Classical mechanics problem (mostly linear algebra). Linearised lagrangian, quadratic forms, eigenvalue decomposition.

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I'm working through Vladamir Arnold's book, Mathematical Methods of Classical Mechanics.

I'm at chapter five on Oscillations. The context of my issue is the linearized lagrangian system (I will attach a photo of the relevant page).

We're considering a system whose kinetic and potential energies are given by

$T = \frac{1}{2}(A\dot{q}, \dot{q})$ and $U = \frac{1}{2}(Bq, q)$ where $q, \dot{q} \in \mathbb{R}^n$

We make a linear change of coordinates $Q = Cq$. And we want $Q$ to be such that our energies in the new coordinate system become

$T = \frac{1}{2} \sum_{i=1}^{n} \dot{Q}_i^2$ and $U = \frac{1}{2} \sum_{i=1}^{n} \lambda_i Q_i^2$

where $\lambda_i$ are called the eigenvalues of the form B with respect to A.

Now, I haven't studied linear algebra for over a year, but with what I recall, it is a relatively triviol matter to decompose $U$ as

$U = \frac{1}{2}q^TBq = \frac{1}{2}Q^T{C^{-1}}^TBC^{-1}Q =: \frac{1}{2}Q^TDQ$

and $D$ is diagonal when $C$ is orthogonal, so that the quadratic form becomes $\frac{1}{2} \sum_{i=1}^{n} \lambda_i Q_i^2$ (with no cross terms) just as required.

Now we arrive to where I am stuck - with the kinetic energy, and the matrix A. The eigenvalues $\lambda_i$ are the eigenvalues of $B$. I'm not sure what arnold means by eigenvalues of B with respect to A. I'm thoroughly stuck as to how to obtain the kinetic energy as it's given by a sum of squares of the velocities $\dot{Q}_i$ with coefficients $\frac{1}{2}$.

Of course, it's no harder than it was for the potential to decompose $T$ into a sum of squares, but in that case we will obtain more eigenvalues (unless there is some way to show that these will all be 1?).

I can see how powerful this coordinate change will be, which is why I am very eager to understand this section. Any help, or a push in the right direction will be greatly appreciated!

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The keyword to search for is simultaneous diagonalization of two quadratic forms $Q_1$ and $Q_2$, one of which is positive definite.

Here's briefly how it works: you can use the positive definite form $Q_1$ to define an inner product on your space, and then use Gram–Schmidt to find an ON basis with respect to that inner product. If you change to that basis, the form $Q_1$ just becomes a sum of squares (its matrix will be $I$), and then you can diagonalize $Q_2$ in the usual way.