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Let $A_{(i_1,i_2),(j_1,j_2)}$ be a double-index matrix for $i_1,i_2,j_1,j_2\ge 0,$ i.e., $i_1,i_2,j_1,j_2\in\mathbf{N}$, with $$ % The '&' signs are alignment markers, the '\\' is a line break. \begin{align} \forall i_1,\,i_2\ge 0 ( \; &A_{(i_1,i_2),(i_1+1,i_2)} =\lambda\;, \\ &A_{(i_1,i_2),(i_1,i_2)} =-(\lambda+i_1\mu_1+i_2\mu_2) \;,\\ &A_{(i_1,i_2),(i_1-1,i_2+1)} =i_1\mu_1 \;, \\ &A_{(i_1,i_2),(i_1,i_2-1)} =i_2\mu_2 \;) \end{align} $$ where $\lambda,\mu_1,\mu_2 \in (0, \infty)$ and $\mu_1\neq\mu_2.$ And all other entries of $A$ equal to 0,
I am trying to find the solution of $X\cdot A=0$ with $\sum_{(i_1,i_2)}X_{(i_1,i_2)}=1.$ Or the value of $\sum_{i_1+i_2=k}X_{(i_1,i_2)}$ of the solution for all natural numbers $k$ is enough. $X\cdot A=0$ I mean the product of matrices, i.e., $(X\cdot A)_{(j_1,j_2)}=\sum_{(i_1,i_2)}X_{(i_1,i_2)}A_{(i_1,i_2),(j_1,j_2)}=0,\forall(j_1,j_2),$ where $X$ can be viewed as an infinite row vector.

I tried to find some iterative relation between the augments, but I don't find an easy one.

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    $\begingroup$ Can you explicitly state what the base-index set is? I assume $\mathbb{N}$? What is $\lambda$? Probably, $\lambda \in \mathbb{K} \in \{\mathbb{R},\,\mathbb{C}\}$? It might seem obvious, but it's better to be clear... since this isn't tagged more specifically, you might also want to state how $X\cdot A$ is defined... $\endgroup$ – polynomial_donut Apr 19 '17 at 18:52
  • $\begingroup$ @polynomial_donut Thanks for the comment. I revised the problem. $\endgroup$ – Connor Apr 19 '17 at 19:05
  • $\begingroup$ If it's helpful, I would be delighted to receive an upvote on the comment ; ) It's good that you added this information, but it would be even better, if you could format everything to look tidier (it's still a bit messy to read). I'll give you a sample edit. $\endgroup$ – polynomial_donut Apr 19 '17 at 19:19

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