Prove that $\vec a_i^T T \vec a_j=\delta_{ij}$

Question

Given a solution of the small oscillation equations (called normal mode)

$$\vec q = \sum_i c_i \vec a_i \cos (\omega_i t - \gamma_i), \ \ \ 1 \leq i \leq n \tag1$$

Where

$$\vec q= \begin{pmatrix} q_1 \\ . \\ . \\ q_n \\ \end{pmatrix} , \ \ \ \ \vec a= \begin{pmatrix} a_1 \\ . \\ . \\ a_n \\ \end{pmatrix}\tag2$$

And given the following equation for $\vec a$

$$(\omega_i^2T-V) \vec a_i=0\tag3$$

Show that

$$\vec a_i^T T \vec a_j=\delta_{ij}\tag4$$

Where $T$ and $V$ are real, positive definite and symmetric matrices.

I am convinced $(4)$ can be shown by means of $(3)$

$$\omega_j^2T\vec a_j=V \vec a_j$$

$$\omega_j^2\vec a_i^T T\vec a_j=\vec a_i^T V \vec a_j$$

But how to advance?

Answer 1

Alright, I finally got it!

Let's have a look at equation $(3)$, where $1 \leq i,j \leq n$

$$(\omega_j^2T-V) \cdot \vec a_j=0\tag3 \Rightarrow V\cdot \vec a_j=\omega_j^2T\cdot \vec a_j$$

We notice that $(3)$ is basically the eigenvalue equation, where $\vec a_j$ is an eigenvector of $V$ with respect to $T$.

There're $\vec a_j$ eigenvectors with $n$ corresponding distinct eigenvalues (let us set $\lambda_j:= \omega^2_j$)

$$V\cdot \vec a_1=\lambda_1 T\cdot \vec a_1 \tag{5.1}$$

$$V\cdot \vec a_2=\lambda_2 T\cdot \vec a_2 \tag{5.2}$$

$$\text{....................}$$

$$V\cdot \vec a_n=\lambda_n T\cdot \vec a_n \tag{5.n}$$

All we have to do is show orthogonality of eigenvectors. For the proof we only need two eigenvectors so let us left-multiply $(5.2)$ by $\vec a_1^T$ both sides to get

$$\vec a_1^T \cdot V \cdot \vec a_2=\lambda_2 \vec a_1^T \cdot T \cdot \vec a_2 \tag6$$

OK now we get another equation for $\vec a_1^T \cdot V \cdot \vec a_2$ exploiting the fact that $\lambda_1$ is an eigenvalue of $V$

$$\vec a_1^T \cdot V \cdot \vec a_2 = (V \cdot \vec a_1)^T \cdot \vec a_2 = (\lambda_1 T\cdot \vec a_1)^T \cdot \vec a_2 = \lambda_1 \vec a_1^T \cdot T \cdot \vec a_2 \tag7$$

Equating $(6)$ and $(7)$

$$(\lambda_1-\lambda_2)(\vec a_1^T \cdot T \cdot \vec a_2)=0 \tag8$$

As $\lambda_1 \neq \lambda_2$, we indeed need

$$\vec a_1^T \cdot T \cdot \vec a_2=0 \tag9$$

OK but what if $\vec a_i^T \cdot T \cdot \vec a_i$? Well as $T$ is positive definite $ \Rightarrow \vec a_i^T \cdot T \cdot \vec a_i \geq 0$. It follows that we can always scale the amplitude vectors such that

$$\vec a_i^T \cdot T \cdot \vec a_i=1 \tag{10}$$

Combining $(9)$ and $(10)$ we indeed get orthonormality

$$\vec a_i^T T \vec a_j=\delta_{ij}\tag4$$