Projection onto the span of linearly independent vectors in Hilbert spaces
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Let $H$ be a Hilbert space and $Ssubseteq H$ a closed subspace. Moreover, let $s_n_n=1^inftysubseteq S$ a complete and linear independent sequence in $S$, i.e.
- $S=overlinetextSpanbig(s_n_n=1^inftybig)$ and
- for $Ngeq 1$ and $lambdainmathbbR^N$, $sum_n=1^Nlambda_n,s_n=0$ implies $lambda=0$.
Denote by $mathcalP$ the orthogonal projection from $H$ onto $S$ and by $mathcalP_N$ the orthogonal projection from $H$ onto $textSpanbig(s_n_n=1^Nbig)$.
Is it true that, for all $xin H$,
$$mathcalP_N(x) quadrightarrowquad mathcalP(x)$$ for
$Nrightarrowinfty$ ?
This is used in a paper without proof or comment and I am wondering how to show this rigorously.
Any help or comment is highly appreciated!
Thanks.
vector-spaces hilbert-spaces orthogonality
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up vote
1
down vote
favorite
Let $H$ be a Hilbert space and $Ssubseteq H$ a closed subspace. Moreover, let $s_n_n=1^inftysubseteq S$ a complete and linear independent sequence in $S$, i.e.
- $S=overlinetextSpanbig(s_n_n=1^inftybig)$ and
- for $Ngeq 1$ and $lambdainmathbbR^N$, $sum_n=1^Nlambda_n,s_n=0$ implies $lambda=0$.
Denote by $mathcalP$ the orthogonal projection from $H$ onto $S$ and by $mathcalP_N$ the orthogonal projection from $H$ onto $textSpanbig(s_n_n=1^Nbig)$.
Is it true that, for all $xin H$,
$$mathcalP_N(x) quadrightarrowquad mathcalP(x)$$ for
$Nrightarrowinfty$ ?
This is used in a paper without proof or comment and I am wondering how to show this rigorously.
Any help or comment is highly appreciated!
Thanks.
vector-spaces hilbert-spaces orthogonality
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $H$ be a Hilbert space and $Ssubseteq H$ a closed subspace. Moreover, let $s_n_n=1^inftysubseteq S$ a complete and linear independent sequence in $S$, i.e.
- $S=overlinetextSpanbig(s_n_n=1^inftybig)$ and
- for $Ngeq 1$ and $lambdainmathbbR^N$, $sum_n=1^Nlambda_n,s_n=0$ implies $lambda=0$.
Denote by $mathcalP$ the orthogonal projection from $H$ onto $S$ and by $mathcalP_N$ the orthogonal projection from $H$ onto $textSpanbig(s_n_n=1^Nbig)$.
Is it true that, for all $xin H$,
$$mathcalP_N(x) quadrightarrowquad mathcalP(x)$$ for
$Nrightarrowinfty$ ?
This is used in a paper without proof or comment and I am wondering how to show this rigorously.
Any help or comment is highly appreciated!
Thanks.
vector-spaces hilbert-spaces orthogonality
Let $H$ be a Hilbert space and $Ssubseteq H$ a closed subspace. Moreover, let $s_n_n=1^inftysubseteq S$ a complete and linear independent sequence in $S$, i.e.
- $S=overlinetextSpanbig(s_n_n=1^inftybig)$ and
- for $Ngeq 1$ and $lambdainmathbbR^N$, $sum_n=1^Nlambda_n,s_n=0$ implies $lambda=0$.
Denote by $mathcalP$ the orthogonal projection from $H$ onto $S$ and by $mathcalP_N$ the orthogonal projection from $H$ onto $textSpanbig(s_n_n=1^Nbig)$.
Is it true that, for all $xin H$,
$$mathcalP_N(x) quadrightarrowquad mathcalP(x)$$ for
$Nrightarrowinfty$ ?
This is used in a paper without proof or comment and I am wondering how to show this rigorously.
Any help or comment is highly appreciated!
