relation between orthogonal projections? [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

favorite

Let $mathbbH$ be a Hilbert space with respect to two different inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Also let $W$ be a closed subspace of $mathbbH$. Also let us assume that $pi_W$ and $pi^1_W$ respectively be the orthogonal projections onto $W$ with respect to the inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Is there exist any relation between $pi_W$ and $pi^1_W$?

functional-analysis hilbert-spaces

share|cite|improve this question

asked Jul 25 at 14:29

SHIBASHIS

636

closed as off-topic by amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister Jul 26 at 0:21

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
1
down vote

favorite

Let $mathbbH$ be a Hilbert space with respect to two different inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Also let $W$ be a closed subspace of $mathbbH$. Also let us assume that $pi_W$ and $pi^1_W$ respectively be the orthogonal projections onto $W$ with respect to the inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Is there exist any relation between $pi_W$ and $pi^1_W$?

functional-analysis hilbert-spaces

share|cite|improve this question

asked Jul 25 at 14:29

SHIBASHIS

636

closed as off-topic by amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister Jul 26 at 0:21

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let $mathbbH$ be a Hilbert space with respect to two different inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Also let $W$ be a closed subspace of $mathbbH$. Also let us assume that $pi_W$ and $pi^1_W$ respectively be the orthogonal projections onto $W$ with respect to the inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Is there exist any relation between $pi_W$ and $pi^1_W$?

functional-analysis hilbert-spaces

share|cite|improve this question

asked Jul 25 at 14:29

SHIBASHIS

636

Let $mathbbH$ be a Hilbert space with respect to two different inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Also let $W$ be a closed subspace of $mathbbH$. Also let us assume that $pi_W$ and $pi^1_W$ respectively be the orthogonal projections onto $W$ with respect to the inner products $[cdot,cdot]$ and $[cdot,cdot]^1$. Is there exist any relation between $pi_W$ and $pi^1_W$?

functional-analysis hilbert-spaces

share|cite|improve this question

asked Jul 25 at 14:29

SHIBASHIS

636

share|cite|improve this question

share|cite|improve this question

asked Jul 25 at 14:29

SHIBASHIS

636

asked Jul 25 at 14:29

SHIBASHIS

636

asked Jul 25 at 14:29

SHIBASHIS

636

636

closed as off-topic by amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister Jul 26 at 0:21

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister Jul 26 at 0:21

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, José Carlos Santos, Chris Custer, max_zorn, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote

Well, as they are both projections onto $W$, they both are idempotent, they both have a range of $W$ (which is the eigenspace corresponding to eigenvalue $1$), and their kernels complement $W$, in that they direct sum with $W$ to give all of $mathbbH$.

Other than that, there's not a lot. If you can find a closed subspace $U$ that complements $W$, then you can define an inner product that that make $U = W^perp$ (or at least you can for separable Hilbert spaces, and I think it holds more generally), and hence $U = operatornameker pi_W$.

share|cite|improve this answer

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

Well, as they are both projections onto $W$, they both are idempotent, they both have a range of $W$ (which is the eigenspace corresponding to eigenvalue $1$), and their kernels complement $W$, in that they direct sum with $W$ to give all of $mathbbH$.

Other than that, there's not a lot. If you can find a closed subspace $U$ that complements $W$, then you can define an inner product that that make $U = W^perp$ (or at least you can for separable Hilbert spaces, and I think it holds more generally), and hence $U = operatornameker pi_W$.

share|cite|improve this answer

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

add a comment | 

up vote
1
down vote

Well, as they are both projections onto $W$, they both are idempotent, they both have a range of $W$ (which is the eigenspace corresponding to eigenvalue $1$), and their kernels complement $W$, in that they direct sum with $W$ to give all of $mathbbH$.

Other than that, there's not a lot. If you can find a closed subspace $U$ that complements $W$, then you can define an inner product that that make $U = W^perp$ (or at least you can for separable Hilbert spaces, and I think it holds more generally), and hence $U = operatornameker pi_W$.

share|cite|improve this answer

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

add a comment | 

up vote
1
down vote

up vote
1
down vote

Well, as they are both projections onto $W$, they both are idempotent, they both have a range of $W$ (which is the eigenspace corresponding to eigenvalue $1$), and their kernels complement $W$, in that they direct sum with $W$ to give all of $mathbbH$.

Other than that, there's not a lot. If you can find a closed subspace $U$ that complements $W$, then you can define an inner product that that make $U = W^perp$ (or at least you can for separable Hilbert spaces, and I think it holds more generally), and hence $U = operatornameker pi_W$.

share|cite|improve this answer

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

Well, as they are both projections onto $W$, they both are idempotent, they both have a range of $W$ (which is the eigenspace corresponding to eigenvalue $1$), and their kernels complement $W$, in that they direct sum with $W$ to give all of $mathbbH$.

Other than that, there's not a lot. If you can find a closed subspace $U$ that complements $W$, then you can define an inner product that that make $U = W^perp$ (or at least you can for separable Hilbert spaces, and I think it holds more generally), and hence $U = operatornameker pi_W$.

share|cite|improve this answer

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

share|cite|improve this answer

share|cite|improve this answer

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

answered Jul 25 at 15:10

Theo Bendit

11.9k1843

11.9k1843

add a comment | 

add a comment |