A-invariant subspace and reachability

Question

I got some problems while studying the geometric approach to nonlinear systems. I do not understand how to choose the vectors of a transformation matrix T in order to get $T^{-1}AT=\begin{pmatrix}A_{11}&& A_{12}\\ 0 && A_{22}\end{pmatrix}$ and how to relate this with A-invariant subspaces. Moreover, is this similar to a Kalman (controllable) form?

Answer 1

First, you note that Kalman's controllability matrix $K=[B\, AB\,\dots\,A^{n-1}B]$ is A-invariant.

Let ${\rm rank}(K)=m<n$. Then you choose the transformation matrix $T=[v_1,\dots,v_m,\tilde T]$, where $v_1,\dots,v_m$ are the linear independent columns of $K$ and $\tilde T$ is arbitrary as long as $T$ is non-singular.

Finally, we need to show that the similarity transformation $T^{-1}AT$ yields the block-triangular matrix $A^C=\begin{bmatrix}A_{11}&A_{12}\\0&A_{22}\end{bmatrix}$, i.e., $A^C=T^{-1}AT$.

We will show that $AT=TA^C$. Let us rewrite this as follows: $$A\begin{bmatrix}v_1&v_2&\ldots&v_m&| &\tilde T\end{bmatrix}=\begin{bmatrix}v_1&v_2&\ldots&v_m&| &\tilde T\end{bmatrix}\begin{bmatrix}A_{11}&A_{12}\\0&A_{22}\end{bmatrix}.$$ The vectors $v_i$ are $A$-invariant, thus the product $A\begin{bmatrix}v_1&v_2&\ldots&v_m\end{bmatrix}$ can be written as a linear combination of the same vectors $v_i$. This implies that the matrix $A_{21}=0$.

For $B^C$, we need to check that $B=TB^C$. This follows from the fact that ${\rm span}\, (B)\subset {\rm span}(K)$.