The Order of Orthogonality [closed]
$begingroup$
I would like to show that $Bsubset A$ implies $A^{bot}subset B^{bot}$.
Note the meaning behind this: The bigger a subset, the smaller its orthogonal should be.
Let $x$ be in the complement of A. Then $langle x,yrangle =0$ for all $yin A$. But since $Bsubset A$, $langle x,yrangle=0$ for all $yin B$. Is it true?
functional-analysis inner-product-space orthogonality
edited Jan 13 at 19:06
Hanno
2,509629
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
$endgroup$
closed as off-topic by José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho Jan 13 at 21:37
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." – José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I would like to show that $Bsubset A$ implies $A^{bot}subset B^{bot}$.
Note the meaning behind this: The bigger a subset, the smaller its orthogonal should be.
Let $x$ be in the complement of A. Then $langle x,yrangle =0$ for all $yin A$. But since $Bsubset A$, $langle x,yrangle=0$ for all $yin B$. Is it true?
functional-analysis inner-product-space orthogonality
edited Jan 13 at 19:06
Hanno
2,509629
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
$endgroup$
closed as off-topic by José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho Jan 13 at 21:37
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." – José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 13 at 18:27
$begingroup$
Note that meaning behind this, the bigger a subset, the smaller its ortogonal should be. Let x be in the complement of A. Then, <x,y>=0 for all y in A. But since, B is least then A, <x,y>=0 for all y in B. Is it true ?
$endgroup$
– mathsstudent
Jan 13 at 18:33
2
$begingroup$
Be careful about what you mean by "B is less than A": being "smaller" is not sufficient. Literally just write out the definitions and you'll have three quarters of the proof.
$endgroup$
– user3482749
Jan 13 at 18:38
add a comment |
$begingroup$
I would like to show that $Bsubset A$ implies $A^{bot}subset B^{bot}$.
Note the meaning behind this: The bigger a subset, the smaller its orthogonal should be.
Let $x$ be in the complement of A. Then $langle x,yrangle =0$ for all $yin A$. But since $Bsubset A$, $langle x,yrangle=0$ for all $yin B$. Is it true?
functional-analysis inner-product-space orthogonality
edited Jan 13 at 19:06
Hanno
2,509629
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
$endgroup$
I would like to show that $Bsubset A$ implies $A^{bot}subset B^{bot}$.
Note the meaning behind this: The bigger a subset, the smaller its orthogonal should be.
Let $x$ be in the complement of A. Then $langle x,yrangle =0$ for all $yin A$. But since $Bsubset A$, $langle x,yrangle=0$ for all $yin B$. Is it true?
functional-analysis inner-product-space orthogonality
functional-analysis inner-product-space orthogonality
edited Jan 13 at 19:06
Hanno
2,509629
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
edited Jan 13 at 19:06
Hanno
2,509629
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
edited Jan 13 at 19:06
Hanno
2,509629
edited Jan 13 at 19:06
Hanno
2,509629
edited Jan 13 at 19:06
Hanno
2,509629
2,509629
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
asked Jan 13 at 18:25
mathsstudentmathsstudent
536
536
closed as off-topic by José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho Jan 13 at 21:37
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." – José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho Jan 13 at 21:37
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." – José Carlos Santos, anomaly, Morgan Rodgers, Namaste, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 13 at 18:27
$begingroup$
Note that meaning behind this, the bigger a subset, the smaller its ortogonal should be. Let x be in the complement of A. Then, <x,y>=0 for all y in A. But since, B is least then A, <x,y>=0 for all y in B. Is it true ?
$endgroup$
– mathsstudent
Jan 13 at 18:33
2
$begingroup$
Be careful about what you mean by "B is less than A": being "smaller" is not sufficient. Literally just write out the definitions and you'll have three quarters of the proof.
$endgroup$
– user3482749
Jan 13 at 18:38
add a comment |
1
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 13 at 18:27
$begingroup$
Note that meaning behind this, the bigger a subset, the smaller its ortogonal should be. Let x be in the complement of A. Then, <x,y>=0 for all y in A. But since, B is least then A, <x,y>=0 for all y in B. Is it true ?
$endgroup$
– mathsstudent
Jan 13 at 18:33
2
$begingroup$
Be careful about what you mean by "B is less than A": being "smaller" is not sufficient. Literally just write out the definitions and you'll have three quarters of the proof.
$endgroup$
– user3482749
Jan 13 at 18:38
1
1
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 13 at 18:27
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 13 at 18:27
$begingroup$
Note that meaning behind this, the bigger a subset, the smaller its ortogonal should be. Let x be in the complement of A. Then, <x,y>=0 for all y in A. But since, B is least then A, <x,y>=0 for all y in B. Is it true ?
