$K = {A in Bbb M_n (Bbb R) | A=A^T, mathrm {tr} (A) = 1, x^TAx ge 0 text {for all} x in Bbb R^n }$ is compact...
$begingroup$
Let $K subseteq Bbb M_n (Bbb R)$ be such that
$$K = {A in Bbb M_n (Bbb R) | A=A^T, mathrm {tr} (A) = 1, x^TAx ge 0 text {for all} x in Bbb R^n }.$$
Is $K$ compact in $Bbb M_n (Bbb R)$?
Please help me in this regard.
Thank you very much.
general-topology compactness
edited Mar 11 at 22:00
user26857
39.5k124284
asked Jan 17 at 7:31
math maniac.math maniac.
1677
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closed as off-topic by GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos Feb 4 at 22:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $K subseteq Bbb M_n (Bbb R)$ be such that
$$K = {A in Bbb M_n (Bbb R) | A=A^T, mathrm {tr} (A) = 1, x^TAx ge 0 text {for all} x in Bbb R^n }.$$
Is $K$ compact in $Bbb M_n (Bbb R)$?
Please help me in this regard.
Thank you very much.
general-topology compactness
edited Mar 11 at 22:00
user26857
39.5k124284
asked Jan 17 at 7:31
math maniac.math maniac.
1677
$endgroup$
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos Feb 4 at 22:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What you describe here is the set of all (real $ntimes n$) density matrices which is in fact compact. What have you tried so far regarding this problem?
$endgroup$
– Frederik vom Ende
Jan 17 at 7:37
add a comment |
$begingroup$
Let $K subseteq Bbb M_n (Bbb R)$ be such that
$$K = {A in Bbb M_n (Bbb R) | A=A^T, mathrm {tr} (A) = 1, x^TAx ge 0 text {for all} x in Bbb R^n }.$$
Is $K$ compact in $Bbb M_n (Bbb R)$?
Please help me in this regard.
Thank you very much.
general-topology compactness
edited Mar 11 at 22:00
user26857
39.5k124284
asked Jan 17 at 7:31
math maniac.math maniac.
1677
$endgroup$
Let $K subseteq Bbb M_n (Bbb R)$ be such that
$$K = {A in Bbb M_n (Bbb R) | A=A^T, mathrm {tr} (A) = 1, x^TAx ge 0 text {for all} x in Bbb R^n }.$$
Is $K$ compact in $Bbb M_n (Bbb R)$?
Please help me in this regard.
Thank you very much.
general-topology compactness
general-topology compactness
edited Mar 11 at 22:00
user26857
39.5k124284
asked Jan 17 at 7:31
math maniac.math maniac.
1677
edited Mar 11 at 22:00
user26857
39.5k124284
asked Jan 17 at 7:31
math maniac.math maniac.
1677
edited Mar 11 at 22:00
user26857
39.5k124284
edited Mar 11 at 22:00
user26857
39.5k124284
edited Mar 11 at 22:00
user26857
39.5k124284
39.5k124284
asked Jan 17 at 7:31
math maniac.math maniac.
1677
asked Jan 17 at 7:31
math maniac.math maniac.
1677
asked Jan 17 at 7:31
math maniac.math maniac.
1677
1677
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos Feb 4 at 22:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos Feb 4 at 22:01
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – GNUSupporter 8964民主女神 地下教會, Adrian Keister, callculus, Paul Frost, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
What you describe here is the set of all (real $ntimes n$) density matrices which is in fact compact. What have you tried so far regarding this problem?
$endgroup$
– Frederik vom Ende
Jan 17 at 7:37
add a comment |
$begingroup$
What you describe here is the set of all (real $ntimes n$) density matrices which is in fact compact. What have you tried so far regarding this problem?
$endgroup$
– Frederik vom Ende
Jan 17 at 7:37
$begingroup$
What you describe here is the set of all (real $ntimes n$) density matrices which is in fact compact. What have you tried so far regarding this problem?
$endgroup$
– Frederik vom Ende
Jan 17 at 7:37
$begingroup$
What you describe here is the set of all (real $ntimes n$) density matrices which is in fact compact. What have you tried so far regarding this problem?
$endgroup$
– Frederik vom Ende
Jan 17 at 7:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you are familiar with spectral radius you can argue as follows: the given set is obviously closed. To show that it is bounded note that eigen values of any matrix $A$ in this set are between $0$ and $1$ so the spectral radius is $leq 1$. For a positive definite matrix, the spectral radius is same as the norm. So we have $|A| leq 1$ for all $A$ in this set which makes the set bounded. By Heine - Borel Theorem it is compact.
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
$begingroup$
Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem?
$endgroup$
– math maniac.
Jan 17 at 8:51
$begingroup$
@mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 8:53
add a comment |
$begingroup$
As a physicist i'm not a specialist on this topic, but I figured that $K$ is isomorph to $partialDelta_ntimes SO(n)$, where $Delta_n$ is the $n$-simplex. Both of these factors are compact, and so, I guess, is $K$. In case this argument is wrong, please let me know.
