A n. & s. condition for $Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = (a_1 x + a_2 y + a_3) (a_4 x + a_5 y + a_6)$ for each $(x, y) \in \mathbb{C}^2$

Question

(Note: n. & s. is short for necessary and sufficient.)

Suppose that $f(x, y) = Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F$, where $A$, $B$, $C$, $D$, $E$, $F$ are six complex numbers. Prove or disprove that there exist six complex numbers $a_1$, $a_2$, $a_3$, $a_4$, $a_5$, $a_6$ such that $$ Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = (a_1 x + a_2 y + a_3) (a_4 x + a_5 y + a_6) $$ for each ordered pair of complex numbers $(x, y) \in \mathbb{C} \times \mathbb{C}$ if and only if $$ \det {\begin{bmatrix} A & B & D \\ B & C & E \\ D & E & F \\ \end{bmatrix}} = 0. $$

The only if part is not hard. Suppose that $$ Ax^2 + 2Bxy + Cy^2 + 2Dx + 2Ey + F = (a_1 x + a_2 y + a_3) (a_4 x + a_5 y + a_6). $$ Comparing both sides, we yield $$ \begin{aligned} & A = a_1 a_4, \quad C = a_2 a_5, \quad F = a_3 a_6, \\ & 2B = a_1 a_5 + a_2 a_4, \\ & 2D = a_1 a_6 + a_3 a_4, \\ & 2E = a_2 a_6 + a_3 a_5. \end{aligned} $$ Hence $$ \begin{aligned} & \det {\begin{bmatrix} A & B & D \\ B & C & E \\ D & E & F \\ \end{bmatrix}} \\ = {} & ACF + BED + DBE - AEE - BBF - DCD \\ = {} & ACF + 2BED - (AE^2 + CD^2 + FB^2) \\ = {} & \frac{1}{4} (4ACF + 2B \cdot 2E \cdot 2D) - \frac{1}{4} (A (2E)^2 + C (2D)^2 + F(2B)^2) \\ = {} & \frac{1}{4} (4 a_1 a_2 a_3 a_4 a_5 a_6 + a_2 a_3^2 a_5 a_4^2+a_2^2 a_3 a_6 a_4^2+a_1 a_3^2 a_5^2 a_4+a_1 a_2^2 a_6^2 a_4 \\ & \quad \quad +2 a_1 a_2 a_3 a_5 a_6 a_4+a_1^2 a_2 a_5 a_6^2+a_1^2 a_3 a_5^2 a_6) \\ & \quad - \frac{1}{4} (a_1 a_3^2 a_4 a_5^2+2 a_1 a_2 a_3 a_4 a_6 a_5+a_1 a_2^2 a_4 a_6^2 \\ & \quad \quad +a_2 a_3^2 a_5 a_4^2+2 a_1 a_2 a_3 a_5 a_6 a_4+a_1^2 a_2 a_5 a_6^2 \\ & \quad \quad +a_2^2 a_3 a_6 a_4^2+2 a_1 a_2 a_3 a_5 a_6 a_4+a_1^2 a_3 a_5^2 a_6) \\ = {} & 0. \end{aligned} $$

However, the if part is hard; I do not know what I should do.

Answer 1

I finally solved the question.

Here is the if part.

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Suppose that $$ \Delta = \det { \begin{bmatrix} A & B & D \\ B & C & E \\ D & E & F \\ \end{bmatrix} } = ACF + 2BED - (AE^2 + CD^2 + FB^2) = 0. $$

The proof splits into a few cases.

(a) $A = B = C = 0$. Then $$ \begin{aligned} f(x, y) = 2Dx + 2Ey + F = (0x + 0y + 1)(2Dx + 2Ey + F). \end{aligned} $$

(b) $A = C = 0$, but $B \neq 0$. Then $$ \begin{aligned} f(x, y) = {} & 2Bxy + 2Dx + 2Ey + F \\ = {} & \frac{2}{B}(Bx \cdot By + D \cdot Bx + E \cdot By) + F \\ = {} & \frac{2}{B}(Bx \cdot By + D \cdot Bx + E \cdot By + DE - DE) + F \\ = {} & \frac{2}{B}(Bx \cdot By + D \cdot Bx + E \cdot By + DE) - \frac{2}{B} \cdot DE + F \\ = {} & \frac{2}{B}(Bx + E)(By + D) - \frac{2ED - FB}{B} \\ = {} & \left(1x + 0y + \frac{E}{B}\right)(0x + 2By + 2D) - \frac{2BED - FB^2}{B^2}. \end{aligned} $$ Since $\Delta = 0$ and $A = C = 0$, $$ \begin{aligned} \Delta = 0 + 2BED - (0 + 0 + FB^2) = 2BED - FB^2 = 0. \end{aligned} $$ We conclude that $$ \begin{aligned} f(x, y) = \left(1x + 0y + \frac{E}{B}\right)(0x + 2By + 2D). \end{aligned} $$

(c) $A \neq 0$. Then $$ \begin{aligned} & f(x, y) \\ = {} & Ax^2 + 2(By + D)x + (Cy^2 + 2Ey + F) \\ = {} & \frac{(Ax + By + D)^2 - (By + D)^2}{A} + (Cy^2 + 2Ey + F) \\ = {} & \frac{1}{A} (Ax + By + D)^2 + \frac{1}{A}((AC - B^2)y^2 + 2(AE - BD)y) + \frac{AF - D^2}{A}. \end{aligned} $$ The case splits into two subcases.

