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Let $P$ and $C \neq0$ a $q \times q$ matrices. I want to prove that there exists a positive constants $\alpha$ such under some assumptions under $P$ we have the inequality $${\left\| {P\left( {I - C} \right)x} \right\|_{{{\mathbb{R}}^q}}} \leqslant \alpha {\left\| {PCx} \right\|_{{{\mathbb{R}}^q}}}$$ for all $x$ in $\mathbb{R}$ whith $I$ is the identity matrix.

Thank you.

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  • $\begingroup$ Why do you expect such an inequality to hold? Could you give us a bit more context? $\endgroup$ – Omnomnomnom Nov 25 '18 at 17:36
  • $\begingroup$ I need this inequality to prove the controllability of some system of PDE. I have to get rid of the R.H.S of the inequality. $\endgroup$ – Gustave Nov 25 '18 at 17:38
  • $\begingroup$ I would suggest that you post a question about the actual system of PDE. As your question stands, it is difficult to know what kind of conditions on $P$ we should be looking for. $\endgroup$ – Omnomnomnom Nov 25 '18 at 17:41
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    $\begingroup$ Your inequality will be true for some $\alpha > 0$ if and only if $\ker(P(I - C)) \subseteq \ker(PC)$. I don't believe that $\ker(C) \subseteq \ker(P)$ will be a sufficient condition to guarantee this. $\endgroup$ – Omnomnomnom Nov 25 '18 at 19:34
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    $\begingroup$ For instance: your condition cannot hold for $$ C = \pmatrix{1&0\\0&1}, \quad P = \pmatrix{1&0\\0&0} $$ even though we have $\ker(C) \subseteq \ker(P)$ $\endgroup$ – Omnomnomnom Nov 25 '18 at 19:45
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If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.

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  • $\begingroup$ Thank tou for the remark, yes, the inequality does'nt hold for any $P$, I want to find some conditions on $P$ so that the inequality is satisfied. $\endgroup$ – Gustave Nov 25 '18 at 17:37
  • $\begingroup$ I have changed the statement. Thank you for the remark. $\endgroup$ – Gustave Nov 25 '18 at 17:43

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