I am considering a Markov chain with $n$ states with a particularly nice structure. The transition matrix is as follows:
\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}
I already deduced that the eigenvalues of the matrix are $\lambda\_i=(-1)^{i-1}\frac{1}{i}$ for $1\leq i\leq n$. However, I feel that there should also exist closed-form expressions for the eigenvectors of this matrix. Any help in proving or disproving this feeling is appreciated. Thanks.



Eigenvectors of a particular transition matrix
*EDIT2*. **(5th Nov, 2014).** Based on Darij's comments, am editing the answer to improve its clarity. The answer below shows how to get both **eigenvalues** and **eigenvectors** (my original answer was just for eigenvectors).


Eigenvalues
-----------


The key idea is to consider $P^{-1}$. Some (Markovian) guessing leads us to the following subdiagonal matrix:
\begin{equation\*}
 L\_n :=
 \begin{pmatrix}
 0 &&&\\
 -1 & 0 &&\\
 0& -2 & 0 &&\\
 &&\ddots&\ddots\\
 \dots& && 1-n & 0
 \end{pmatrix}.
\end{equation\*}
Compute now the matrix exponential $V=\exp(L\_n)$, which is a lower-triangular matrix with the lower-triangle given by the binomial coefficients:
\begin{equation\*}
 V\_{ij} = (-1)^{(i-j)}\binom{i-1}{j-1}.
\end{equation\*}
A quick experiment shows that (something that can be verified by a moderately tedious induction):
\begin{equation\*}
 VP^{-1}V^{-1} = V\begin{pmatrix}
 &&&1-n & n\\
 &&2-n & n-1&\\
 &\dots&\dots&\\
 -1& 2 &&&\\
 1&&&&
 \end{pmatrix}V^{-1} = \begin{pmatrix}
 1 & \* & \cdots &\*\\
 & -2 & \* &\* \\
 &&\ddots&\*\\
 &&&(-1)^{n+1}n
 \end{pmatrix},
\end{equation\*}
which is upper triangular, so we can immediately read off the eigenvalues of $P^{-1}$ (and hence $P$).


Eigenvectors
------------


Although $V$ does not diagonalize $P^{-1}$, we observe that it turns $P^{-2}$ into the bidiagonal matrix:
\begin{equation\*}
 B := VP^{-2}V^{-1} =
 \begin{pmatrix}
 1 & -2(n-1)\\
 & 4 & -3(n-2)\\
 &&9 & -4(n-3)&\\
 &&&\ddots & \ddots\\
 \\
 &&&&(n-1)^2 & -n(1)\\
 &&&&& n^2
 \end{pmatrix}.
\end{equation\*}


If we succeed in diagonalizing $B$ in closed-form, then we are done. Suppose, $SBS^{-1}=\Lambda$, then we obtain ($\Lambda$ can be read off of the diagonal):
\begin{equation\*}
 S^{-1}\Lambda S = VP^{-2}V^{-1} \implies SVP^{-2}V^{-1}S^{-1} = \Lambda = \text{Diag}([i^2]\_{i=1}^n),
\end{equation\*}
which shows that $SV$ diagonalizes $P^{-2}$, completing the answer.


Technical Lemma
---------------


It remains to find $S$. This reduces to the system of equations:
\begin{equation\*}
 SB = \Lambda S\quad\leftrightarrow\quad B^TS^T=S^T\Lambda.
\end{equation\*}
I prefer to solve the latter formulation. Let us write $U := S^T$. Consider the $j$th eigenvalue $\lambda\_j=j^2$; denote the corresponding column of $U$ by $u$. Then, the equation to solve is
\begin{equation\*}
 \begin{split}
 B^Tu = j^2u,\qquad \implies & u\_1 = j^2u\_1\\
 -(k+1)(n-k)u\_k + (k+1)^2u\_{k+1} = j^2u\_{k+1}.
\end{split}
\end{equation\*}
Examining these equations, we see that $U$ is a lower-triangular matrix, with $1$s on its diagonal (which comes from the equation with $k+1=j$). The subsequent values are nonzero. Solving the recurrences for a few different choices of $j$, we can guess the general solution with some help from Mathematica:
\begin{equation\*}
 u\_k = \frac{(j+1)\_{k-j}(n-k+1)\_{k-j}}{(2j+1)\_{k-j}(k-j)!},\qquad k \ge j.
\end{equation\*}


