I have the following arrangement of hyperplanes: $\mathcal{A}=(x-y)(y+w)(x+z)(z-w)(x+w)(y+z)$. Its a rank-three arrangement but it is in four variables. My professor recommends that I apply a change of coordinates so that the intersection of the six hyperplanes is the $w$ axis and the normal vectors of each lie in the $w=0$ plane.
I am not sure how to do this.
The arrangement can be representing by the following matrix:
\begin{bmatrix} 1 & -1 & 0 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & 1 & 0\\ 0 & 0 & 1 & -1\\ 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0\\ \end{bmatrix}
Any help is appreciated. Thanks!
Start by finding the intersection of the hyperplanes. The rref of your matrix is $$\begin{bmatrix}1&0&0&1\\0&1&0&1\\0&0&1&-1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$ so their intersection is the line spanned by $(1,1,-1,-1)^T$.
Try a reflection that maps this vector onto the $w$-axis. A reflection is handier than a rotation here because you don’t have to make any choices for the remaining degrees of freedom that a rotation would have and a reflection is its own inverse. A convenient reflector is the angle bisector of the intersection line and the $w$-axis, i.e., the hyperplane with normal $\mathbf n = \frac12(1,1,-1,-1)^T-(0,0,0,1)^T = \frac12(1,1,-1,-3)^T$. The corresponding reflection matrix is $$R = I_4-2{\mathbf n\mathbf n^T\over\mathbf n^T\mathbf n} = \frac16\begin{bmatrix}5&-1&1&3\\-1&5&1&3\\1&1&5&-3\\3&3&-3&-3\end{bmatrix}.$$ The rows of your matrix are the normals of the hyperplanes, so right-multiply by $R$ to transform them, producing $$\begin{bmatrix}1&-1&0&0\\\frac13&\frac43&-\frac13&0\\1&0&1&0\\-\frac13&-\frac13&\frac43&0\\\frac43&\frac13&-\frac13&0\\0&1&1&0\end{bmatrix}.$$ Happily, the last coordinate of all of them is zero (which we would expect since the intersection of the hyperplanes is orthogonal to all of their normals), so there’s nothing more to do. You can of course multiply the second, fourth and fifth rows by 3 to eliminate the fractions without changing the corresponding hyperplanes.