How many orthogonal projectors on a given subspace are there?
$begingroup$
This question is probably trivial, for that I apologize.
Let's take a vector space $V$ equipped with a scalar product and a vector subspace $Ssubset V$.
How many different projectors onto S, orthogonal with respect to the scalar product, can be constructed?
And what if $V$ is, more generally, a Hilbert space and $S$ is a finite-dimensional subspace?
Intuition tells me that there should be only one, but I am not sure that is correct.
linear-algebra hilbert-spaces orthogonality projection
edited Dec 31 '18 at 18:11
Hanno
2,274628
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
$endgroup$
add a comment |
$begingroup$
This question is probably trivial, for that I apologize.
Let's take a vector space $V$ equipped with a scalar product and a vector subspace $Ssubset V$.
How many different projectors onto S, orthogonal with respect to the scalar product, can be constructed?
And what if $V$ is, more generally, a Hilbert space and $S$ is a finite-dimensional subspace?
Intuition tells me that there should be only one, but I am not sure that is correct.
linear-algebra hilbert-spaces orthogonality projection
edited Dec 31 '18 at 18:11
Hanno
2,274628
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
$endgroup$
$begingroup$
Assuming onto $S$ then $S$ is point-wise fixed and its orthogonal complement must be in the kernel. Now note that $V=Soplus S^{perp}$.
$endgroup$
– WimC
Sep 30 '18 at 10:15
$begingroup$
I am sorry, what do you mean by point-wise fixed? (I didn't study linear algebra in English, so I probably know the concept but not the term)
$endgroup$
– Angelo Brillante Romeo
Sep 30 '18 at 10:39
add a comment |
$begingroup$
This question is probably trivial, for that I apologize.
Let's take a vector space $V$ equipped with a scalar product and a vector subspace $Ssubset V$.
How many different projectors onto S, orthogonal with respect to the scalar product, can be constructed?
And what if $V$ is, more generally, a Hilbert space and $S$ is a finite-dimensional subspace?
Intuition tells me that there should be only one, but I am not sure that is correct.
linear-algebra hilbert-spaces orthogonality projection
edited Dec 31 '18 at 18:11
Hanno
2,274628
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
$endgroup$
This question is probably trivial, for that I apologize.
Let's take a vector space $V$ equipped with a scalar product and a vector subspace $Ssubset V$.
How many different projectors onto S, orthogonal with respect to the scalar product, can be constructed?
And what if $V$ is, more generally, a Hilbert space and $S$ is a finite-dimensional subspace?
Intuition tells me that there should be only one, but I am not sure that is correct.
linear-algebra hilbert-spaces orthogonality projection
linear-algebra hilbert-spaces orthogonality projection
edited Dec 31 '18 at 18:11
Hanno
2,274628
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
edited Dec 31 '18 at 18:11
Hanno
2,274628
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
edited Dec 31 '18 at 18:11
Hanno
2,274628
edited Dec 31 '18 at 18:11
Hanno
2,274628
edited Dec 31 '18 at 18:11
Hanno
2,274628
2,274628
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
asked Sep 30 '18 at 8:47
Angelo Brillante RomeoAngelo Brillante Romeo
1006
1006
$begingroup$
Assuming onto $S$ then $S$ is point-wise fixed and its orthogonal complement must be in the kernel. Now note that $V=Soplus S^{perp}$.
$endgroup$
– WimC
Sep 30 '18 at 10:15
$begingroup$
I am sorry, what do you mean by point-wise fixed? (I didn't study linear algebra in English, so I probably know the concept but not the term)
$endgroup$
– Angelo Brillante Romeo
Sep 30 '18 at 10:39
add a comment |
$begingroup$
Assuming onto $S$ then $S$ is point-wise fixed and its orthogonal complement must be in the kernel. Now note that $V=Soplus S^{perp}$.
$endgroup$
– WimC
Sep 30 '18 at 10:15
$begingroup$
I am sorry, what do you mean by point-wise fixed? (I didn't study linear algebra in English, so I probably know the concept but not the term)
$endgroup$
– Angelo Brillante Romeo
Sep 30 '18 at 10:39
$begingroup$
Assuming onto $S$ then $S$ is point-wise fixed and its orthogonal complement must be in the kernel. Now note that $V=Soplus S^{perp}$.
$endgroup$
– WimC
Sep 30 '18 at 10:15
$begingroup$
Assuming onto $S$ then $S$ is point-wise fixed and its orthogonal complement must be in the kernel. Now note that $V=Soplus S^{perp}$.