(c.1) $AC - B^2 = 0$. Hence $C = \frac{B^2}{A}$. Hence $$ \begin{aligned} \Delta = {} & A \cdot \frac{B^2}{A} \cdot F + 2BED - AE^2 - \frac{B^2 D^2}{A} - FB^2 \\ = {} & {-\frac{1}{A}} (A^2 E^2 - 2AE BD + B^2 D^2) \\ = {} & {-\frac{1}{A}} (AE - BD)^2. \end{aligned} $$ Since $\Delta = 0$, we have $AE - BD = 0$. Hence $$ \begin{aligned} f(x, y) = \frac{(Ax + By + D)^2 - (D^2 - AF)}{A}. \end{aligned} $$ There exists a complex number $d$ such that $d^2 = D^2 - AF$. Hence $$ \begin{aligned} f(x, y) = \frac{1}{A} (Ax + By + D + d)(Ax + By + D - d), \end{aligned} $$ which is $$ \begin{aligned} f(x, y) = \left( 1x + \frac{B}{A} y + \frac{D+d}{A} \right) (Ax + By + (D - d)). \end{aligned} $$

(c.2) $AC - B^2 \neq 0$. Then $$ \begin{aligned} & (AC - B^2) y^2 + 2(AE - BD)y \\ = {} & \frac{1}{AC - B^2} ((AC - B^2)y + (AE - BD))^2 - \frac{(AE - BD)^2}{AC - B^2}. \end{aligned} $$ Hence $$ \begin{aligned} f(x, y) = {} & \frac{(Ax + By + D)^2}{A} - \frac{((B^2 - AC)y + (BD - AE))^2}{A(B^2 - AC)} \\ & \quad + \frac{AF - D^2}{A} - \frac{(AE - BD)^2}{A(AC - B^2)}, \end{aligned} $$ which is just $$ \begin{aligned} f(x, y) = \frac{(Ax + By + D)^2}{A} - \frac{((B^2 - AC)y + (BD - AE))^2}{A(B^2 - AC)} + \frac{\Delta}{AC - B^2}. \end{aligned} $$ Since $\Delta = 0$, $$ \begin{aligned} f(x, y) = \frac{(Ax + By + D)^2}{A} - \frac{((B^2 - AC)y + (BD - AE))^2}{A(B^2 - AC)}. \end{aligned} $$ There exists a complex number $e$ such that $e^2 = B^2 - AC$. Put $f = \frac{BD-AE}{e}$. Then $$ \begin{aligned} f(x, y) = {} & \frac{(Ax + By + D)^2}{A} - \frac{(e^2 y + ef)^2}{Ae^2} \\ = {} & \frac{(Ax + By + D)^2}{A} - \frac{(ey + f)^2}{A} \\ = {} & \frac{1}{A}(Ax + By + D + ey + f)(Ax + By + D - ey - f) \\ = {} & \left( 1x + \frac{B+e}{A}y + \frac{D+f}{A} \right) (Ax + (B-e)y + (D-f)). \end{aligned} $$

(d) $C \neq 0$. We rewrite $f(x, y)$ as $$ \begin{aligned} Cy^2 + 2Byx + Ax^2 + 2Ey + 2Dx + F. \end{aligned} $$ Put $$ \begin{aligned} g(x, y) = Cx^2 + 2Bxy + Ay^2 + 2Ex + 2Dy + F, \end{aligned} $$ which means that for any ordered pair of complex numbers $(z, w)$, $$ \begin{aligned} g(z, w) = f(w, z). \end{aligned} $$ A straightforward computation show that $$ \begin{aligned} \Delta' = {} & \det {\begin{bmatrix} C & B & E \\ B & A & D \\ E & D & F \\ \end{bmatrix}} \\ = {} & CAF + 2BDE - (CD^2 + AE^2 + FB^2) \\ = {} & ACF + 2BED - (AE^2 + CD^2 + FB^2) \\ = {} & \Delta. \end{aligned} $$ By (c), there exist six complex numbers $a_1$, $a_2$, $a_3$, $a_4$, $a_5$, $a_6$ such that $$ \begin{aligned} g(x, y) = (a_1 x + a_2 y + a_3) (a_4 x + a_5 y + a_6). \end{aligned} $$ Hence $$ \begin{aligned} & f(x, y) = g(y, x) = (a_2 x + a_1 y + a_3) (a_5 x + a_4 y + a_6). \end{aligned} $$