The system of equations is the following: $$ \Gamma_i^{\ -1} = \sum_{i=1}^nA_{ij}\Gamma_j, $$ where $\Gamma = (\Gamma_i)$ is a vector of size $n$ and $A$ is a matrix of size $n\times n$, with $n \gt 100$.

So there is a paper (Numerical and computational aspects of cosmo-based activity coefficient models, Brazilian Journal of Chemical Engineering vol.36 no.1) showing, that successive substitution is faster than Newton–Raphson if solved as mentioned above.

I was wondering if through some kind of linear algebra change the system of equations becomes easier/faster to solve?

I don't know if it helps, but here is how the matrix $A$ is calculated (hadamard product): $$ A = B \circ D $$ Where $B$ is symmetric, dense with only positive (and negative) entries and $D$ is dense with only positive entries and all rows are the same and their sum is 1.

I just corrected that $B$ is positive symmetric, and since $D$ is also positive, this would make $A$ positive. So: $$ A_{ij}>0 $$

The system of equations is the following: $$ \Gamma_i^{\ -1} = \sum_{i=1}^nA_{ij}\Gamma_j, $$ where $\Gamma = (\Gamma_i)$ is a vector of size $n$ and $A$ is a matrix of size $n\times n$, with $n \gt 100$.

So there is a paper (Numerical and computational aspects of cosmo-based activity coefficient models, Brazilian Journal of Chemical Engineering vol.36 no.1) showing, that successive substitution is faster than Newton–Raphson if solved as mentioned above.

I was wondering if through some kind of linear algebra change the system of equations becomes easier/faster to solve?

I don't know if it helps, but here is how the matrix $A$ is calculated (Hadamard product): $$ A = B \circ D $$ Where $B$ is symmetric, dense with only positive (and negative) entries and $D$ is dense with only positive entries and all rows are the same and their sum is 1.

I just corrected that $B$ is positive symmetric, and since $D$ is also positive, this would make $A$ positive. So: $$ A_{ij}>0 $$

The system of equations is the following: $$ \Gamma_i^{\ -1} = \sum_{i=1}^nA_{ij}\Gamma_j, $$ where $\Gamma = (\Gamma_i)$ is a vector of size $n$ and $A$ is a matrix of size $n\times n$, with $n \gt 100$.

So there is a paper (Numerical and computational aspects of cosmo-based activity coefficient models, Brazilian Journal of Chemical Engineering vol.36 no.1) showing, that successive substitution is faster than Newton–Raphson if solved as mentioned above.

I was wondering if through some kind of linear algebra change the system of equations becomes easier/faster to solve?

I don't know if it helps, but here is how the matrix $A$ is calculated: $$ A = B \circ D $$ Where $B$ is symmetric, dense with only positive (and negative) entries and $D$ is dense with only positive entries and all rows are the same and their sum is 1.

I just corrected that $B$ is positive symmetric, and since $D$ is also positive, this would make $A$ positive. So: $$ A_{ij}>0 $$

The system of equations is the following: $$ \Gamma_i^{\ -1} = \sum_{i=1}^nA_{ij}\Gamma_j, $$ where $\Gamma = (\Gamma_i)$ is a vector of size $n$ and $A$ is a matrix of size $n\times n$, with $n \gt 100$.

So there is a paper (Numerical and computational aspects of cosmo-based activity coefficient models, Brazilian Journal of Chemical Engineering vol.36 no.1) showing, that successive substitution is faster than Newton–Raphson if solved as mentioned above.

I was wondering if through some kind of linear algebra change the system of equations becomes easier/faster to solve?

I don't know if it helps, but here is how the matrix $A$ is calculated (hadamard product): $$ A = B \circ D $$ Where $B$ is symmetric, dense with only positive (and negative) entries and $D$ is dense with only positive entries and all rows are the same and their sum is 1.

I just corrected that $B$ is positive symmetric, and since $D$ is also positive, this would make $A$ positive. So: $$ A_{ij}>0 $$

The system of equations is the following: $$ \Gamma_i^{\ -1} = \sum_{i=1}^nA_{ij}\Gamma_j, $$ where $\Gamma = (\Gamma_i)$ is a vector of size $n$ and $A$ is a matrix of size $n\times n$, with $n \gt 100$.

So there is a paper (Numerical and computational aspects of cosmo-based activity coefficient models, Brazilian Journal of Chemical Engineering vol.36 no.1) showing, that successive substitution is faster than Newton–Raphson if solved as mentioned above.

I was wondering if through some kind of linear algebra change the system of equations becomes easier/faster to solve?

I don't know it it helps, but here is how the matrix $A$ is calculated: $$ A = B \circ D $$ Where $B$ is symmetric, dense with only positive (and negative entries) and $D$ is dense with only positive entries and all rows are the same and their sum is 1.

I just corrected that $B$ is positive symmetric, and since $D$ is also positive, this would make $A$ positive. So: $$ A_{ij}>0 $$

The system of equations is the following: $$ \Gamma_i^{\ -1} = \sum_{i=1}^nA_{ij}\Gamma_j, $$ where $\Gamma = (\Gamma_i)$ is a vector of size $n$ and $A$ is a matrix of size $n\times n$, with $n \gt 100$.

So there is a paper (Numerical and computational aspects of cosmo-based activity coefficient models, Brazilian Journal of Chemical Engineering vol.36 no.1) showing, that successive substitution is faster than Newton–Raphson if solved as mentioned above.

I was wondering if through some kind of linear algebra change the system of equations becomes easier/faster to solve?

I don't know if it helps, but here is how the matrix $A$ is calculated: $$ A = B \circ D $$ Where $B$ is symmetric, dense with only positive (and negative) entries and $D$ is dense with only positive entries and all rows are the same and their sum is 1.

I just corrected that $B$ is positive symmetric, and since $D$ is also positive, this would make $A$ positive. So: $$ A_{ij}>0 $$