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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?

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  • $\begingroup$ Why do you impose $-\alpha^2 > 0$? The computation of $A_2$ tells you exactly that the upper-left $2 \times 2$ minor defines a quadratic form of signature $(1, 1)$ (or a degenerate, nonzero positive semidefinite quadratic form iff $\alpha = 0$). $\endgroup$ – Travis Jan 8 at 10:39
  • $\begingroup$ @Travis Thanks for your answer. I thought that $>0$ was the default sign from which to start the analysis. But if not, what I have to impose for $-\alpha^2$? $\endgroup$ – Marco Pittella Jan 8 at 10:54
  • $\begingroup$ There's no notion of "default sign", and there's nothing to impose. Rather, $-\alpha^2$ controls the signature of the resulting form. $\endgroup$ – Travis Jan 8 at 11:21
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    $\begingroup$ Note, too, by the way, that when determining signature using eigenvalues, you don't actually need to determine the eigenvalues themselves (which in this case would involve solving a cubic), you only need their signs, and you can extract that information from Descartes' Rule of Signs. If it would be useful, I'm happy to write up a short solution explaining this method. $\endgroup$ – Travis Jan 8 at 11:23
  • $\begingroup$ @Travis Thanks again for your help. $-\alpha^2$ controls the sign of the form, of course. So, how can I conclude in this case? Any value I put on place of $\alpha$ I obtain an indefinite form. Could you show me the steps to follow? Instead, for the method that you mentioned well, ok, thank you so much for your time. $\endgroup$ – Marco Pittella Jan 8 at 11:40
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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.

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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.

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