# mapping from one invariant subspace to another

• Dec 1st 2010, 04:51 AM
ronan
mapping from one invariant subspace to another
Hello to everybody,

I am trying to find a constructive proof for the solution of a problem in control engineering, and I got stuck at some point for the last few days. The problem is as follows:

------------------
For a system defined in state space as:

$
\dot{x}(t)=Ax(t)+Bu(t)
$

$
y(t)=Cx(t)
$

where

$
A \in \mathbb{R}^{nxn}
$

$
B \in \mathbb{R}^{nxm}
$

$
C \in \mathbb{R}^{pxn}
$

There are defined subspaces $\mathcal{S}$ and $\mathcal{V}
$
of $\mathbb{R}^n$ that the rest of the problem deals with, with properties:

$
\mathcal{S} \subset \mathcal{V} \subset \mathbb{R}^n \\
$

$
\exists \text{ a } G \text{ st. } (A+GC)\mathcal{S} \subset \mathcal{S} \\
$

$
\exists \text{ an } F \text{ st. } (A+BF)\mathcal{V} \subset \mathcal{V} \\
$

The problem is to find an $N$ st.

$
(A+BNC)\mathcal{S} \subset \mathcal{V}
$

------------------

Of course, the first thing that comes to mind is to find $N$ st.

$
(A+BNC)\mathcal{S} \subset \mathcal{S}
$

which satisfies the problem requirement as $\mathcal{S} \subset \mathcal{V}$. However, such a method $N$ appears to be too "tight" (meaning that more freedom on $N$ is needed) for many examples from application.

Another idea is to find $N$ st.

$
(A+BNC)\mathcal{V} \subset \mathcal{V}
$

then find a restriction of the resulting $N$ on subspace $\mathcal{S}$. But then my mathematical knowledge is limited on restrictions on linear operators.

In short, I am stuck at finding an elegant approach for the construction of $N$. Any help and ideas on the topic would be of great help. Thanks in advance.
• Dec 1st 2010, 06:13 AM
Ackbeet
Your second idea is definitely better, and it is sufficient, I believe. If you compare the statement

$\exists\,F\;\text{s.t.}\;(A+BF)\mathcal{V}\subset\ mathcal{V}$ with

Find an $N$ s.t.

$(A+BNC)\mathcal{V} \subset \mathcal{V},$

then why not just solve

$F=NC$ for $N?$

Dimensions:

$A$ is n x n,

$B$ is n x m,

$C$ is p x n

$F$ must be m x n, and so

$N$ must be m x p.

As for the restriction, I wouldn't worry about it at all. If an operator takes $\mathcal{V}$ into $\mathcal{V},$ then it must, of necessity, take a subset $\mathcal{S}\subset\mathcal{V}$ into $\mathcal{V}.$
• Dec 7th 2010, 01:45 AM
ronan
thank you for the answer, ackbeet!
• Dec 7th 2010, 03:41 AM
Ackbeet
You're welcome. Have a good one!