Matries inequality with norms
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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
$${left| {Pleft( {I - C} right)x} right|_{{{mathbb{R}}^q}}} leqslant alpha {left| {PCx} right|_{{{mathbb{R}}^q}}}$$ for all $x$ in $mathbb{R}$
whith $I$ is the identity matrix.
Thank you.
matrices inequality operator-theory normed-spaces functional-inequalities
asked Nov 25 at 17:31
Gustave
694211
|
show 3 more comments
up vote
-1
down vote
favorite
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
asked Nov 25 at 17:31
Gustave
694211
Why do you expect such an inequality to hold? Could you give us a bit more context?
– Omnomnomnom
Nov 25 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 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 at 17:41
1
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 at 19:34
1
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 at 19:45
|
show 3 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
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
asked Nov 25 at 17:31
Gustave
694211
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
asked Nov 25 at 17:31
Gustave
694211
asked Nov 25 at 17:31
Gustave
694211
edited Nov 25 at 17:42
asked Nov 25 at 17:31
Gustave
694211
asked Nov 25 at 17:31
Gustave
694211
asked Nov 25 at 17:31
Gustave
694211
694211
Why do you expect such an inequality to hold? Could you give us a bit more context?
– Omnomnomnom
Nov 25 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 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 at 17:41
1
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 at 19:34
1
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 at 19:45
|
show 3 more comments
Why do you expect such an inequality to hold? Could you give us a bit more context?
– Omnomnomnom
Nov 25 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 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 at 17:41
1
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 at 19:34
1
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 at 19:45
Why do you expect such an inequality to hold? Could you give us a bit more context?
– Omnomnomnom
Nov 25 at 17:36
Why do you expect such an inequality to hold? Could you give us a bit more context?
– Omnomnomnom
Nov 25 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 at 17:38
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 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 at 17:41
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 at 17:41
1
1
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 at 19:34
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 at 19:34
1
1
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 at 19:45
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 at 19:45
|
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
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 at 17:37
I have changed the statement. Thank you for the remark.
– Gustave
Nov 25 at 17:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
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 at 17:37
I have changed the statement. Thank you for the remark.
– Gustave
Nov 25 at 17:43
add a comment |
up vote
0
down vote
If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
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 at 17:37
I have changed the statement. Thank you for the remark.
– Gustave
Nov 25 at 17:43
add a comment |
up vote
0
down vote
up vote
0
down vote
If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
If $P=I$ and $C=0$, then there is no such positive constant, so that the inequality holds.
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
answered Nov 25 at 17:34
Yiorgos S. Smyrlis
62.2k1383162
62.2k1383162
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 at 17:37
I have changed the statement. Thank you for the remark.
– Gustave
Nov 25 at 17:43
add a comment |
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 at 17:37
I have changed the statement. Thank you for the remark.
– Gustave
Nov 25 at 17:43
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 at 17:37
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 at 17:37
I have changed the statement. Thank you for the remark.
– Gustave
Nov 25 at 17:43
I have changed the statement. Thank you for the remark.
– Gustave
Nov 25 at 17:43
add a comment |
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Why do you expect such an inequality to hold? Could you give us a bit more context?
– Omnomnomnom
Nov 25 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 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 at 17:41
1
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 at 19:34
1
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 at 19:45