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)
