Orthogonal projection of the form I - V where I is identity matrix and V is also matrix of the same size as I
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I have come across a few papers which talk about creating an orthogonal projection and use the following algebra to create one:
$textbf{U} = textbf{I} - textbf{V} $
where $textbf{I}$ is the identity matrix and $textbf{V}$ is the same size as I (obviously).
I was wondering if someone can clearly explain the geometry behind the above equation. Assume a background of the reader of early college level linear algebra.
A small example perhaps in R2 or R3 would be helpful to explain.
linear-algebra
asked Dec 14 '18 at 15:52
AvedisAvedis
647
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show 2 more comments
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I have come across a few papers which talk about creating an orthogonal projection and use the following algebra to create one:
$textbf{U} = textbf{I} - textbf{V} $
where $textbf{I}$ is the identity matrix and $textbf{V}$ is the same size as I (obviously).
I was wondering if someone can clearly explain the geometry behind the above equation. Assume a background of the reader of early college level linear algebra.
A small example perhaps in R2 or R3 would be helpful to explain.
linear-algebra
asked Dec 14 '18 at 15:52
AvedisAvedis
647
$endgroup$
$begingroup$
Is $U$ or $V$ a projection? In other words, is $V^2=V$ or $U^2=U$ (the defining property of projection)? See en.wikipedia.org/wiki/…
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– jobe
Dec 14 '18 at 16:45
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@jobe V is a projection. You can reference step 17 in Algorithm 1 of this paper if you're very curious. lx.it.pt/~bioucas/files/ieeegrsVca04.pdf
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– Avedis
Dec 14 '18 at 18:10
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Without chasing the link to the paper, I would guess that $V$ is an orthogonal projection onto the orthogonal complement of the space onto which you’re projecting. If the complement has a smaller dimension, it’s often easier to construct that projection directly.
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– amd
Dec 14 '18 at 20:04
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@amd what does this mean geometrically?
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– Avedis
Dec 14 '18 at 20:45
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You can decompose any vector into the sum of its orthogonal projection onto a subspace and a vector orthogonal to that subspace. You can therefore compute the projection by subtracting this orthogonal component from the original vector.
$endgroup$
– amd
Dec 14 '18 at 21:00
|
show 2 more comments
$begingroup$
I have come across a few papers which talk about creating an orthogonal projection and use the following algebra to create one:
$textbf{U} = textbf{I} - textbf{V} $
where $textbf{I}$ is the identity matrix and $textbf{V}$ is the same size as I (obviously).
I was wondering if someone can clearly explain the geometry behind the above equation. Assume a background of the reader of early college level linear algebra.
A small example perhaps in R2 or R3 would be helpful to explain.
linear-algebra
asked Dec 14 '18 at 15:52
AvedisAvedis
647
$endgroup$
I have come across a few papers which talk about creating an orthogonal projection and use the following algebra to create one:
$textbf{U} = textbf{I} - textbf{V} $
where $textbf{I}$ is the identity matrix and $textbf{V}$ is the same size as I (obviously).
I was wondering if someone can clearly explain the geometry behind the above equation. Assume a background of the reader of early college level linear algebra.
A small example perhaps in R2 or R3 would be helpful to explain.
linear-algebra
linear-algebra
asked Dec 14 '18 at 15:52
AvedisAvedis
647
asked Dec 14 '18 at 15:52
AvedisAvedis
647
edited Dec 14 '18 at 20:46
Avedis
asked Dec 14 '18 at 15:52
AvedisAvedis
647
asked Dec 14 '18 at 15:52
AvedisAvedis
647
asked Dec 14 '18 at 15:52
AvedisAvedis
647
647
$begingroup$
Is $U$ or $V$ a projection? In other words, is $V^2=V$ or $U^2=U$ (the defining property of projection)? See en.wikipedia.org/wiki/…
$endgroup$
– jobe
Dec 14 '18 at 16:45
$begingroup$
@jobe V is a projection. You can reference step 17 in Algorithm 1 of this paper if you're very curious. lx.it.pt/~bioucas/files/ieeegrsVca04.pdf
$endgroup$
– Avedis
Dec 14 '18 at 18:10
$begingroup$
Without chasing the link to the paper, I would guess that $V$ is an orthogonal projection onto the orthogonal complement of the space onto which you’re projecting. If the complement has a smaller dimension, it’s often easier to construct that projection directly.
$endgroup$
– amd
Dec 14 '18 at 20:04
$begingroup$
@amd what does this mean geometrically?
$endgroup$
– Avedis
Dec 14 '18 at 20:45
$begingroup$
You can decompose any vector into the sum of its orthogonal projection onto a subspace and a vector orthogonal to that subspace. You can therefore compute the projection by subtracting this orthogonal component from the original vector.
