Canonical form of a Quadratic form. I've been given the following quadratic form to find the canonical form of:
$$
Q(\bf{z})= z_1z_2 + 2z_2z_3 − 3z_3z_4
$$
through the method of forming perfect squares.
The method I've been taugh/show is to look at terms consisting of a specific variable, say $z_1$ and form the perfect square. We then set the canonical basis as the inside of the square (hopefully that makes sense...)
Until we get the quadratic form to look like:
$$
Q(z)=\alpha_1(\eta^1)^2+\alpha_2(\eta^2)^2+\alpha_3(\eta^3)^2+\alpha_4(\eta^4)^2
$$
where the alphas are the canonical coefficients.
Now I am stuck with the above problem as there is no square term. Typically my approach in these problems has been to start with a term that has a square term, and go from there. In this case, whenever I get to the final $\eta$ to find, I am left with two square terms, say $z_3$ and $z_4$
Is there something im missing?

Without spelling out all the computations; your quadratic form is given by the symmetric matrix
$${\bf z}\begin{pmatrix}
0&\tfrac12&0&\hphantom{-}0\\
\tfrac12&0&1&\hphantom{-}0\\
0&1&0&-\tfrac32\\
0&0&-\tfrac32&\hphantom{-}0
\end{pmatrix}{\bf z}^{\top}=0,$$
and you want to find a linear transformation $T$ such that
$$(T{\bf z})\begin{pmatrix}
\alpha_1&0&0&0\\
0&\alpha_2&0&0\\
0&0&\alpha_3&0\\
0&0&0&\alpha_4
\end{pmatrix}(T{\bf z})^{\top}=0,$$
so it suffices to diagonalize the symmetric matrix above.
