Are there non-orthogonal "normal" modes for non-identical coupled oscillators? The question is broad, I will specify an example to elaborate what I'm asking.
Suppose I have two different LC circuits inductively coupled (or capacitively, but the question I have will be relevant to both so for simplicity lets consider inductive coupling) as below.

Starting from Kirchoff's Voltage and Current Laws, and using the constitutive relations one can arrive at the equations of motion which look something like $$ \boldsymbol{A}\ddot{\vec{v}}= \boldsymbol{B}\vec{v}$$
here $\vec{v}=(v_1 \, v_5)^{T}$ where $v_1$ is the voltage across $C_1$ and $v_5$ is the voltage across $C_5$. One can easily check that this circuit has 2 degrees of freedom. Now I left $\boldsymbol{A},\boldsymbol{B}$ without explicitly stating them because the question is more general than this particular equation. Now I'm trying to find the normal modes of this circuit. But it occurs to me that $\boldsymbol{A},\boldsymbol{B}$ are guaranteed to be of rank $2$ and diagonalizable so we can find the eigenvectors for this generalized eigenvalue problem and be guaranteed they form a basis for $\mathbb{R}^2$. So we can decouple the equations of motion into 2 independent second order ODEs.
The problem is that normal modes are defined to be the orthogonal, that is the resulting eigenvectors are orthogonal and hence it is a different decoupling. It specifically requires $\boldsymbol{A},\boldsymbol{B}$ to be symmetric such that the resulting eigenvectors form an orthogonal basis.
Check this discussion on transforming second order linear systems of ODEs of same form as this.
So this leads me to think that if the matrices are not symmetric, the eigenvectors will not be orthogonal therefore no normal modes. But a decoupling is possible it seems. What is this decoupling and is this true - non-identical coupled oscillators don't have normal modes?
One note is that the lack of symmetry in the oscillators could justify not being able to exploit symmetric and anti symmetric normal modes (for a basic example like this) so perhaps not too surprising, but I was under the impression all coupled oscillators have normal modes.
Any help is appreciated!
 A: The math in the link or post is not very clear. When done carefully, the prescription to find the modes becomes intuitive: it is equation $(*)$ below.
Start by writing $\boldsymbol{A}\ddot{\vec{v}}= \boldsymbol{B}\vec{v}$ as
$$
 -\boldsymbol{A}\omega^2 \vec{v}_0= \boldsymbol{B}\vec{v}_0 
\quad \Rightarrow  \quad   (\boldsymbol{A}\omega^2+\boldsymbol{B}) \vec{v}_0= 0,
$$
thanks to the usual substitution $\vec{v}(t) = \vec{v}_0 e^{-i \omega t}$.
Now, call $\boldsymbol{C}(\omega) =  \boldsymbol{A}\omega^2+\boldsymbol{B}$. The above system only has the trivial solution $ \vec{v}_0=0$ if $\boldsymbol{C}(\omega)$ has maximum rank (in this case 2). Therefore, to have non-trivial oscillatory solutions you have to require that $\omega$ takes some special value so that a solution with a non-zero amplitude $\vec{v}_0$ can exist. This amounts to require that the determinant of $ \boldsymbol{C}(\omega)$ is zero.
To summarize, you can find the frequencies of the natural modes of the system by solving the equation
$$
det[\boldsymbol{C}(\omega)]  = 0
\quad \quad (*)
$$
that is a polynomial equation in $\omega$. The solutions are the frequencies of the mode: for you 2D case you will find two (complex) solutions. As usual, the real part tells you the frequency of the oscillation, the imaginary part tells you how fast the mode decays or grows (depending on its sign).