Thanks.
vector-spaces hilbert-spaces orthogonality
vector-spaces hilbert-spaces orthogonality
asked Sep 4 at 9:31
Mark
13310
13310
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1 Answer
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Apply Gram-Schmidt process on the sequence $s_n_n=1^infty$ to obtain an orthonormal sequence $e_n_n=1^infty$ in $S$ such that $operatornamespan e_1, ldots, e_n = operatornamespan s_1, ldots, s_n, forall n in mathbbN$ and $operatornamespan e_n_n=1^infty = operatornamespan s_n_n=1^infty$.
Then clearly $e_n_n=1^infty$ is an orthonormal basis for $S$ so
$$P_nx = sum_k=1^n langle x, e_krangle e_k xrightarrowntoinfty sum_k=1^infty langle x, e_krangle e_k = Px$$
for all $x in H$.
Clear and concise. Thanks!
â Mark
Sep 5 at 10:08
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Apply Gram-Schmidt process on the sequence $s_n_n=1^infty$ to obtain an orthonormal sequence $e_n_n=1^infty$ in $S$ such that $operatornamespan e_1, ldots, e_n = operatornamespan s_1, ldots, s_n, forall n in mathbbN$ and $operatornamespan e_n_n=1^infty = operatornamespan s_n_n=1^infty$.
Then clearly $e_n_n=1^infty$ is an orthonormal basis for $S$ so
$$P_nx = sum_k=1^n langle x, e_krangle e_k xrightarrowntoinfty sum_k=1^infty langle x, e_krangle e_k = Px$$
for all $x in H$.
Clear and concise. Thanks!
â Mark
Sep 5 at 10:08
add a comment |Â
up vote
1
down vote
accepted
Apply Gram-Schmidt process on the sequence $s_n_n=1^infty$ to obtain an orthonormal sequence $e_n_n=1^infty$ in $S$ such that $operatornamespan e_1, ldots, e_n = operatornamespan s_1, ldots, s_n, forall n in mathbbN$ and $operatornamespan e_n_n=1^infty = operatornamespan s_n_n=1^infty$.
Then clearly $e_n_n=1^infty$ is an orthonormal basis for $S$ so
$$P_nx = sum_k=1^n langle x, e_krangle e_k xrightarrowntoinfty sum_k=1^infty langle x, e_krangle e_k = Px$$
for all $x in H$.
Clear and concise. Thanks!
â Mark
Sep 5 at 10:08
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Apply Gram-Schmidt process on the sequence $s_n_n=1^infty$ to obtain an orthonormal sequence $e_n_n=1^infty$ in $S$ such that $operatornamespan e_1, ldots, e_n = operatornamespan s_1, ldots, s_n, forall n in mathbbN$ and $operatornamespan e_n_n=1^infty = operatornamespan s_n_n=1^infty$.
Then clearly $e_n_n=1^infty$ is an orthonormal basis for $S$ so
$$P_nx = sum_k=1^n langle x, e_krangle e_k xrightarrowntoinfty sum_k=1^infty langle x, e_krangle e_k = Px$$
for all $x in H$.
Apply Gram-Schmidt process on the sequence $s_n_n=1^infty$ to obtain an orthonormal sequence $e_n_n=1^infty$ in $S$ such that $operatornamespan e_1, ldots, e_n = operatornamespan s_1, ldots, s_n, forall n in mathbbN$ and $operatornamespan e_n_n=1^infty = operatornamespan s_n_n=1^infty$.
Then clearly $e_n_n=1^infty$ is an orthonormal basis for $S$ so
$$P_nx = sum_k=1^n langle x, e_krangle e_k xrightarrowntoinfty sum_k=1^infty langle x, e_krangle e_k = Px$$
for all $x in H$.
answered Sep 5 at 10:04
mechanodroid
24.4k62245
24.4k62245
Clear and concise. Thanks!
â Mark
Sep 5 at 10:08
add a comment |Â
Clear and concise. Thanks!
â Mark
Sep 5 at 10:08
Clear and concise. Thanks!
â Mark
Sep 5 at 10:08
Clear and concise. Thanks!
â Mark
Sep 5 at 10:08
add a comment |Â
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