$endgroup$
– mathsstudent
Jan 13 at 18:33
$begingroup$
Note that meaning behind this, the bigger a subset, the smaller its ortogonal should be. Let x be in the complement of A. Then, <x,y>=0 for all y in A. But since, B is least then A, <x,y>=0 for all y in B. Is it true ?
$endgroup$
– mathsstudent
Jan 13 at 18:33
2
2
$begingroup$
Be careful about what you mean by "B is less than A": being "smaller" is not sufficient. Literally just write out the definitions and you'll have three quarters of the proof.
$endgroup$
– user3482749
Jan 13 at 18:38
$begingroup$
Be careful about what you mean by "B is less than A": being "smaller" is not sufficient. Literally just write out the definitions and you'll have three quarters of the proof.
$endgroup$
– user3482749
Jan 13 at 18:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$xin A^botiff xnotin Aoverbrace{Longrightarrow}^{Bsubset A} xnotin Biff xin B^bot$$
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Thanks your help.
$endgroup$
– mathsstudent
Jan 13 at 19:04
$begingroup$
Wish you luck!.
$endgroup$
– Mostafa Ayaz
Jan 13 at 19:05
$begingroup$
Why do you say that $xnotin ARightarrow xin A^{perp}$?
$endgroup$
– Lorenzo Quarisa
Jan 13 at 20:20
$begingroup$
Since the definition of $A^bot$ we are able to say so$$A^bot={xin U| xnot A}$$where $U$ is the Universal Set.
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:22
add a comment |
$begingroup$
Yes, your argument is correct and straightforward.
Although true on the level of set inclusions, one should be aware
that the implication $:A^perpsubset B^perp:$ merely depends on the vector subspace $,operatorname{span}(B),$ being a subspace of $,operatorname{span}(A),$ (as implied by $Bsubset A$), and not on the "bigness" of $A$ and $B$ in the sense of sets.
A note regarding terminology: $V^perp$ is usually called "orthogonal complement" only if $V$ is a subspace, else $Voplus V^perp$ won't yield the entire ambient vector space.
answered Jan 13 at 20:18
HannoHanno
2,509629
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$xin A^botiff xnotin Aoverbrace{Longrightarrow}^{Bsubset A} xnotin Biff xin B^bot$$
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Thanks your help.
$endgroup$
– mathsstudent
Jan 13 at 19:04
$begingroup$
Wish you luck!.
$endgroup$
– Mostafa Ayaz
Jan 13 at 19:05
$begingroup$
Why do you say that $xnotin ARightarrow xin A^{perp}$?
$endgroup$
– Lorenzo Quarisa
Jan 13 at 20:20
$begingroup$
Since the definition of $A^bot$ we are able to say so$$A^bot={xin U| xnot A}$$where $U$ is the Universal Set.
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:22
add a comment |
$begingroup$
$$xin A^botiff xnotin Aoverbrace{Longrightarrow}^{Bsubset A} xnotin Biff xin B^bot$$
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$begingroup$
Thanks your help.
$endgroup$
– mathsstudent
Jan 13 at 19:04
$begingroup$
Wish you luck!.
$endgroup$
– Mostafa Ayaz
Jan 13 at 19:05
$begingroup$
Why do you say that $xnotin ARightarrow xin A^{perp}$?
$endgroup$
– Lorenzo Quarisa
Jan 13 at 20:20
$begingroup$
Since the definition of $A^bot$ we are able to say so$$A^bot={xin U| xnot A}$$where $U$ is the Universal Set.
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:22
add a comment |
$begingroup$
$$xin A^botiff xnotin Aoverbrace{Longrightarrow}^{Bsubset A} xnotin Biff xin B^bot$$
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
$$xin A^botiff xnotin Aoverbrace{Longrightarrow}^{Bsubset A} xnotin Biff xin B^bot$$
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
edited Jan 13 at 21:05
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
answered Jan 13 at 18:48
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
$begingroup$
Thanks your help.
$endgroup$
– mathsstudent
Jan 13 at 19:04
$begingroup$
Wish you luck!.
$endgroup$
– Mostafa Ayaz
Jan 13 at 19:05
$begingroup$
Why do you say that $xnotin ARightarrow xin A^{perp}$?
$endgroup$
– Lorenzo Quarisa
Jan 13 at 20:20
$begingroup$
Since the definition of $A^bot$ we are able to say so$$A^bot={xin U| xnot A}$$where $U$ is the Universal Set.
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:22
add a comment |
$begingroup$
Thanks your help.
$endgroup$
– mathsstudent
Jan 13 at 19:04
$begingroup$
Wish you luck!.
$endgroup$
– Mostafa Ayaz
Jan 13 at 19:05
$begingroup$
Why do you say that $xnotin ARightarrow xin A^{perp}$?
