What is modal analysis? I know what modal analysis is, and I know how to conduct one. I can get the eigenvalues and vectors (modes). Not a problem.
However, I am lost at trying to understand the philosophy of what it is.
By that, I mean ...
You take a two mass oscillator, for example (two masses connected by springs).  I use either Newtonian or analytical mechanics (Hamilton's Principle) to get the equations.  They are coupled and I cannot solve them.  So I assume that there is a solution such that each mass oscillates with the same natural frequency and I find the frequency and the normalized shapes.
OK.
What did I just do?
What is the philosophy of this?
How did they think this up?
If I look at a two mass system, what tells me to set the determinant to zero (and please do not just say it allows for non-trivial solutions--tell me what is motivating me.
Also, it seems that we discuss this topic when discussing vibrations.
We solve for harmonic oscillator response and, suddenly, when the equations become coupled, we switch to modal analysis (I am focused on DISCRETE mass systems here, NOT continuaa, like beams).  Why?  Is the solution the same?  Different?
If you were to write a textbook on Vibrations, how would you introduce modal analysis.  Most books just do it, and provide NO historical context or motivation, and that is what I am looking for.
 A: I don't know any of the history behind modal analysis but I can give you some intuition. The main reason modal analysis is useful is because eigenvalue problems are well understood and straightforward to calculate. They may still be hard to calculate but at least it's straightforward. Let's call all the positions in our system $\mathbf x=(x_1,\dots,x_n)^T$. If we assume all the accelerations are a linear combination of positions (so also no friction) then our coupled system of equations can be written as
$$\frac{d^2}{dt^2}\mathbf x=A\mathbf x$$
where $A$ is a $n\times n$ matrix. This already looks similar to an eigenvalue problem but the time derivative spoils the fun.
Let's take a little side tour. Derivatives are linear operators: $\frac{d}{dt}(x(t)+y(t))=\frac{dx(t)}{dt}+\frac{dy(t)}{dt}$. This means that we can define something similar to an eigenvalue problem:
$$\frac{d}{dt}x(t)=\lambda x(t)$$
This means $\frac{d}{dt}$ has 'eigenfunctions' $e^{\lambda t}$. Why would we want this? Eigenfunctions often make differential equations easier and turn them into algebraic equations. For example
\begin{align}
\left(\frac{d^2}{dt^2}+b \frac{d}{dt}+c\right)f(t)&=0\\
\left(\frac{d^2}{dt^2}+b \frac{d}{dt}+c\right)e^{\lambda t}&=0\\
\left(\lambda^2+b \lambda+c\right)e^{\lambda t}&=0\\
\implies  \left(\lambda^2+b \lambda+c\right)&=0
\end{align}
which means we can solve for $\lambda$ in terms of $b$ and $c$. The operator $\left(\frac{d^2}{dt^2}+b \frac{d}{dt}+c\right)$ is still linear so the general solution is a sum over these eigenfunctions where we substitute our solved $\lambda$ for the eigenvalues.
Our little side tour now motivates us to take $\mathbf x(t)=e^{i\omega t}\mathbf x_0$. This turns our initial equation into a proper eigenvalue equation
$$A\mathbf x_0=-\omega^2 \mathbf x_0$$
As you know, this can be written as
$$(A+\omega^2 I)\mathbf x_0=0$$
I stress that $\omega$ and $x_0$ are our unknowns. This equation has only solutions for particular values of $\omega$. Generally a matrix will stretch or shrink vectors but it will not give the zero vector. Similar to how $ax=0$ only has solution $x=0$. But when a matrix has determinant zero the output of the matrix has less dimensions than its output. For example the entire output of a  2x2 matrix with determinant zero lies on a line. This means there are non-trivial vectors that get sent to zero by the matrix. These are the eigenvectors of the matrix.
I hope this answered some of your questions.
