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Given $a,b>0$ such that $a\ne b$ but nothing else is given (e.g. $a,b$ are not known to be coprime), I want to understand the structure of the set of solutions to the following system of two linear equations:

$a(x+y)+b(z+w)=0$

$a(z-w)+b(x-y)=0$

How can I approach such a problem? Are additional assumptions on $a,b$ helpful in this case?

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2 Answers 2

up vote 1 down vote accepted

You can look at this as finding the null space of the matrix

$\begin{pmatrix} x+y & z+w \\ z-w & x-y \end{pmatrix}$

To have a non-trivial null space, the determinant of this matrix must be zero, which tells you something about x, y, w and z

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Elaborating a bit on Chris' post, the determinant condition is $x^{2} + w^{2} = y^{2} + z^{2}$. Adding and subtracting the original system of equations gives \begin{eqnarray} a(x + y + z - w) + b(x - y + z + w) = 0 \quad \text{and} \quad a(x + y - z + w) + b(w - x + y + z) = 0. \end{eqnarray} which places more constraints on the possible solutions $(x,y,z,w)$ if $a$ and $b$ are non-zero.

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