Matries inequality with norms












-1












$begingroup$


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
$${left| {Pleft( {I - C} right)x} right|_{{{mathbb{R}}^q}}} leqslant alpha {left| {PCx} right|_{{{mathbb{R}}^q}}}$$ for all $x$ in $mathbb{R}$
whith $I$ is the identity matrix.



Thank you.






matrices inequality operator-theory normed-spaces functional-inequalities







share|cite|improve this question











$endgroup$












  • $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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
















-1












$begingroup$


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
$${left| {Pleft( {I - C} right)x} right|_{{{mathbb{R}}^q}}} leqslant alpha {left| {PCx} right|_{{{mathbb{R}}^q}}}$$ for all $x$ in $mathbb{R}$
whith $I$ is the identity matrix.



Thank you.






matrices inequality operator-theory normed-spaces functional-inequalities







share|cite|improve this question











$endgroup$












  • $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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














-1












-1








-1





$begingroup$


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
$${left| {Pleft( {I - C} right)x} right|_{{{mathbb{R}}^q}}} leqslant alpha {left| {PCx} right|_{{{mathbb{R}}^q}}}$$ for all $x$ in $mathbb{R}$
whith $I$ is the identity matrix.



Thank you.






matrices inequality operator-theory normed-spaces functional-inequalities







share|cite|improve this question











$endgroup$




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
$${left| {Pleft( {I - C} right)x} right|_{{{mathbb{R}}^q}}} leqslant alpha {left| {PCx} right|_{{{mathbb{R}}^q}}}$$ for all $x$ in $mathbb{R}$
whith $I$ is the identity matrix.



Thank you.






matrices inequality operator-theory normed-spaces functional-inequalities




matrices inequality operator-theory normed-spaces functional-inequalities






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 17:42







Gustave

















asked Nov 25 '18 at 17:31









GustaveGustave

725211




725211












  • $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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


















  • $begingroup$
    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








  • 1




    $begingroup$
    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








  • 1




    $begingroup$
    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
















$begingroup$
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






$begingroup$
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 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






$begingroup$
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






1




1




$begingroup$
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






$begingroup$
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






1




1




$begingroup$
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




$begingroup$
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










1 Answer
1






active

oldest

votes


















0












$begingroup$

If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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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active

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$begingroup$

If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















0












$begingroup$

If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












0








0





$begingroup$

If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.






share|cite|improve this answer









$endgroup$



If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 25 '18 at 17:34









Yiorgos S. SmyrlisYiorgos S. Smyrlis

62.9k1384163




62.9k1384163












  • $begingroup$
    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




















  • $begingroup$
    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


















$begingroup$
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






$begingroup$
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






$begingroup$
I have changed the statement. Thank you for the remark.
$endgroup$
– Gustave
Nov 25 '18 at 17:43




















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