I am trying to find a stationary distribution of some matrix in probability, so I am looking at a solution for $\pi$ in the equation $\pi A = 0$ where $\pi$ is a row vector and $A$ is an infinite matrix as follows. $\pi$ represents a probability distribution over the infinite state space and hence if a non-zero solution to $\pi A$ exists, then I am looking forward to a solution such that the sum of the elements of $\pi = 1$.
$$A = \begin{bmatrix} -6 & 6 & 0 \cdots & \, & \, & \, & 0\\ 1 & -7 & 6 & 0 &\cdots & \, & 0 \\ 0 & 2 & -8 & 6 & 0 &\cdots & 0 \\ \vdots & \ddots & \\ 0 & \dots & \dots & \, && n-1 & -n-5 \\ \vdots & \cdots & & & & & \vdots \end{bmatrix}$$
So I have $-k-5$ as the diagonal element for the $k^\text{th}$ row, $6$ above the diagonal and $k$ as the element below the diagonal element for the $k^\text{th}$ row.
May I know how I can solve for $\pi$? If $A$ were a finite matrix of size $n$, I understand that then $A$ is invertible and $\pi = 0$ is the unique solution.
$\pi$ is parallel to the row $R=[a_0=1,a_1,a_2,\cdots]$ with
$a_{k}=\dfrac{6^k}{k!}$.
Then $\pi=e^{-6}R$.
EDIT. Answer to the OP.
Let $B_n$ be the matrix composed with the first $n$ rows of $A^T$ and the first $n+1$ columns of $A^T$.
Then $\ker(B_n)$ has dimension $1$ and admits the basis $[a_0,\cdots,a_n]^T$.