Sign of quadratic form with parameter 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?
 A: 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.
A: Here, it is an elementary solution.
After completing the square
$$
Q(x,y,z)=(x+\alpha y +z)^2 -\alpha^2(y+\frac{1}{\alpha}z)^2+z^2
$$
assuming $\alpha\neq 0$.
It means that $Q$ indefinite quadratic form if $\alpha\neq 0$, otherwise it is positive semi definite only.