$endgroup$
– Lorenzo Quarisa
Jan 13 at 20:20
$begingroup$
Since the definition of $A^bot$ we are able to say so$$A^bot={xin U| xnot A}$$where $U$ is the Universal Set.
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:22
$begingroup$
Thanks your help.
$endgroup$
– mathsstudent
Jan 13 at 19:04
$begingroup$
Thanks your help.
$endgroup$
– mathsstudent
Jan 13 at 19:04
$begingroup$
Wish you luck!.
$endgroup$
– Mostafa Ayaz
Jan 13 at 19:05
$begingroup$
Wish you luck!.
$endgroup$
– Mostafa Ayaz
Jan 13 at 19:05
$begingroup$
Why do you say that $xnotin ARightarrow xin A^{perp}$?
$endgroup$
– Lorenzo Quarisa
Jan 13 at 20:20
$begingroup$
Why do you say that $xnotin ARightarrow xin A^{perp}$?
$endgroup$
– Lorenzo Quarisa
Jan 13 at 20:20
$begingroup$
Since the definition of $A^bot$ we are able to say so$$A^bot={xin U| xnot A}$$where $U$ is the Universal Set.
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:22
$begingroup$
Since the definition of $A^bot$ we are able to say so$$A^bot={xin U| xnot A}$$where $U$ is the Universal Set.
$endgroup$
– Mostafa Ayaz
Jan 13 at 20:22
add a comment |
$begingroup$
Yes, your argument is correct and straightforward.
Although true on the level of set inclusions, one should be aware
that the implication $:A^perpsubset B^perp:$ merely depends on the vector subspace $,operatorname{span}(B),$ being a subspace of $,operatorname{span}(A),$ (as implied by $Bsubset A$), and not on the "bigness" of $A$ and $B$ in the sense of sets.
A note regarding terminology: $V^perp$ is usually called "orthogonal complement" only if $V$ is a subspace, else $Voplus V^perp$ won't yield the entire ambient vector space.
answered Jan 13 at 20:18
HannoHanno
2,509629
$endgroup$
add a comment |
$begingroup$
Yes, your argument is correct and straightforward.
Although true on the level of set inclusions, one should be aware
that the implication $:A^perpsubset B^perp:$ merely depends on the vector subspace $,operatorname{span}(B),$ being a subspace of $,operatorname{span}(A),$ (as implied by $Bsubset A$), and not on the "bigness" of $A$ and $B$ in the sense of sets.
A note regarding terminology: $V^perp$ is usually called "orthogonal complement" only if $V$ is a subspace, else $Voplus V^perp$ won't yield the entire ambient vector space.
answered Jan 13 at 20:18
HannoHanno
2,509629
$endgroup$
add a comment |
$begingroup$
Yes, your argument is correct and straightforward.
Although true on the level of set inclusions, one should be aware
that the implication $:A^perpsubset B^perp:$ merely depends on the vector subspace $,operatorname{span}(B),$ being a subspace of $,operatorname{span}(A),$ (as implied by $Bsubset A$), and not on the "bigness" of $A$ and $B$ in the sense of sets.
A note regarding terminology: $V^perp$ is usually called "orthogonal complement" only if $V$ is a subspace, else $Voplus V^perp$ won't yield the entire ambient vector space.
answered Jan 13 at 20:18
HannoHanno
2,509629
$endgroup$
Yes, your argument is correct and straightforward.
Although true on the level of set inclusions, one should be aware
that the implication $:A^perpsubset B^perp:$ merely depends on the vector subspace $,operatorname{span}(B),$ being a subspace of $,operatorname{span}(A),$ (as implied by $Bsubset A$), and not on the "bigness" of $A$ and $B$ in the sense of sets.
A note regarding terminology: $V^perp$ is usually called "orthogonal complement" only if $V$ is a subspace, else $Voplus V^perp$ won't yield the entire ambient vector space.
answered Jan 13 at 20:18
HannoHanno
2,509629
answered Jan 13 at 20:18
HannoHanno
2,509629
answered Jan 13 at 20:18
HannoHanno
2,509629
answered Jan 13 at 20:18
HannoHanno
2,509629
2,509629
add a comment |
add a comment |
1
$begingroup$
What have you tried so far?
$endgroup$
– user3482749
Jan 13 at 18:27
$begingroup$
Note that meaning behind this, the bigger a subset, the smaller its ortogonal should be. Let x be in the complement of A. Then, <x,y>=0 for all y in A. But since, B is least then A, <x,y>=0 for all y in B. Is it true ?
$endgroup$
– mathsstudent
Jan 13 at 18:33
2
$begingroup$
Be careful about what you mean by "B is less than A": being "smaller" is not sufficient. Literally just write out the definitions and you'll have three quarters of the proof.
$endgroup$
– user3482749
Jan 13 at 18:38