$endgroup$
– amd
Dec 14 '18 at 21:00
|
show 2 more comments
$begingroup$
Is $U$ or $V$ a projection? In other words, is $V^2=V$ or $U^2=U$ (the defining property of projection)? See en.wikipedia.org/wiki/…
$endgroup$
– jobe
Dec 14 '18 at 16:45
$begingroup$
@jobe V is a projection. You can reference step 17 in Algorithm 1 of this paper if you're very curious. lx.it.pt/~bioucas/files/ieeegrsVca04.pdf
$endgroup$
– Avedis
Dec 14 '18 at 18:10
$begingroup$
Without chasing the link to the paper, I would guess that $V$ is an orthogonal projection onto the orthogonal complement of the space onto which you’re projecting. If the complement has a smaller dimension, it’s often easier to construct that projection directly.
$endgroup$
– amd
Dec 14 '18 at 20:04
$begingroup$
@amd what does this mean geometrically?
$endgroup$
– Avedis
Dec 14 '18 at 20:45
$begingroup$
You can decompose any vector into the sum of its orthogonal projection onto a subspace and a vector orthogonal to that subspace. You can therefore compute the projection by subtracting this orthogonal component from the original vector.
$endgroup$
– amd
Dec 14 '18 at 21:00
$begingroup$
Is $U$ or $V$ a projection? In other words, is $V^2=V$ or $U^2=U$ (the defining property of projection)? See en.wikipedia.org/wiki/…
$endgroup$
– jobe
Dec 14 '18 at 16:45
$begingroup$
Is $U$ or $V$ a projection? In other words, is $V^2=V$ or $U^2=U$ (the defining property of projection)? See en.wikipedia.org/wiki/…
$endgroup$
– jobe
Dec 14 '18 at 16:45
$begingroup$
@jobe V is a projection. You can reference step 17 in Algorithm 1 of this paper if you're very curious. lx.it.pt/~bioucas/files/ieeegrsVca04.pdf
$endgroup$
– Avedis
Dec 14 '18 at 18:10
$begingroup$
@jobe V is a projection. You can reference step 17 in Algorithm 1 of this paper if you're very curious. lx.it.pt/~bioucas/files/ieeegrsVca04.pdf
$endgroup$
– Avedis
Dec 14 '18 at 18:10
$begingroup$
Without chasing the link to the paper, I would guess that $V$ is an orthogonal projection onto the orthogonal complement of the space onto which you’re projecting. If the complement has a smaller dimension, it’s often easier to construct that projection directly.
$endgroup$
– amd
Dec 14 '18 at 20:04
$begingroup$
Without chasing the link to the paper, I would guess that $V$ is an orthogonal projection onto the orthogonal complement of the space onto which you’re projecting. If the complement has a smaller dimension, it’s often easier to construct that projection directly.
$endgroup$
– amd
Dec 14 '18 at 20:04
$begingroup$
@amd what does this mean geometrically?
$endgroup$
– Avedis
Dec 14 '18 at 20:45
$begingroup$
@amd what does this mean geometrically?
$endgroup$
– Avedis
Dec 14 '18 at 20:45
$begingroup$
You can decompose any vector into the sum of its orthogonal projection onto a subspace and a vector orthogonal to that subspace. You can therefore compute the projection by subtracting this orthogonal component from the original vector.
$endgroup$
– amd
Dec 14 '18 at 21:00
$begingroup$
You can decompose any vector into the sum of its orthogonal projection onto a subspace and a vector orthogonal to that subspace. You can therefore compute the projection by subtracting this orthogonal component from the original vector.
$endgroup$
– amd
Dec 14 '18 at 21:00
|
show 2 more comments
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$begingroup$
Is $U$ or $V$ a projection? In other words, is $V^2=V$ or $U^2=U$ (the defining property of projection)? See en.wikipedia.org/wiki/…
$endgroup$
– jobe
Dec 14 '18 at 16:45
$begingroup$
@jobe V is a projection. You can reference step 17 in Algorithm 1 of this paper if you're very curious. lx.it.pt/~bioucas/files/ieeegrsVca04.pdf
$endgroup$
– Avedis
Dec 14 '18 at 18:10
$begingroup$
Without chasing the link to the paper, I would guess that $V$ is an orthogonal projection onto the orthogonal complement of the space onto which you’re projecting. If the complement has a smaller dimension, it’s often easier to construct that projection directly.
$endgroup$
– amd
Dec 14 '18 at 20:04
$begingroup$
@amd what does this mean geometrically?
$endgroup$
– Avedis
Dec 14 '18 at 20:45
$begingroup$
You can decompose any vector into the sum of its orthogonal projection onto a subspace and a vector orthogonal to that subspace. You can therefore compute the projection by subtracting this orthogonal component from the original vector.
$endgroup$
– amd
Dec 14 '18 at 21:00