Mixing
Projection Formula
$begingroup$
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 Jan 5 at 11:55
Matt Samuel
37.8k63665
asked Jan 5 at 11:00
KaiKai
446
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 11:55
Matt Samuel
37.8k63665
asked Jan 5 at 11:00
KaiKai
446
$endgroup$
2
$begingroup$
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.
$endgroup$
– Michael Burr
Jan 5 at 11:06
add a comment |
$begingroup$
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 Jan 5 at 11:55
Matt Samuel
37.8k63665
asked Jan 5 at 11:00
KaiKai
446
$endgroup$
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 Jan 5 at 11:55
Matt Samuel
37.8k63665
asked Jan 5 at 11:00
KaiKai
446
edited Jan 5 at 11:55
Matt Samuel
37.8k63665
asked Jan 5 at 11:00
KaiKai
446
edited Jan 5 at 11:55
Matt Samuel
37.8k63665
edited Jan 5 at 11:55
Matt Samuel
37.8k63665
edited Jan 5 at 11:55
Matt Samuel
37.8k63665
37.8k63665
asked Jan 5 at 11:00
KaiKai
446
asked Jan 5 at 11:00
KaiKai
446
asked Jan 5 at 11:00
KaiKai
446
446
2
$begingroup$
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.
$endgroup$
– Michael Burr
Jan 5 at 11:06
add a comment |
2
$begingroup$
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.
$endgroup$
– Michael Burr
Jan 5 at 11:06
2
2
$begingroup$
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.
$endgroup$
– Michael Burr
Jan 5 at 11:06
$begingroup$
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.
$endgroup$
– Michael Burr
Jan 5 at 11:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
$endgroup$
$begingroup$
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 ?
$endgroup$
– Kai
Jan 5 at 16:12
$begingroup$
@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.
$endgroup$
– Matt Samuel
Jan 5 at 16:18
add a comment |
Your Answer
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1 Answer
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$begingroup$
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 Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
$endgroup$
$begingroup$
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 ?
$endgroup$
– Kai
Jan 5 at 16:12
$begingroup$
@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.
$endgroup$
– Matt Samuel
Jan 5 at 16:18
add a comment |
$begingroup$
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 Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
$endgroup$
$begingroup$
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 ?
$endgroup$
– Kai
Jan 5 at 16:12
$begingroup$
@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.
$endgroup$
– Matt Samuel
Jan 5 at 16:18
add a comment |
$begingroup$
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 Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
$endgroup$
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 Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
answered Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
answered Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
answered Jan 5 at 11:58
Matt SamuelMatt Samuel
37.8k63665
37.8k63665
$begingroup$
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 ?
$endgroup$
– Kai
Jan 5 at 16:12
$begingroup$
@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.
$endgroup$
– Matt Samuel
Jan 5 at 16:18
add a comment |
$begingroup$
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 ?
$endgroup$
– Kai
Jan 5 at 16:12
$begingroup$
@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.
$endgroup$
– Matt Samuel
Jan 5 at 16:18
$begingroup$
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 ?
$endgroup$
– Kai
Jan 5 at 16:12
$begingroup$
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 ?
$endgroup$
– Kai
Jan 5 at 16:12
$begingroup$
@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.
$endgroup$
– Matt Samuel
Jan 5 at 16:18
$begingroup$
@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.
$endgroup$
– Matt Samuel
Jan 5 at 16:18
add a comment |
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$begingroup$
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.
$endgroup$
– Michael Burr
Jan 5 at 11:06