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?
A-invariant subspace and reachability
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linear-algebra
control-theory
invariant-subspace
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0https://en.wikipedia.org/wiki/Invariant_subspace#Matrix_representation – 2017-01-12
1 Answers
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First, you note that Kalman's controllability matrix $K=[B\, AB\,\dots\,A^{n-1}B]$ is A-invariant.
Let ${\rm rank}(K)=m 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 ${\rm span}\, (B)\subset {\rm span}(K)$.