$endgroup$
– WimC
Sep 30 '18 at 10:15
$begingroup$
I am sorry, what do you mean by point-wise fixed? (I didn't study linear algebra in English, so I probably know the concept but not the term)
$endgroup$
– Angelo Brillante Romeo
Sep 30 '18 at 10:39
$begingroup$
I am sorry, what do you mean by point-wise fixed? (I didn't study linear algebra in English, so I probably know the concept but not the term)
$endgroup$
– Angelo Brillante Romeo
Sep 30 '18 at 10:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Start by ignoring the scalar (or inner) product, and fix a subspace $Ssubset V$.
All the projectors onto $S,$ (which means that the image equals $S$) are in 1-to-1 correspondence with subspaces $Ksubset V$ such that that $Soplus K=V$. The
mutual relationship between projector and complementing subspace $K$ is determined by $K=ker(text{ projector })$. This holds true in every vector space $V$.
In the lucky case where a scalar product is available for $V$, a distinguished choice for the complementing subspace can be made, namely the orthogonal complement $S^perp$. It is uniquely determined by $S$.
It follows that there is exactly one orthogonal projector onto $S$, thus confirming your intuition.
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
$endgroup$
add a comment |
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$begingroup$
Start by ignoring the scalar (or inner) product, and fix a subspace $Ssubset V$.
All the projectors onto $S,$ (which means that the image equals $S$) are in 1-to-1 correspondence with subspaces $Ksubset V$ such that that $Soplus K=V$. The
mutual relationship between projector and complementing subspace $K$ is determined by $K=ker(text{ projector })$. This holds true in every vector space $V$.
In the lucky case where a scalar product is available for $V$, a distinguished choice for the complementing subspace can be made, namely the orthogonal complement $S^perp$. It is uniquely determined by $S$.
It follows that there is exactly one orthogonal projector onto $S$, thus confirming your intuition.
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
$endgroup$
add a comment |
$begingroup$
Start by ignoring the scalar (or inner) product, and fix a subspace $Ssubset V$.
All the projectors onto $S,$ (which means that the image equals $S$) are in 1-to-1 correspondence with subspaces $Ksubset V$ such that that $Soplus K=V$. The
mutual relationship between projector and complementing subspace $K$ is determined by $K=ker(text{ projector })$. This holds true in every vector space $V$.
In the lucky case where a scalar product is available for $V$, a distinguished choice for the complementing subspace can be made, namely the orthogonal complement $S^perp$. It is uniquely determined by $S$.
It follows that there is exactly one orthogonal projector onto $S$, thus confirming your intuition.
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
$endgroup$
add a comment |
$begingroup$
Start by ignoring the scalar (or inner) product, and fix a subspace $Ssubset V$.
All the projectors onto $S,$ (which means that the image equals $S$) are in 1-to-1 correspondence with subspaces $Ksubset V$ such that that $Soplus K=V$. The
mutual relationship between projector and complementing subspace $K$ is determined by $K=ker(text{ projector })$. This holds true in every vector space $V$.
In the lucky case where a scalar product is available for $V$, a distinguished choice for the complementing subspace can be made, namely the orthogonal complement $S^perp$. It is uniquely determined by $S$.
It follows that there is exactly one orthogonal projector onto $S$, thus confirming your intuition.
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
$endgroup$
Start by ignoring the scalar (or inner) product, and fix a subspace $Ssubset V$.
All the projectors onto $S,$ (which means that the image equals $S$) are in 1-to-1 correspondence with subspaces $Ksubset V$ such that that $Soplus K=V$. The
mutual relationship between projector and complementing subspace $K$ is determined by $K=ker(text{ projector })$. This holds true in every vector space $V$.
In the lucky case where a scalar product is available for $V$, a distinguished choice for the complementing subspace can be made, namely the orthogonal complement $S^perp$. It is uniquely determined by $S$.
It follows that there is exactly one orthogonal projector onto $S$, thus confirming your intuition.
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
answered Dec 31 '18 at 17:54
HannoHanno
2,274628
2,274628
add a comment |
add a comment |
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$begingroup$
Assuming onto $S$ then $S$ is point-wise fixed and its orthogonal complement must be in the kernel. Now note that $V=Soplus S^{perp}$.
$endgroup$
– WimC
Sep 30 '18 at 10:15
$begingroup$
I am sorry, what do you mean by point-wise fixed? (I didn't study linear algebra in English, so I probably know the concept but not the term)
$endgroup$
– Angelo Brillante Romeo
Sep 30 '18 at 10:39