Is there any way to prove stability for a discrete state space model by using Singular Value Decomposition

control theorylinear-controloptimal controlstability-theorysvd

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?

Best Answer

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.

Related Question