. The question is asked in complete generality, however.

Background

Let $$\boldsymbol Q$$ be a $$4 \times 4$$ such that

$$\boldsymbol Q = \begin{bmatrix} -( ap_2 + bp_3 + cp_4) & ap_2 & bp_3 & cp_4\\ ap_1 & -(ap_1 + dp_3 + ep_4 ) & dp_3& ep_4 \\ bp_1 & dp_2 & -( bp_1 + dp_2 + p_4 )& p_4 \\ cp_1 & ep_2 & p_3 & -(cp_1 + ep_2 + p_3) \\ \end{bmatrix}$$ with all off-diagonal entries are positive, with rates $$a \ldots e > 0$$ and equilibrium probabilities $$0 < p_i < 1, \, i = 1, 2, 3, 4$$ and $$\sum_i p_i = 1$$. For $$t > 0$$, we can then construct the stochastic matrix $$\boldsymbol P(t) := \exp(t\boldsymbol Q)$$.

Example: as a special case, consider $$p_i = 1/4$$ for all $$i$$ and $$a = b = \ldots = e = \lambda$$. Then $$\boldsymbol P(t)$$ is of the form

$$\boldsymbol P(t) = \begin{bmatrix} p_0(t) & p_1(t) & p_1(t) & p_1(t)\\ p_1(t) & p_0(t) & p_1(t)& p_1(t) \\ p_1(t) & p_1(t) & p_0(t)& p_1(t) \\ p_1(t) & p_1(t) &p_1(t) & p_0(t) \\ \end{bmatrix}$$ whence $$p_0(t) = \frac{1}{4} + \frac{3}{4}\exp(-4\lambda t)$$ and $$p_1(t)= \frac{1}{4} - \frac{1}{4}\exp(-4\lambda t)$$.

Question

I am aware it is not possible to obtain $$\boldsymbol P(t)$$ in closed-form for the general formulation. For computation, diagonalisation is the way to go.

My question is: is it possible to at least know the general functional form the entries in $$\boldsymbol P(t)$$ will take?

For instance, in the example above the general form of the solution(s) is $$p(t) = \alpha + \beta \exp(-\gamma t)$$. For more general models the pattern seems to sort of carry on, with solutions looking like $$p(t) = \sum_{j = 1}^K w_j\exp(-a_j t)$$. Are there any tools/(simple) facts I can use to prove this, should it be true?

• As currently written, the matrix $Q$ doesn't depend on $f$. Did you mean to include $f$ in the $(3,4)$, $(4,3)$, and $(4,4)$ entries? Nov 10, 2019 at 5:59
• @JimmyK4542 Nah, it was a typo. Edited to clarify, thanks. Nov 10, 2019 at 9:24

Indeed, the entries in the exponential $$P(t)=e^{tQ}$$ will be of the form $$\sum w_j\exp(a_jt)$$ (i.e. linear combinations of exponentials), so long as the matrix $$Q$$ is diagonalizable. In this case, there exists an invertible matrix $$B\in M_{4\times 4}(\mathbb R)$$ such that

$$B^{-1}QB=D=\mathrm{Diag}\{\lambda_1,\lambda_2,\lambda_3,\lambda_4\}$$

and therefore

$$(tQ)=B(tD)B^{-1}=B\cdot \mathrm{Diag}\{t\lambda_1,t\lambda_2,t\lambda_3,t\lambda_4\}\cdot B^{-1}.$$

As you may know, the exponential can then be calculated by the power series:

$$e^{tQ}=\sum_{k=0}^\infty \dfrac{t^kQ^k}{k!} =B\left( \sum_{k=0}^\infty \dfrac{t^kD^k}{k!} \right)B^{-1}=B\cdot \mathrm{Diag}\{e^{t\lambda_1},e^{t\lambda_2},e^{t\lambda_3},e^{t\lambda_4}\}\cdot B^{-1}.$$

Therefore, if $$Q$$ is diagonalizable, the entries of $$e^{tQ}=(a_{ij}(t))$$ are linear combinations of the form

$$a_{ij}(t)=w_{ij,1}\,e^{t\lambda_1}+w_{ij,2}\,e^{t\lambda_2}+w_{ij,3}\,e^{t\lambda_3}+w_{ij,4}\,e^{t\lambda_4},$$

for some $$w_{ij,k}$$ ($$k=1,2,3,4$$) which are determined by $$B$$ and $$B^{-1}$$. If one of the eigenvalues $$\lambda_i$$ is $$0$$ you might get a constant term like in your example.

