Proving that motion of an $n$ dimensional oscillator can be written as a linear combination of “sine waves”

Here is a related question which might provide some context: LINK.

Let's consider an oscillator with equation of motion in $$n$$ dimensions: $$\frac{d^2}{dt^2} \vec{x} = K \vec{x}.$$

Given that $$\vec x=0$$ is a stable equilibrium, how can I show that the system will oscillate in sine waves? In other words, how to prove that the system will not behave like $$x=\sinh t$$? More precisely, how can I show that the $$y_i$$ in the answer in the link above will be sine waves?

As the answer in the link above suggest, to prove the solution to the DE has sine wave pattern, I need to prove that $$K$$ is symmetric with negative eigenvalues (see also the comment below the answer).

What about the case that $$\frac{d^2}{dt^2} \vec{x} = f(\vec{x})$$, where $$f(\vec{x})$$ doesn't have to be linear, but can be approximated linearly by $$K \vec{x}$$? If $$\vec x=0$$ is a stable equilibrium, must the eigenvalues of $$K$$ be negative OR zero?

I hope I am clear. Please tell me if I am not expressing myself clealy.

In the case that $$K\vec{x}$$ is exact rather than an approximation, note that if the equilibrium is stable, work done of a virtual displacement $$\delta \mathbf x$$ must be negative. So, $$\text{work done}=F.d\propto (K\delta \mathbf x).\delta \mathbf x=\delta \mathbf x^T K\delta \mathbf x<0.$$ So $$K$$ is negative definite, and all eigenvalues are negative.
If $$K$$ is just an approximation of $$f$$, then we may only have $$\leq$$ instead of $$<$$. So you can still show that it is non-positive.
But why $$K$$ must be symmetric?