Given $Q(x,y,z;\alpha)=x^2+z^2+2\alpha xy+2xz$, i have to study the sign of quadratic form. Obviously i can use eigenvalues or studying the sign of minors, but in this case i have a hard time to understand how to act. In fact, for $A=\begin{bmatrix} 1 & \alpha & 1\\ \alpha & 0&0\\ 1& 0&1 \end{bmatrix}$ I have:
$A_1=1$,
$A_2=\det \begin{bmatrix} 1&\alpha \\ \alpha & 0 \end{bmatrix}=-\alpha^2>0\Rightarrow \alpha^2<0$,
$A_3=\det \begin{bmatrix} 1& \alpha &1\\ \alpha &0 &0\\ 1 &0 &1 \end{bmatrix}=-\alpha^2>0\Rightarrow \alpha^2<0$
So, $\left\{\begin{matrix} 1>0 \\ \alpha^2<0 \\ \alpha^2<0 \end{matrix}\right.=\left\{\begin{matrix} 1>0 \\ \alpha^2<0 \end{matrix}\right.$. Now, since for $\alpha^2<0$ no solutions exist, how can I conclude? Is the matrix positive-definite or what?
Following a remark in the comments, here's an efficient way to calculate the sign using the characteristic polynomial---recall that the signature is essentially a count of the positive, zero negative eigenvalues of the representative matrix---but without actually computing its roots.
The characteristic polynomial of the matrix representation $A$ of the quadratic form is $$\det \left(t I_3 - \pmatrix{1&\alpha&1\\ \alpha&0&0\\1&0&1}\right) = t^3 - 2 t^2 - \alpha^2 t + \alpha^2 .$$ For $\alpha \neq 0$, we have $\alpha^2 > 0$, in which case the number of sign changes of the coefficients is $2$, and thus by Descartes' Rule of Signs (and the fact that all of the eigenvalues of a real, symmetric matrix are real) the polynomial has (1) two positive roots, and, (2) since $0$ is not a root, one negative root. Thus, the signature is $(2, 1)$ (Lorentzian). NB we avoided computing explicitly the roots of the characteristic polynomial, which would have been difficult.
On the other hand, if $\alpha = 0$, the constant term of the characteristic polynomial is zero, so in that case the quadratic form is degenerate.