Decomposing bivariate quadratic form into sum of two squares

Question

I would like to decompose

$$ax_1^2 + bx_2^2 + 2cx_1x_2$$

into two expressions, each involving only one variable. I'm trying to use a transform like $x_1 = x_+ + x_-$ and $x_2 = x_+ - x_-$ to hopefully get something like

$$\left( bx_+ + \frac ac \right)^2 + \left(ax_ - + \frac bc \right)^2$$

or something of that form. Is this possible?

Answer 1

You can write quadratic expressions as $x^T Q x$; in this case $x^T = (x_1, x_2), Q = \begin{pmatrix}a & c \\ c & b\end{pmatrix}$.

Then, if you write $Q$ as $Q = A^TDA$, with $D$ diagonal, you have $x^TQx = x^TA^TDAx = (Ax)^TD(Ax)$: $Ax = (y_1, y_2)^T$ are the new variables you are searching, $D = \begin{pmatrix}d_1 & 0 \\ 0 & d_2\end{pmatrix}$ are the new coefficients; so you can rewrite $ax_1^2 + bx_2^2 + 2cx_1x_2 = d_1 y_1^2 + d_2y_2^2$.

EDIT: another (somewhat equivalent) trick is to "complete the squares":

$$\begin{aligned} ax_1^2 + bx_2^2 + 2cx_1x_2 &= ax_1^2 + 2cx_1x_2 + \frac{c^2}{a}x_2^2 + \left( b - \frac{c^2}{a} \right)x_2^2 \\ &= \left( \sqrt{a}x_1 + \frac{c}{\sqrt{a}}x_2 \right)^2 + \left(b - \frac{c^2}{a} \right)x_2^2 \end{aligned}$$