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


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 $p_1(t)= \frac{1}{4} - \frac{1}{4}\exp(-4\lambda t)$.


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?

  • 1
    $\begingroup$ 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? $\endgroup$
    – JimmyK4542
    Nov 10, 2019 at 5:59
  • $\begingroup$ @JimmyK4542 Nah, it was a typo. Edited to clarify, thanks. $\endgroup$ Nov 10, 2019 at 9:24

1 Answer 1


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


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


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).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .