Assume that we have this state model representation
$$x(k+1) = Ax(k) + Bu(k)$$
To check stability of this model, we need to check the eigenvalues. In discrete mode, $\lambda$ will result both a real part and an imaginary part. $$det(\lambda I - A) = 0$$
If the $$1 > \sqrt{\lambda_{real} ^2 + \lambda_{img}^2}$$
Then the system is stable. But is it possible to prove that by using Singular Value Decomposition?
$$A = USV^T$$
Where $S$ contains the singular values (real positive) in a diagonal order, where the largest is at the top and the smallest is on the bottom.
I wonder this because
$$det(\lambda I - AA^T) = 0$$
Is equivalent to $$A = USV^T$$ $$S \equiv \sqrt \lambda$$
Let's say that I compute the eigenvalues $\lambda$ from
$$det(\lambda I - AA^T) = 0$$
Can I then determine if $A$ is stable?
Sure. Since $\rho(A) \leq \lVert A \rVert_2 = \sigma_\max(A)$ you can immediately deduce stability if $\sigma_\max(A)<1$, where $\sigma_\max$ is the maximum singular value. However, converse is not true and this is a very conservative way for checking stability. Most likely you will get $\sigma_\max(A) > 1$ even if the system is stable, unless $A$ is in a special form.