I am studying control theory and I am starting the concept of geometric control theory. As a prerequisite to this, I am studying the concet of invariant subspaces, and I having some troubles understanding some concepts.
To try to understand my doubts I think I have to start from the beginning, but please correct me if I say something wrong. So, $V$ is an invariant subspace under $A$ if:
$AV\subset V$
and in this context, we can find a coordinate transformation such that :
$TAT^{-1}=\begin{pmatrix}
A_{11} &A_{12} \\
0 & A_{22}
\end{pmatrix}$
at this point the notes of my professor say that this implies that the invariant subspace is an eigenspace (but I don't understand why). And so this should implies that the evolutions that start in $V$ remains in $V$, and this can be seen from the system in the new coordinates:
$\dot{z_1}=A_{11}z_1 + A_{12}z_2+B_1u$
$\dot{z_2}=A_{22}z_2+B_2u$
Moreover, it says that if I consider two generic initial conditions whose difference belong to $V$, their evolution remain if an affine variety of the same class (these last few are the literal words in my note, which I don't understand). And it says that the structure induced by translation from $V$ is called foliation.
I am very confused from this arguments, especially I cannot understand the concept of foliation. To give more context, I am studying this in order to arrive to characterize the reachability and observability in control theory, but don't know if it matters.
Can somebody please help me make clarity?
Best Answer
I think you're asking this in the context of the controlled dynamical system $(\mathcal{U},\Sigma): \dot{x}(t)=Ax(t)+Bu(t)$, but i'll try to keep the context general. First of all, your definition of invariant subspace doesn't seem right to me (it's correct if you denote a proper subset by the symbol $\subset$).
If $A \in \text{Hom}(V,V)=\mathcal{L}(V,V)$ is a linear map, a subspace $\mathcal{S} \subseteq V$ is called $A$-invariant if $A\mathcal{S} \subseteq \mathcal{S}$. For now let's consider $V:=\mathbb{R}^n$, then we can also restate $A$-invariance as: A subspace $\mathcal{S} \subseteq \mathbb{R}^n$ is $A$-invariant if the matrix $T_0 \in \mathbb{R}^{n \times r}$ consisting of basis elements of $\mathcal{S}$ satisfies $AT_0=T_0M$, for some $M\in \mathbb{R}^{r \times r}$, $r \le n~,~r,n \in \mathbb{Z}_{>0}$. It literally means under the action of operator $A$ vectors in $\mathcal{S}$ remains in $\mathcal{S}$.
Some exercises for you : If $A$ is an homeomorphism i.e algebraic isomorphism then show $\text{(a)}\{0\}~\text{(b)}V ~\text{(c)}\text{Null}(A)~\text{(d)}\text{Im}(A)$ are $A$-invariant.
Now, to show the connection between invariant subspaces and eigenspace, i'll take the simplest case, the rest is upto you to figure out. Consider a one-dimensional invariant subspace, suppose $0 \neq v \in V$ and let $U:=\left\{ \lambda v : \lambda \in \mathbb{K}\right\}=\text{Span}(v)$, where $\mathbb{K}:=\mathbb{R} ~\text{or}~\mathbb{C}$. Now if $U$ is invariant under $A$ then from definition we have $Av=\lambda v~,~\lambda \in \mathbb{K}$, well this gives us the motivation to define a quantity called eigenvalue, i.e if $A \in \mathcal{L}(V,V)$, $\mathbb{K} \ni \lambda$ is called an eigenvalue of operator-$A$ if $\exists v \in V~,~v \neq 0$ : $Av=\lambda v$, and $(\lambda,v)$ is called the eigenspace. Hope this makes things clear.
Further, we know that $\mathcal{S}=\text{Span}\{v_1,v_2,\ldots,v_r\}$, and define $T_0:=\left[ v_1 \cdots v_r \right]$, now consider the case when the eigenvalue matrix $M$ is not diagonalizable (diagonalizable case can be handled similarly), then for some matrix $W$ s.t $\text{det}(W) \neq 0$ we have $MW=W \Lambda$, where $\Lambda$ is the Jordan matrix of eigenvalues. Now define $V:=T_0W$, to this end, we look at $T_0MW=T_0W\Lambda=V\Lambda$, but we know that $T_0M=AT_0$ which gives us $T_0MW=AT_0W=AV$, so from these two equations we see that $AV=\Lambda V$, which is the eigenvalue equation implying that columns of $V$ actually forms an eigenspace of $A$, associated with $\Lambda$. Now let $T_1$ is a matrix whose colums are the eigenvectors of $\mathcal{S}$-perp i.e $\mathcal{S}^{\perp}$. Then the matrix formed by stacking columns of $T_0$ and $T_1$ is say, $T:=\left[T_0 ~~T_1 \right]$ and $\text{det}(T) \neq 0$. Then we have $T_{i}:=T^{-1}=\begin{pmatrix}T_{i1} \\T_{i2} \end{pmatrix}$ and $TT_{i}=T_0T_{i1}+T_1T_{i2}=\mathbb{I}$. Also $$T_iT=\begin{pmatrix} T_{i1}T_0 &T_{i1}T_1 \\T_{i2}T_0&T_{i2}T_1\end{pmatrix}=\begin{pmatrix}\mathbb{I}_r &0\\0&\mathbb{I}_{n-r}\end{pmatrix} $$ Finally $$\begin{align}T^{-1}AT&=T_{i}AT=T_i\left[AT_0~~AT_1\right]=\begin{pmatrix}T_{i1} \\T_{i2} \end{pmatrix}\left[T_0M~~AT_1\right]\\&=\begin{pmatrix}T_{i1}T_0M&T_{i1}AT_1\\T_{i2}T_0M &T_{i2}AT_1\end{pmatrix}=\begin{pmatrix}M&T_{i1}AT_1\\0&T_{i2}AT_1\end{pmatrix}\end{align}$$ This is the block-triangular matrix that comes out. Similar matrices arise in the context of controllability analysis as a consequence of rechability space i.e $$\mathfrak{R}_0\underbrace{=}_{[1]}\langle A|B\rangle:=B+AB+\cdots+A^{n-1}B$$ is $A$-invarint i.e $A\mathfrak{R}_0 \subseteq \mathfrak{R}_0$, in fact it's the smallest $A$-invarint subspace. Similarly the unobservable subspace is a (biggest)$A$-invarint subspace.
I guess foliation is a bit over-kill in this context. Foliation is basically an equivalence relation in a $n$-manifold, which in this case is $\mathbb{R}^n$, surely it's a smooth manifold with a single chart atlas i.e $\left(\mathbb{R}^n,(\text{id}:\mathbb{R}^n \to \mathbb{R}^n) \right)$, take any subset $Y\subseteq \mathbb{R}^n $ define an equivalence relation: $$\left( x \sim y ~\text{if}~x-y \in Y ~\forall x,y \in \mathbb{R}^n \right)$$and define $\left[x\right]:=\left\{y \in \mathbb{R}^n:y \sim x\right\}$ which are connected and injectively immersed submanifolds, i think this is how the equivalence classes has been defined on the initial condition which makes the solutions stay in the same algebraic structure.
[1]: Note this is not inner-product or "bra-ket" notation, Wonham, Murray: Linear Multivariable control uses this abundantly.