Generic rotation to remove Quadratic Cross-product

Question

Show that if $b\neq 0$, then the cross-product term can be eliminated from the quadratic $ax^2 + 2bxy + cy^2$ by rotating the coordinate axes through an angle $\theta$ that satisfies the equation $$ \cot{2\theta}=\frac{a-c}{2b}. $$

I know that $ax^2 + 2bxy + cy^2$ can be rewritten as $$ \begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix}a&b\\b&c\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix}, $$ and I am familiar with how to rotate it using unitary matrices. However, it is very tedious to find the eigenvectorss of $$ \begin{bmatrix}a&b\\b&c\end{bmatrix}. $$ Is there a method for solving this problem that does not require finding the eigenvectors?

Answer 1

Working directly with the rotation substitution $x=cx'-sy',\ y=sx'+cy',$ where $c=\cos \theta,\ s=\sin \theta,$ we need only compute the resulting coefficient of $x'y'$ and set it to zero. From the form $ax^2+2bxy+cy^2$ this coefficient is $$a(-2sc)+2b(c^2-s^2)+c(2sc),$$ and then (assuming $b \neq 0$) we have $$\frac{a-c}{2b}=\frac{c^2-s^2}{2sc}=\frac{\cos 2\theta}{\sin 2\theta}=\cot(2 \theta),$$ the desired formula.

Note: maybe it's a bad choice to use $s,c$ for the sine and cosine of the rotation angle, since $c$ also appears in the form $ax^2+2bxy+cy^2.$ However I think it's clear which is which in the above so I'll leave it as it is.

Answer 2

Let $u$ and $v$ be the rotated coordinates. That is, $$ \begin{pmatrix} u \\ v \end{pmatrix} = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}. $$ You want to be able to substitute a quadratic with no $uv$ term for the quadratic $ax^2 + bxy + cy^2.$ That is, you want there to be fixed coefficients $p$ and $q$ such that $$ pu^2 + qv^2 = ax^2 + bxy + cy^2 \tag{1}$$ for every pair $(x,y).$

It's slightly tedious, but not difficult, to write $u$ and $v$ each in terms of $x,$ $y,$ and $\theta,$ and to plug these substitutions into $pu^2 + qv^2.$ If you collect the terms in $x^2,$ in $xy,$ and in $y^2,$ knowing that the result must equal $ax^2 + bxy + cy^2,$ you can deduce the values of $a$, $b$, and $c$ in terms of these other parameters.

Now see if you can take $a - c$ in those terms and make it look like $(\mathit{something})\times \cos 2\theta.$ If you can then make $2b$ look like $(\mathit{the\ same\ thing})\times \sin 2\theta,$ you will have proved your desired result.

You can avoid carrying around all those factors of $x^2,$ $xy,$ and $y^2$ if you rewrite Equation $(1)$ using the matrices $\begin{pmatrix} p&0 \\ q&0 \end{pmatrix}$ and $\begin{pmatrix} a&b \\ x&d \end{pmatrix}$ and the vectors $\begin{pmatrix} u \\ v \end{pmatrix}$ and $\begin{pmatrix} x \\ y \end{pmatrix},$ and peform the obvious substitution for $\begin{pmatrix} x \\ y \end{pmatrix}$ immediately; you can then derive a matrix equation that doesn't involve $x$ or $y$ at all. That notation may be a bit more agreeable with your initial thoughts on the problem.

Answer 3

if you are diagonalizing the matrix $A,$ then there is no way you can avoid the eigenvectors and eigenvalues. the diagonal entries are the eigenvalues.