linear transformation to change $xy+xz+yz$ into $u^2+v^2+w^2$ Like the title says I have expression : $xy+xz+yz$.
With $x\,\,,\,\,y\,\,,\,\,z\,\,\in \mathbb{C}$.
I found this linear transformation : $(x\,\,,\,\,y\,\,,\,\,z)\mapsto (u\,\,,\,\,v\,\,,\,\,w\,\,) : \mathbb{C} \mapsto \mathbb{C}$ :
$ x=-u-i\frac{1}{2}\sqrt{2}v $ 
$ y=-i\frac{1}{2}\sqrt{2}v-w $
$ z=\frac{1}{2}u +\frac{1}{2}w $ .
which gives me : $xy+xz+yz \mapsto u^2+v^2+w^2$
My question: What is the most general form of linear transformations that map $xy+xz+yz$ to $u^2+v^2+w^2$?
Hope someone can help me with this, I don't see how to proceed. Thanks in advance!
NB: Edited question after receiving comments and answers to reflect the fact that we're working in $\mathbb{C}$.
 A: Here is not the most general solution but a way to determine logically an adequate transformation.
In a first step, a linear transformation with real coefficients can be used on the initial quadratic form (reduction to proper axes), using a kind of Gauss reduction :
$$\left(\frac{x+y+2z}{2}\right)^2-\left(\frac{x-y}2\right)^2-z^2\tag{1}$$
which has the form $$U^2-V^2-W^2\tag{2}$$ exhibiting a signature (+,-,-) (see an identical development in (https://math.stackexchange.com/q/1714385)).
It is important to remark that (2) is as general as (1), therefore can be substituted to it.
In a second step, if we set $U=u$, $V=iv$ and $W=iw$ in (2); we get the desired expression:
$$u^2+v^2+w^2\tag{3}$$
Remark : (1) shows that level sets $xy+yz+zx=k$ are hyperboloids of 2 sheets.
A: The most general linear transformation is $x = au + bv + cw$, for some constants $a,b,c$ (similarly for $y,z$). So, just use these to get $xy+yz+zx$ in terms of $u,v,w$, and compare coefficients with $u^2+v^2+w^2$ to get the appropriate values of the constants.