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
answered Jan 17 at 7:53
denklodenklo
4457
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you are familiar with spectral radius you can argue as follows: the given set is obviously closed. To show that it is bounded note that eigen values of any matrix $A$ in this set are between $0$ and $1$ so the spectral radius is $leq 1$. For a positive definite matrix, the spectral radius is same as the norm. So we have $|A| leq 1$ for all $A$ in this set which makes the set bounded. By Heine - Borel Theorem it is compact.
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
$begingroup$
Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem?
$endgroup$
– math maniac.
Jan 17 at 8:51
$begingroup$
@mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 8:53
add a comment |
$begingroup$
If you are familiar with spectral radius you can argue as follows: the given set is obviously closed. To show that it is bounded note that eigen values of any matrix $A$ in this set are between $0$ and $1$ so the spectral radius is $leq 1$. For a positive definite matrix, the spectral radius is same as the norm. So we have $|A| leq 1$ for all $A$ in this set which makes the set bounded. By Heine - Borel Theorem it is compact.
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
$begingroup$
Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem?
$endgroup$
– math maniac.
Jan 17 at 8:51
$begingroup$
@mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 8:53
add a comment |
$begingroup$
If you are familiar with spectral radius you can argue as follows: the given set is obviously closed. To show that it is bounded note that eigen values of any matrix $A$ in this set are between $0$ and $1$ so the spectral radius is $leq 1$. For a positive definite matrix, the spectral radius is same as the norm. So we have $|A| leq 1$ for all $A$ in this set which makes the set bounded. By Heine - Borel Theorem it is compact.
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
If you are familiar with spectral radius you can argue as follows: the given set is obviously closed. To show that it is bounded note that eigen values of any matrix $A$ in this set are between $0$ and $1$ so the spectral radius is $leq 1$. For a positive definite matrix, the spectral radius is same as the norm. So we have $|A| leq 1$ for all $A$ in this set which makes the set bounded. By Heine - Borel Theorem it is compact.
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
edited Jan 18 at 8:39
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
answered Jan 17 at 7:51
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
74.4k53270
$begingroup$
Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem?
$endgroup$
– math maniac.
Jan 17 at 8:51
$begingroup$
@mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 8:53
add a comment |
$begingroup$
Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem?
$endgroup$
– math maniac.
Jan 17 at 8:51
$begingroup$
@mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 8:53
$begingroup$
Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem?
$endgroup$
– math maniac.
Jan 17 at 8:51
$begingroup$
Do you try to say by Heine-Borel theorem instead of Bolzano-Weierstrass theorem?
$endgroup$
– math maniac.
Jan 17 at 8:51
$begingroup$
@mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 8:53
$begingroup$
@mathmaniac. Yes. I have made the same mistake several times! Just absent-mindedness.
$endgroup$
– Kavi Rama Murthy
Jan 17 at 8:53
add a comment |
$begingroup$
As a physicist i'm not a specialist on this topic, but I figured that $K$ is isomorph to $partialDelta_ntimes SO(n)$, where $Delta_n$ is the $n$-simplex. Both of these factors are compact, and so, I guess, is $K$. In case this argument is wrong, please let me know.
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
answered Jan 17 at 7:53
denklodenklo
4457
$endgroup$
add a comment |
$begingroup$
As a physicist i'm not a specialist on this topic, but I figured that $K$ is isomorph to $partialDelta_ntimes SO(n)$, where $Delta_n$ is the $n$-simplex. Both of these factors are compact, and so, I guess, is $K$. In case this argument is wrong, please let me know.
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
answered Jan 17 at 7:53
denklodenklo
4457
$endgroup$
add a comment |
$begingroup$
As a physicist i'm not a specialist on this topic, but I figured that $K$ is isomorph to $partialDelta_ntimes SO(n)$, where $Delta_n$ is the $n$-simplex. Both of these factors are compact, and so, I guess, is $K$. In case this argument is wrong, please let me know.
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
answered Jan 17 at 7:53
denklodenklo
4457
$endgroup$
As a physicist i'm not a specialist on this topic, but I figured that $K$ is isomorph to $partialDelta_ntimes SO(n)$, where $Delta_n$ is the $n$-simplex. Both of these factors are compact, and so, I guess, is $K$. In case this argument is wrong, please let me know.
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
answered Jan 17 at 7:53
denklodenklo
4457
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
edited Feb 4 at 16:26
GNUSupporter 8964民主女神 地下教會
14.1k82651
14.1k82651
answered Jan 17 at 7:53
denklodenklo
4457
answered Jan 17 at 7:53
denklodenklo
4457
answered Jan 17 at 7:53
denklodenklo
4457
4457
add a comment |
add a comment |
$begingroup$
What you describe here is the set of all (real $ntimes n$) density matrices which is in fact compact. What have you tried so far regarding this problem?
$endgroup$
– Frederik vom Ende
Jan 17 at 7:37