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Suppose you have a differential equation with n distinct functions of $t$ where

$fracd^2x_1dt^2=k_11x_1+...k_1nx_n$

.

.

.

$fracd^2x_ndt^2=k_n1x_1+...k_nnx_n$

I want to show that any set of solutions of this differential equation $(x_1,x_2,...,x_n) $

can be written as a linear combination of solutions of the form $(e^iw_1t,...,e^iw_1t), (e^iw_2t,...e^iw_2t), ...,(e^iw_mt,...e^iw_mt)$ where each $w_j$ is a real number.

i.e. I want to know why the motion of any oscillator can be written as a linear combination of its normal modes. I would also appreciate it if you could tell me if a proof of this fact has to do with eigenvalues and eigenvectors in general.

The system of differential equations you wrote could be written as,

$$ fracd^2dt^2 left[beginarrayc x_1 \ vdots \ x_n endarrayright] = left[beginarrayccc k_11 & cdots & k_1n \ vdots & ddots & vdots \ k_1n & cdots & k_nn endarrayright] left[ beginarrayc x_1 \ vdots \ x_nendarrayright]$$

$$ fracd^2dt^2 vecx = K vecx$$

The matrix $K=[k_ij]$ acts on the indices of the functions $x_1,dots,x_n$.

We will suppose that $K$ is diagonalizable with eigenvalues $lambda_1, dots, lambda_n$.

Let $Lambda$ be the diagonal form of $K$.

Let $V$ be the matrix of eigenvectors of $K$. Note that $Lambda = V^-1 K V$.

We can now write the system of differential equations as,

$$
fracd^2dt^2 vecx = V Lambda V^-1 vecx
$$

$$
V^-1fracd^2dt^2 vecx = Lambda V^-1 vecx
$$

$$
fracd^2dt^2 V^-1vecx = Lambda V^-1 vecx
$$

Let $vecy = V^-1 vecx$, then we have $
fracd^2dt^2 vecy = Lambda vecy
$. This corresponds to the following system of equations.

$$
fracd^2 y_1dt^2 = lambda_1 y_1
$$
$$
fracd^2 y_2dt^2 = lambda_2 y_2
$$
$$
vdots
$$
$$
fracd^2 y_ndt^2 = lambda_n y_n
$$

Clearly the solutions are of the form,

$$y_j(t) = C_1 e^sqrtlambda_j t + C_2 e^-sqrtlambda_j t,$$

to obtain the $x_j$'s we just multiply by the $V$ matrix.

$$ x_j(t) = sum_i V_ji y_i(t)$$

Whether or not the solutions are oscillators depends on whether the eigenvalues are positive, negative, or complex. In physical applications it wouldn't be uncommon for $K$ to be a symmetric matrix with negative eigenvalues.