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Suppose the matrices $A_N$ which in dimension $N$ have non-zero elements given by

$(A_N)_{i,i+1}=i$ and $(A_N)_{i+1,i}=N-i$, for $i=1,...,N-1$

For example,

$A_4=\left[ \begin {array}{cccc} 0&1&0&0\\ 3&0&2&0 \\0&2&0&3\\ 0&0&1&0\end {array} \right] $, $A_5=\left[ \begin {array}{ccccc} 0&1&0&0&0\\ 4&0&2&0&0\\ 0&3&0&3&0\\ 0&0&2&0&4\\ 0&0&0&1&0\end {array} \right] $

Matrix $A_N$ is equal to $N-1$ times a stochastic matrix, so the Perron-Frobenius theorem guarantees that $N-1$ is the largest eigenvalue, with left eigenvector constant.

However, these matrices are very particular. For example, the eigenvalues are precisely the integers from $-(N-1)$ to $(N-1)$. Moreover, the right eigenvectors $v_i$ are such that the generating functions $f_i(x)=\sum_{k=1}^N(v_i)_k x^k$ are given by $f_i(x)=x(1+x)^{N-i}(x-1)^{i-1}$.

I could not get a handle on the left eigenvectors.

My questions are: Has anyone seen these matrices before? Does anyone have a suggestion on how to understand the left eigenvectors?

thedude
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    $P_n=\frac 1{n}A_{n+1}^t$ is a transition for a symmetric random walk on a hypercube ${0,1}^n$ indexed by manhanttan distance from $0^n$. – A.S. Dec 09 '15 at 12:55
  • @A.S. What is $t$ in your answer? Also, do you have a reference? Thanks – thedude Dec 09 '15 at 13:25
  • $t$ is transpose as it's common to use row-stochastic matrices rather than column-stochastic matrices. The connection is easy to see - just think of a string of $n$ bits and flipping a random bit on each step - the state being number of $1$'s. I've encountered this formulation here where an expected time to walk from one corner to another is computed. – A.S. Dec 09 '15 at 13:28
  • @A.S. Following your reference, I found out that this matrix is known as the Kac matrix, and was already studied by Sylvester. Thanks – thedude Dec 09 '15 at 18:00
  • Glad to help. Could you add some references/main results you found? I found this which directly addresses the question of eigenvectors in Theorem 2.1. – A.S. Dec 09 '15 at 18:15
  • I found that one, and reference [18] in there is very good. Beyond what I already mentioned, it is known that the coefficients of the left eigenvectors are given by the corresponding coefficients in the right ones, divided by binomial numbers. Kac himself was intrigued that he could not find a generating polynomial for the left eigenvectors (which was indeed my own problem) – thedude Dec 10 '15 at 00:45

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