Projection Formula
Does someone know if in the problem the projection of x onto U is defined like that :
$x_u = displaystyle frac{langle x,urangle}{u. u}$ $u$
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
Show that every $xinmathbb C^n$ can be written as $x=x_u+x_v$ with $x_uin U$ and $x_vin V$ and that this decomposition is unique.
projection
edited 2 days ago
Matt Samuel
37.5k63565
asked 2 days ago
Kai
256
add a comment |
Does someone know if in the problem the projection of x onto U is defined like that :
$x_u = displaystyle frac{langle x,urangle}{u. u}$ $u$
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
Show that every $xinmathbb C^n$ can be written as $x=x_u+x_v$ with $x_uin U$ and $x_vin V$ and that this decomposition is unique.
projection
edited 2 days ago
Matt Samuel
37.5k63565
asked 2 days ago
Kai
256
2
That formula is for a projection onto a vector, not a subspace. So you need vectors to make it work (hint: use a basis). This, however, might not be the best approach to this problem.
– Michael Burr
2 days ago
add a comment |
Does someone know if in the problem the projection of x onto U is defined like that :
$x_u = displaystyle frac{langle x,urangle}{u. u}$ $u$
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
Show that every $xinmathbb C^n$ can be written as $x=x_u+x_v$ with $x_uin U$ and $x_vin V$ and that this decomposition is unique.
projection
edited 2 days ago
Matt Samuel
37.5k63565
asked 2 days ago
Kai
256
Does someone know if in the problem the projection of x onto U is defined like that :
$x_u = displaystyle frac{langle x,urangle}{u. u}$ $u$
Problem:
Let $U,Vsubsetmathbb C^n$ be two subspaces, such that $mathbb C^n = U+V$ and further assume $Ucap V = {0}$.
Show that every $xinmathbb C^n$ can be written as $x=x_u+x_v$ with $x_uin U$ and $x_vin V$ and that this decomposition is unique.
projection
projection
edited 2 days ago
Matt Samuel
37.5k63565
asked 2 days ago
Kai
256
edited 2 days ago
Matt Samuel
37.5k63565
asked 2 days ago
Kai
256
edited 2 days ago
Matt Samuel
37.5k63565
edited 2 days ago
Matt Samuel
37.5k63565
edited 2 days ago
Matt Samuel
37.5k63565
37.5k63565
asked 2 days ago
Kai
256
asked 2 days ago
Kai
256
asked 2 days ago
Kai
256
256
2
That formula is for a projection onto a vector, not a subspace. So you need vectors to make it work (hint: use a basis). This, however, might not be the best approach to this problem.
– Michael Burr
2 days ago
add a comment |
2
That formula is for a projection onto a vector, not a subspace. So you need vectors to make it work (hint: use a basis). This, however, might not be the best approach to this problem.
– Michael Burr
2 days ago
2
2
That formula is for a projection onto a vector, not a subspace. So you need vectors to make it work (hint: use a basis). This, however, might not be the best approach to this problem.
– Michael Burr
2 days ago
That formula is for a projection onto a vector, not a subspace. So you need vectors to make it work (hint: use a basis). This, however, might not be the best approach to this problem.
– Michael Burr
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
If $xin mathbb{C}^n$, then since $mathbb{C}^n = U+V$ there must exist $x_uin U$ and $x_vin v$ such that
$$x=x_u+x_v$$
Suppose now that there exist alternative $y_u,y_v$ with
$$x=y_u+y_v$$
Then
$$x-x=0=(x_u-y_u)+(x_v-y_v)$$
Hence
$$y_u-x_u=x_v-y_v$$
Since $U$ and $V$ are subspaces, $y_u-x_uin U$ and $x_v-y_vin V$. Thus these vectors are in $Ucap V={0}$, hence $y_u-x_u=x_v-y_v=0$, and hence the decomposition is unique.
Clearly we didn't need projection here. Indeed this proof works for infinite dimensional vector spaces as well, as well as vector spaces over an arbitrary field (not necessarily real or complex numbers, or even characteristic $0$), and an inner product is not required.
answered 2 days ago
Matt Samuel
37.5k63565
Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ?
– Kai
yesterday
@Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection.
– Matt Samuel
yesterday
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $xin mathbb{C}^n$, then since $mathbb{C}^n = U+V$ there must exist $x_uin U$ and $x_vin v$ such that
$$x=x_u+x_v$$
Suppose now that there exist alternative $y_u,y_v$ with
$$x=y_u+y_v$$
Then
$$x-x=0=(x_u-y_u)+(x_v-y_v)$$
Hence
$$y_u-x_u=x_v-y_v$$
Since $U$ and $V$ are subspaces, $y_u-x_uin U$ and $x_v-y_vin V$. Thus these vectors are in $Ucap V={0}$, hence $y_u-x_u=x_v-y_v=0$, and hence the decomposition is unique.
