# Matries inequality with norms

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.

• Why do you expect such an inequality to hold? Could you give us a bit more context? – Omnomnomnom Nov 25 '18 at 17:36
• 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. – Gustave Nov 25 '18 at 17:38
• 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. – Omnomnomnom Nov 25 '18 at 17:41
• 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. – Omnomnomnom Nov 25 '18 at 19:34
• For instance: your condition cannot hold for $$C = \pmatrix{1&0\\0&1}, \quad P = \pmatrix{1&0\\0&0}$$ even though we have $\ker(C) \subseteq \ker(P)$ – Omnomnomnom Nov 25 '18 at 19:45

If $$P=I$$ and $$C=0$$, then there is no such positive constant, so that the inequality holds.
• 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. – Gustave Nov 25 '18 at 17:37