For a homogeneous polynomial with real coefficients:$f(x,y,z)=ax^2+by^2+cz^2+dxy+exz+fyz$, suppose we know $f$ factors into products of linear forms$f=(p_1x+p_2y+p_3z)(q_1x+q_2y+q_3z)$, are there anyway to determinate whether the coefficients of the linear forms are real or complex?
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$\begingroup$ Test in terms of what criteria? If you have the 4 factors, can't you just check to see whether the ratios of the coefficients in each factor are real? If so, then each factor is a complex multiple of a real linear form. This is necessary and sufficient. Since this answer is so easy, I suspect that you must have some other restrictions in mind on what you can test. $\endgroup$– Robert BryantCommented Mar 2, 2012 at 16:45
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$\begingroup$ I edited my question to make it more clear. The situation is I know it factors but don't know what the factorization is. All I want to know is whether the factorization is in real or complex. $\endgroup$– QiuryaqCommented Mar 2, 2012 at 20:16
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The coefficients are real if and only if (depending on whether or not $p_1x+p_2y+p_3z$ and $q_1x+q_2y+q_3z$ are proportional or not) such a quadratic form can be, by a coordinate change, brought to $\pm x^2$ or to $xy$ (this is an obvious re-phrasing, take the factor(s) for new coordinate(s)). The second one can be brought to $x^2-y^2$ also. In other words, the signature of your forms(number of pluses/minuses when brought to sum of squares) should be (1,0) or (0,1) or (1,1) in order for a factorisation over reals to exist.
The ways to bring a quadratic form to sum of squares over reals are explained in most linear algebra textbooks. In fact, to compute the signature, you don't even need that. Form the matrix $$\begin{pmatrix}a&\frac12d&\frac12e\\ \frac12d&b&\frac12f\\ \frac12e&\frac12f&c\end{pmatrix}.$$ You want the characteristic polynomial to have two zero roots, or one zero root, one positive root, and one negative root.
answered Mar 2, 2012 at 21:02