Finding the rank and signature of a hermitian form I am trying to find the rank and signature of the Hermitian form for $V=C^3$
$$\psi(x,x)=|x_1+ix_2|^2+|x_2+ix_3|^2+|x_3+ix_1|^2-|x_1+x_2+x_3|^2$$
I've tried considering the subspaces for which this is positive or negative but I'm not really getting anywhere. 
 A: There is a much easier way to do this. If you know the Polarization Identity (see the Symemtric Bilinear Forms section of https://en.wikipedia.org/wiki/Polarization_identity), you can use this to find $\psi(x,y)$. Note that the matrix is symmetric. What can you say about symmetric matrices over a complex vector space? 
Hint:

 What are they congruent to?

A: The linear forms $$\begin{cases}
l_1(x) &=x_1+i x_2\\
l_2(x) &=x_2+i x_3\\
l_3(x) &=x_3+i x_1\\
\end{cases}$$ are linearly independent and you have $$l_1(x)+l_2(x)+l_3(x)=(1+i)(x_1+x_2+x_3)$$ Hence $$\begin{aligned}
\psi(x) &= \vert l_1(x) \vert^2 + \vert l_2(x) \vert^2 + \vert l_3(x) \vert^2 - \frac{1}{4}\vert l_1(x) + l_2(x) + l_3(x) \vert^2\\
&=\frac{3}{4}(\vert l_1(x) \vert^2 + \vert l_2(x) \vert^2 + \vert l_3(x) \vert^2)-\frac{1}{2}(\Re(l_1 \bar l_2)+\Re(l_2 \bar l_3)+\Re(l_3 \bar l_1))
\end{aligned}$$ using Gauss reduction algorithm (sorry I haven't find the equivalent English Wikipedia page) you get
$$\psi(x)=\frac{3}{4} \vert l_1 -\frac{1}{3} l_2 -\frac{1}{3} l_2 \vert^2
+ \frac{2}{3} \vert l_2- \frac{3}{8} l_3 \vert^2 + \frac{2}{3} \frac{55}{64} \vert l_3 \vert^2$$ (except my potential miscalculations...)
Which means that the rank is equal to $3$ and the signature to $(3,0)$.