However, real matrices are not always diagonalizable -- in the language of topology, they're not a dense set in $$M_{n\times n}(\mathbb R)$$, which means that in the set of matrices $$Q$$ given by your definition (with $$9$$ parameters $$p_1,...,p_4$$ and $$a,...,e$$) you're very likely to stumble upon a non-diagonalizable matrix, even if it's an stochastic matrix.

Luckily, in the case where $$Q$$ is not diagonalizable we can take the real Jordan form, which always exists. That is, we can always find an invertible matrix $$B$$ such that $$B^{-1}QB=J$$, where $$J$$ is a real, block diagonal matrix of the form

$$J=\begin{pmatrix}J_1\end{pmatrix}, \begin{pmatrix}J_1 & 0 \\ 0 & J_2\end{pmatrix}, \begin{pmatrix}J_1 & 0 & 0 \\ 0 & J_2 & 0 \\ 0 & 0 & J_3\end{pmatrix} \text{or} \begin{pmatrix}J_1 & 0 & 0 & 0 \\ 0 & J_2 & 0 & 0 \\ 0 & 0 & J_3 & 0 \\ 0 & 0 & 0 & J_4\end{pmatrix}$$

where each $$J_i$$ is a square matrix of size $$1\times 1, ..., 4\times 4$$. (If there are four $$1\times 1$$ blocks then $$J$$ is a diagonal matrix and $$Q$$ is diagonalizable.)

Furthermore, the exponential of a Jordan form is very well known, so calculating $$e^{tQ}=Be^{tJ}B^{-1}$$ is relatively easy. Since $$J$$, and therefore $$tJ$$ is block diagonal, the exponential $$e^{tJ}$$ is also block diagonal, where the blocks are the corresponding exponentials of the blocks in $$tJ$$.

Since listing all the blocks and the respective exponentials that could possibly appear would lengthen this already long answer considerably (and is a topic already covered in countless sources), I'll leave you with the following result from Lawrence Perko's Differential equations and dynamical systems (2006, p.42):

Corollary. Each coordinate in the solution $$x(t)$$ of the initial value problem $$x'=Ax$$ is a linear combination of functions of the form $$t^ke^{at}\cos bt ~~~\text{or}~~~ t^ke^{at}\sin bt$$ where $$\lambda=a+ib$$ is an eigenvalue of the matrix $$A$$ and $$0\leq k \leq n-1$$.

Since here $$x(t)=e^{At}x_0$$ for some initial condition $$x_0\in \mathbb R^n$$, this applies directly to your problem: we can conclude that the entries of the exponential $$e^{Qt}$$, where $$Q$$ is defined as in your question, are linear combinations of terms of the form

$$e^{a_kt}\cos b_kt, e^{a_kt}\sin b_kt, te^{a_kt}\cos b_kt, ..., t^3e^{a_kt}\sin b_kt$$

where $$\lambda_k= a_k+ib_k$$ is one of the eigenvalues of $$Q$$ ($$k=1,2,3,4$$ with some $$\lambda_k$$ possibly equal, i.e. of multiplicity $$\geq 2$$). Moreover, if $$Q$$ is an stochastic matrix, then each of its eigenvalues has a norm lesser than or equal to $$1$$, and $$1$$ is always an eigenvalue of $$Q$$. I believe this is as far as you can get without imposing excessive restrictions on the parameters $$p_1,...,p_4$$ and $$a,...,e$$.

For further reading on the Jordan form and matrix exponentials, section 1.8 of Perko's book is a good source, but it's also covered in some linear algebra and differential equations textbooks (particularly those that treat linear systems like Perko does).