Clearly we didn't need projection here. Indeed this proof works for infinite dimensional vector spaces as well, as well as vector spaces over an arbitrary field (not necessarily real or complex numbers, or even characteristic $0$), and an inner product is not required.
answered 2 days ago
Matt Samuel
37.5k63565
Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ?
– Kai
yesterday
@Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection.
– Matt Samuel
yesterday
add a comment |
If $xin mathbb{C}^n$, then since $mathbb{C}^n = U+V$ there must exist $x_uin U$ and $x_vin v$ such that
$$x=x_u+x_v$$
Suppose now that there exist alternative $y_u,y_v$ with
$$x=y_u+y_v$$
Then
$$x-x=0=(x_u-y_u)+(x_v-y_v)$$
Hence
$$y_u-x_u=x_v-y_v$$
Since $U$ and $V$ are subspaces, $y_u-x_uin U$ and $x_v-y_vin V$. Thus these vectors are in $Ucap V={0}$, hence $y_u-x_u=x_v-y_v=0$, and hence the decomposition is unique.
Clearly we didn't need projection here. Indeed this proof works for infinite dimensional vector spaces as well, as well as vector spaces over an arbitrary field (not necessarily real or complex numbers, or even characteristic $0$), and an inner product is not required.
answered 2 days ago
Matt Samuel
37.5k63565
Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ?
– Kai
yesterday
@Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection.
– Matt Samuel
yesterday
add a comment |
If $xin mathbb{C}^n$, then since $mathbb{C}^n = U+V$ there must exist $x_uin U$ and $x_vin v$ such that
$$x=x_u+x_v$$
Suppose now that there exist alternative $y_u,y_v$ with
$$x=y_u+y_v$$
Then
$$x-x=0=(x_u-y_u)+(x_v-y_v)$$
Hence
$$y_u-x_u=x_v-y_v$$
Since $U$ and $V$ are subspaces, $y_u-x_uin U$ and $x_v-y_vin V$. Thus these vectors are in $Ucap V={0}$, hence $y_u-x_u=x_v-y_v=0$, and hence the decomposition is unique.
Clearly we didn't need projection here. Indeed this proof works for infinite dimensional vector spaces as well, as well as vector spaces over an arbitrary field (not necessarily real or complex numbers, or even characteristic $0$), and an inner product is not required.
answered 2 days ago
Matt Samuel
37.5k63565
If $xin mathbb{C}^n$, then since $mathbb{C}^n = U+V$ there must exist $x_uin U$ and $x_vin v$ such that
$$x=x_u+x_v$$
Suppose now that there exist alternative $y_u,y_v$ with
$$x=y_u+y_v$$
Then
$$x-x=0=(x_u-y_u)+(x_v-y_v)$$
Hence
$$y_u-x_u=x_v-y_v$$
Since $U$ and $V$ are subspaces, $y_u-x_uin U$ and $x_v-y_vin V$. Thus these vectors are in $Ucap V={0}$, hence $y_u-x_u=x_v-y_v=0$, and hence the decomposition is unique.
Clearly we didn't need projection here. Indeed this proof works for infinite dimensional vector spaces as well, as well as vector spaces over an arbitrary field (not necessarily real or complex numbers, or even characteristic $0$), and an inner product is not required.
answered 2 days ago
Matt Samuel
37.5k63565
answered 2 days ago
Matt Samuel
37.5k63565
answered 2 days ago
Matt Samuel
37.5k63565
answered 2 days ago
Matt Samuel
37.5k63565
37.5k63565
Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ?
– Kai
yesterday
@Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection.
– Matt Samuel
yesterday
add a comment |
Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ?
– Kai
yesterday
@Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection.
– Matt Samuel
yesterday
Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ?
– Kai
yesterday
Thank you very much for your answer,but I still cannot understand how this proves that the sum of the two projections equals x . That's why I am asking what actually is x onto U ?
– Kai
yesterday
@Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection.
– Matt Samuel
yesterday
@Kai Because the whole space is equal to $U+V$, by definition every vector is the sum of a vector in $U$ and a vector in $V$. This proof shows that it's unique. It's not a projection onto $U$. Protection onto $U$ essentially removes the orthogonal complement of $U$, which might not be $V$, so the component in $U$ wouldn't necessarily be the projection.
– Matt Samuel
yesterday
add a comment |
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That formula is for a projection onto a vector, not a subspace. So you need vectors to make it work (hint: use a basis). This, however, might not be the best approach to this problem.
– Michael Burr
2 days ago