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Math Help - mapping from one invariant subspace to another

  1. #1
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    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:

    <br />
\dot{x}(t)=Ax(t)+Bu(t)<br />
    <br />
y(t)=Cx(t)<br />

    where

    <br />
A \in \mathbb{R}^{nxn}<br />
    <br />
B \in \mathbb{R}^{nxm}<br />
    <br />
C \in \mathbb{R}^{pxn}<br />

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

    <br />
\mathcal{S} \subset \mathcal{V} \subset \mathbb{R}^n \\<br />

    <br />
\exists \text{ a } G \text{ st. } (A+GC)\mathcal{S} \subset \mathcal{S} \\<br />

    <br />
\exists \text{ an } F \text{ st. } (A+BF)\mathcal{V} \subset \mathcal{V} \\<br />

    The problem is to find an N st.

    <br />
(A+BNC)\mathcal{S} \subset \mathcal{V}<br />
    ------------------

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

    <br />
(A+BNC)\mathcal{S} \subset \mathcal{S}<br />

    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.

    <br />
(A+BNC)\mathcal{V} \subset \mathcal{V}<br />

    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.
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  2. #2
    A Plied Mathematician
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    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}.
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  3. #3
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    thank you for the answer, ackbeet!
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  4. #4
    A Plied Mathematician
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    You're welcome. Have a good one!
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