The basic quantity you build when performing a phonon calculation is the matrix of force constants, given by:
$$
D_{i\alpha,i^{\prime}\alpha^{\prime}}(\mathbf{R}_p,\mathbf{R}_{p^{\prime}})=\frac{\partial^2 E}{\partial u_{p\alpha i}\partial u_{p^{\prime}\alpha^{\prime}i^{\prime}}},
$$
where $E$ is the potential energy surface in which the nuclei move, $u_{p\alpha i}$ is the displacement of atom $\alpha$ (of all atoms in the basis), in Cartesian direction $i$ ($x$, $y$, $z$), and located in the cell within the supercell at $\mathbf{R}_p$. Mathematically speaking, this matrix of force constants is the Hessian matrix of the potential energy function $E$. As for any Hessian matrix, its eigenvalues give the curvature of the function $E$ along the direction of the associated eigenvectors, and we can have:
- Positive eigenvalue: the curvature is positive, which means that the function $E$ is a minimum along that direction.
- Negative eigenvalue: the curvature is negative, which means that the function $E$ is a maximum along that direction.
To relate this to your question, note that the phonon frequencies are the square root of the eigenvalues above, so positive eigenvalues give positive phonon frequencies and negative eigenvalues give imaginary phonon frequencies (your soft modes).
This means that imaginary frequencies are associated with maxima of $E$, so that displacing that atoms along the corresponding eigenvectors will lower the energy (make $E$ smaller). As a dynamically stable structure is one defined at a (local) minimum of $E$, then whenever you have imaginary frequencies you are not at a local minimum (you are at a saddle point), so the structure is unstable. By distorting the structure along the eigenvector associated with the imaginary mode, you will be able to find a lower energy structure.
The addition of temperature can stabilize an otherwise unstable structure, but for that one needs to include anharmonic terms in the calculation (i.e. phonon-phonon interactions), so the harmonic calculations you are performing cannot resolve this question of stability at finite temperature.
