Consider the following $n \times n$ matrix with a particularly nice structure: \begin{equation}\mathbf{P}=\begin{pmatrix} 0 & 0& \dots&0 & 0 &1\\ 0 & 0& \dots&0 & \frac{1}{2}&\frac{1}{2}\\ \vdots& & & & & \vdots \\ 0 &\frac{1}{n-1}& \dots&\frac{1}{n-1}&\frac{1}{n-1}&\frac{1}{n-1}\\ \frac{1}{n} &\frac{1}{n} &\dots&\frac{1}{n} &\frac{1}{n} &\frac{1}{n} \end{pmatrix} , \end{equation} and let \begin{equation} \mathbf{\pi}=(1,2, \text{ ..., }n). \end{equation} Is there any hope in fiding closed-form expressions for the elements of the $n$-dimensional vector $\mathbb{w}$ (not in terms of matrix inverses) that solve the following set of linear equations: \begin{equation} \begin{pmatrix} \mathbf{I}-\mathbf{P}\\ \mathbf{\pi} \end{pmatrix}\mathbb{w}=\begin{pmatrix} \mathbf{a}\\ 0 \end{pmatrix}, \end{equation} where $a_i=\frac{1}{i+1}-C$ and $C=\frac{2(n+1-\sum_{j=1}^n\frac{1}{j})}{n(n+1)}$, or is that a hopeless endeavor?

(See also: Eigenvectors of a particular transition matrix)

  • $\begingroup$ Is $\pi$ a diagonal matrix with the said entries? Also $I-P$ is singular, so anyways we cannot write $(I-P)^{-1}$... $\endgroup$ – Suvrit Feb 6 '14 at 15:04
  • $\begingroup$ I guess $\pi$ is a vector, so it's a pseudo-inverse + a normalization condition on $w$. If I am correct, that system is called Poisson equation for a Markov chain. $\endgroup$ – Federico Poloni Feb 6 '14 at 16:06
  • $\begingroup$ But Federico then it's a syntax error, because how does $\pi$ operator on the $w$? or maybe it is $\langle pi, w\rangle$, so $\pi$ is a "row"-vector. Ok, got it. $\endgroup$ – Suvrit Feb 6 '14 at 17:25
  • $\begingroup$ But this system need not always have a solution...or am I missing something? $\endgroup$ – Suvrit Feb 6 '14 at 17:30
  • $\begingroup$ Well, apparently it should, still unclear why for me, but I'm not very strong in linear algebra. And I already deduced that $C=\frac{2}{n(n+1)}\sum_{i=1}^n\frac{i}{i+1}=\frac{2(n+1-\sum_{j=1}^n\frac{1}{j})}{n(n+1)}$. $\endgroup$ – MthQ Feb 6 '14 at 19:36

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.