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For small oscillations, my textbook equation for amplitude says:

$(V-\omega^2T) \cdot a=0$ where $a$ is a column vector in which each component $a_i$ is related to $q_i$ as $q_i=a_i\cos(\omega t-\gamma)$. Now it says for a solution to exist $\det(V-\omega^2T)=0$ must be satisfied.

My objection is: if it's non-zero it has an inverse and simple matrix algebra would do the trick. Although the problem we might face here is that the RHS is 0, which would give $a=0$. But if we make $\det(V-\omega^2T)=0$, doesn't it imply that the system has infinitely many solutions, or no solution- again as useless as previous case. I somehow get a feeling that the author is trying to use the argument that $x \cdot y=0$ and we want some value of $x$, hence $y=0$ but I am not sure can we extend this line of reasoning to matrices.

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Say $A$ is some square matrix. If we have an equation $Ax=0$, then there is a nonzero solution if and only if $\det(A)=0$. We can prove this as follows:

If $\det(A)\neq 0$, then $A$ is invertible, and so only one element $x$ can map to $0$. We know $A 0=0$, so no other element can map to zero.

If $\det(A)=0$, then we know $A$ is not invertible, and so somewhere, two elements map to the same element. Say $Ax=Ay$, where $x\neq y$. Then $0=Ax-Ay=A(x-y)$, where $x-y\neq 0$. We thus have a nonzero solution to $Ax=0$.

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  • $\begingroup$ Just to add one thing: "infinitely many" solutions typically means that you have "all possible amplitudes" of oscillation, including zero. But the frequencies of the oscillations do not change - they are determined from solving for the zero determinant. $\endgroup$ – Floris Oct 18 '15 at 20:11
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What math leads to this expression?

The argument is a lot simpler, I think. Suppose we start from a Lagrangian $$L = \frac12 T_{ab}~\dot q^a~\dot q^b - \frac12 V_{ab}~q^a~q^b,$$with normal Einstein-summation rules when we see a repeated lower-and-upper index. The equations of motion of such a system are simply$$ \left({dp_c \over dt} =\right)\;\; {d\over dt}{\partial L \over \partial \dot q^c} = {\partial L \over \partial q^c} \;\;\Big(= F_c\Big),$$ relating a generalized momentum to a generalized force. Calculating them outright we see: $$\frac12 \big(T_{cb}~\ddot q^b + T_{ac}~\ddot q^a\big) = -\frac12\big( V_{cb}~q^b, - V_{ac}~q^a\big),$$ and assuming both matrices are symmetric yields $$T_{ca}~ \ddot q^a = -V_{ca}~ q^a.$$This very clearly "looks" like the 1D harmonic oscillator situation and we can just assume $q^a = \exp(i\omega t)~p^a$ (or you can use $\sin$ and $\cos$ for some $q^a = \sin(\omega t)~A^a + \cos(\omega t)~B^a$ if you prefer) to find that:$$\big(V_{ca} - T_{ca} \omega^2\big)~p^a = 0,$$where the extra $-1$ comes from the $i^2$ of the derivatives.

What does this mean in your case?

This formula is essentially saying $k - m \omega^2 = 0$ in the most simple 1D case: the only frequency of oscillation that the equations of motion "allow" is this one. This is simply a generalization to all of the directions in space, with the $T_{ca}$ matrix taking the role of mass and $V_{ca}$ taking the role of spring constant.

Saying $\det(V - \omega^2 T) = 0$ in turn means that we don't care about the directions in space $p^a$ that we are calculating this for, but rather just the frequencies.

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