I've got the following quadratic form
$T(x_1,x_2)=x^TQx$, with $$ x=\begin{pmatrix}x_1\\x_2\end{pmatrix}, Q=\begin{pmatrix}\frac{1}{2}(m_1+m_2)L_1^2 & \frac{1}{2} m_2L_1L_2\cos(\theta_1-\theta_2)\\\frac{1}{2} m_2L_1L_2\cos(\theta_1-\theta_2) & \frac{1}{2}m_2 L_2^2\end{pmatrix}. $$ $Q$ is positive definit.
(This all is deduced from a double pendulum which upper bob has mass $m_1$ and the angle $\theta_1$ with length $L_1$ and lower bob has mass $m_2$, angle $\theta_2$ and length $L_2$.)
Now the task is to make a "Hauptachsentransformation" (sorry, I do not know the correct english word for that!). Maybe "mean axis transformation"? I think the result is that one can see better that it is an ellipse?
I think the first step is to determine the eigenvalues of $Q$, right?
I got $$ \lambda_1=-\frac{1}{2}(\sqrt{c^2-4d}-c), \lambda_2=\frac{1}{2}(\sqrt{c^2-4d}+c) $$ with $$ c:=\frac{1}{2}(m_1+m_2)L_2^2+\frac{1}{2}m_2L_2^2, $$ $$ d:=\frac{1}{4}m_2(m_1+m_2)L_2^4-\frac{1}{4}m_2^2L_1^2L_2^2\cos^2(\theta_1-\theta_2). $$
I think now in the next step I have to determine the eigenvectors but to be honest I do not know how to do it here...
I never did a "Hauptachsentransformation" before...
The matrix you have is of the form $$ Q = \left[\begin{array}{cc} a & c \\ c & b\end{array}\right]. $$ So the quadratic form $x^{T}Qx$ has the form $$ T(x_{1},x_{2})=ax_{1}^{2}+2cx_{1}x_{2}+bx_{2}^{2}. $$ There's a big short-cut you can take when dealing with a $2\times 2$ matrix. What you're trying to do is rotate the coordinate system until the cross term disappears. The standard technique is a rotation where you set $$ \begin{align} x_{1}+ix_{2} & = (x_{1}'+ix_{2}')(\cos\theta+i\sin\theta) \\ & = (x_{1}'\cos\theta-x_{2}'\sin\theta)+i(x_{1}'\sin\theta+x_{2}'\cos\theta) \end{align} $$ That is, $x_{1}=x_{1}'\cos\theta-x_{2}'\sin\theta$ and $x_{2}=x_{1}'\sin\theta+x_{2}'\cos\theta$. Plug this in to get $$ \begin{align} T(x_{1},x_{2}) & =a(x_{1}'\cos\theta-x_{2}'\sin\theta)^{2} \\ & +2c(x_{1}'\cos\theta-x_{2}'\sin\theta)(x_{1}'\sin\theta+x_{2}'\cos\theta) \\ & +b(x_{1}'\sin\theta+x_{2}'\cos\theta)^{2}. \end{align} $$ The cross terms $x_{1}'x_{2}'$ disappear when $\theta$ is chosen to satisfy $$ -a\cos\theta\sin\theta +c(\cos^{2}\theta-\sin^{2}\theta)+b\sin\theta\cos\theta = 0. $$ Then the matrix $$ P=\left[\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{array}\right] $$ is orthogonal ($P^{T}=P^{-1}$) and gives $P^{T}QP$ equal to a diagonal matrix.