Post Closed as "off-topic" by Daniel Lichtblau, MarcoB, Alex Trounev, José Antonio Díaz Navas, bbgodfrey
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$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$. where $-N<=i,j<=N$$-N\le i,j\le N$ Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.

$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$. where $-N<=i,j<=N$ Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.

$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$. where $-N\le i,j\le N$ Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.

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$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$.

Here where $-N<=i,j<=N$ Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.

$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$.

Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.

$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$. where $-N<=i,j<=N$ Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.

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How to write the equation into matrix form

$[(a+k_i)^2+(b+k_j)^2]X_{i,j}-\sum_{m,n}V_{m,n}X_{i-m,j-n}=\mu X_{i,j}$.

Here we can set $ N=10,a =1, b=1$ and $V_{m,n}$ is the matrix element of $V$. Once I write the coefficient matrix of $X$, I can solve the eigenvalue $\mu$.