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Recently, I read a book on cluster algebra and come across a problem that could finally be reduced to a problem of solving a system of nonlinear equations.

The question is:

Give $b_{12},b_{13} , b_{14},b_{23},b_{24},b_{34}\in \mathbb{C}$ satisfying $b_{12}b_{34}+b_{14}b_{23}-b_{13}b_{24}=0$.

Solve the system of nonlinear equations:

$\begin{cases} a_{11}a_{22}-a_{12}a_{21}=b_{12}\\ a_{11}a_{23}-a_{13}a_{21}=b_{13}\\ a_{11}a_{24}-a_{14}a_{21}=b_{14}\\ a_{12}a_{23}-a_{13}a_{22}=b_{23}\\ a_{12}a_{24}-a_{14}a_{22}=b_{24}\\ a_{13}a_{24}-a_{14}a_{23}=b_{34}\\ \end{cases}$

I want to solve the above system of nonlinear equations, but I don't know anything about systems of nonlinear equations.

Is there a solution to the system of equations? If so, what is the relationship between the solutions?

If anyone can give me some reference books or methods, I would be very grateful!

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  • $\begingroup$ Your issue looks connected with grassmannian space G(2,4) (set of 2D planes in the 4D space) as can be seen here, the first constraint being known as Klein surface. But with real (not complex) coordinates. $\endgroup$
    – Jean Marie
    Mar 6, 2023 at 11:20
  • $\begingroup$ Thanks! It is just the problem. I will read it. $\endgroup$
    – fusheng
    Mar 6, 2023 at 12:26
  • $\begingroup$ I had never heard about "cluster algebra". In the Wikipedia article, they mention indeed connections with grassmannian spaces and Plücker coordinates. $\endgroup$
    – Jean Marie
    Mar 6, 2023 at 17:11
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    $\begingroup$ @Jean Marie I am a new graduate student. I've just started studying cluster algebra. In fact , this problem is about the motivation of cluseter algebra. Thank you! $\endgroup$
    – fusheng
    Mar 7, 2023 at 12:31

1 Answer 1

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Suppose that $b_{12}$ is nonzero.

If you assume that $$\left[\begin{array}{cc} a_{11}&a_{12}\\ a_{21}&a_{22}\end{array}\right]$$ is $b_{12}$ times an identity matrix, then you will find that you get linear equations to solve for the other entries.

Thinking of the entries $a_{ij}$ as a $2\times 4$ matrix, you can multiply it on the left by a two-by-two matrix with determinant 1, and this will give you all the other possible solutions.

If $b_{12}$ is zero, you can use a somewhat similar approach, first assuming that $a_{21}$ and $a_{22}$ are zero, and then, again, getting general solutions by multiplying by an invertible two-by-two matrix.

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