General Projections
I have the following problem, however I cannot understand what I exactly have to show, as I am not sure what $x_u$ means.
So can someone tell me if it is the projection of $x$ onto $U$, or is it the orthogonal projection of $x$ onto $U$?
Or is just every projection orthogonal :D ?
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.
vector-spaces projection
edited Jan 4 at 15:43
amWhy
192k28225439
asked Jan 4 at 13:41
KaiKai
206
add a comment |
I have the following problem, however I cannot understand what I exactly have to show, as I am not sure what $x_u$ means.
So can someone tell me if it is the projection of $x$ onto $U$, or is it the orthogonal projection of $x$ onto $U$?
Or is just every projection orthogonal :D ?
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.
vector-spaces projection
edited Jan 4 at 15:43
amWhy
192k28225439
asked Jan 4 at 13:41
KaiKai
206
There is the recent post "How many orthogonal projectors on a given subspace are there?", and the answer is "one". This is explained there. The problem you are going to solve deals with Direct sum decompositions $Uoplus V$ of $,mathbb C^n$.
– Hanno
Jan 4 at 15:30
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:31
This holds for all $x$ if $U$ is 1-dimensional and $uin U$ is a unit vector.
– Hanno
Jan 5 at 9:36
oh should then be $x_u =( <$x,$u>/ $u times $u) * $u
– Kai
Jan 5 at 9:38
add a comment |
I have the following problem, however I cannot understand what I exactly have to show, as I am not sure what $x_u$ means.
So can someone tell me if it is the projection of $x$ onto $U$, or is it the orthogonal projection of $x$ onto $U$?
Or is just every projection orthogonal :D ?
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.
vector-spaces projection
edited Jan 4 at 15:43
amWhy
192k28225439
asked Jan 4 at 13:41
KaiKai
206
I have the following problem, however I cannot understand what I exactly have to show, as I am not sure what $x_u$ means.
So can someone tell me if it is the projection of $x$ onto $U$, or is it the orthogonal projection of $x$ onto $U$?
Or is just every projection orthogonal :D ?
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.
vector-spaces projection
vector-spaces projection
edited Jan 4 at 15:43
amWhy
192k28225439
asked Jan 4 at 13:41
KaiKai
206
edited Jan 4 at 15:43
amWhy
192k28225439
asked Jan 4 at 13:41
KaiKai
206
edited Jan 4 at 15:43
amWhy
192k28225439
edited Jan 4 at 15:43
amWhy
192k28225439
edited Jan 4 at 15:43
amWhy
192k28225439
192k28225439
asked Jan 4 at 13:41
KaiKai
206
asked Jan 4 at 13:41
KaiKai
206
asked Jan 4 at 13:41
KaiKai
206
206
There is the recent post "How many orthogonal projectors on a given subspace are there?", and the answer is "one". This is explained there. The problem you are going to solve deals with Direct sum decompositions $Uoplus V$ of $,mathbb C^n$.
– Hanno
Jan 4 at 15:30
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:31
This holds for all $x$ if $U$ is 1-dimensional and $uin U$ is a unit vector.
– Hanno
Jan 5 at 9:36
oh should then be $x_u =( <$x,$u>/ $u times $u) * $u
– Kai
Jan 5 at 9:38
add a comment |
There is the recent post "How many orthogonal projectors on a given subspace are there?", and the answer is "one". This is explained there. The problem you are going to solve deals with Direct sum decompositions $Uoplus V$ of $,mathbb C^n$.
– Hanno
Jan 4 at 15:30
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:31
This holds for all $x$ if $U$ is 1-dimensional and $uin U$ is a unit vector.
– Hanno
Jan 5 at 9:36
oh should then be $x_u =( <$x,$u>/ $u times $u) * $u
– Kai
Jan 5 at 9:38
There is the recent post "How many orthogonal projectors on a given subspace are there?", and the answer is "one". This is explained there. The problem you are going to solve deals with Direct sum decompositions $Uoplus V$ of $,mathbb C^n$.
– Hanno
Jan 4 at 15:30
There is the recent post "How many orthogonal projectors on a given subspace are there?", and the answer is "one". This is explained there. The problem you are going to solve deals with Direct sum decompositions $Uoplus V$ of $,mathbb C^n$.
– Hanno
Jan 4 at 15:30
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:31
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:31
This holds for all $x$ if $U$ is 1-dimensional and $uin U$ is a unit vector.
– Hanno
Jan 5 at 9:36
This holds for all $x$ if $U$ is 1-dimensional and $uin U$ is a unit vector.
– Hanno
Jan 5 at 9:36
oh should then be $x_u =( <$x,$u>/ $u times $u) * $u
– Kai
Jan 5 at 9:38
oh should then be $x_u =( <$x,$u>/ $u times $u) * $u
– Kai
Jan 5 at 9:38
add a comment |
1 Answer
1
active
oldest
votes
Technically, $x_U$ is just some vector in $U$ which, together with $x_V$ happens to make up $x$. You don't need to interpret it as anything special to do the problem.
But yes, it is a projection of $x$ onto $U$. But not necessarily orthogonal. It is the projection "along" $V$. So if $U$ and $V$ are orthogonal, then yes, $x_U$ is the orthogonal projection of $x$ onto $U$, but not in general.
For instance, if $n = 2$, with $U$ being the span of $(1, 0)$ and $V$ being the span of $(1, 1)$, then the projection is not orthogonal. Given $x = (0, 1)$, we actually have $x_U = (-1, 0)$ and $x_V = (1, 1)$.
answered Jan 4 at 13:51
ArthurArthur
111k7105186
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28
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
Technically, $x_U$ is just some vector in $U$ which, together with $x_V$ happens to make up $x$. You don't need to interpret it as anything special to do the problem.
But yes, it is a projection of $x$ onto $U$. But not necessarily orthogonal. It is the projection "along" $V$. So if $U$ and $V$ are orthogonal, then yes, $x_U$ is the orthogonal projection of $x$ onto $U$, but not in general.
For instance, if $n = 2$, with $U$ being the span of $(1, 0)$ and $V$ being the span of $(1, 1)$, then the projection is not orthogonal. Given $x = (0, 1)$, we actually have $x_U = (-1, 0)$ and $x_V = (1, 1)$.
answered Jan 4 at 13:51
ArthurArthur
111k7105186
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28
add a comment |
Technically, $x_U$ is just some vector in $U$ which, together with $x_V$ happens to make up $x$. You don't need to interpret it as anything special to do the problem.
But yes, it is a projection of $x$ onto $U$. But not necessarily orthogonal. It is the projection "along" $V$. So if $U$ and $V$ are orthogonal, then yes, $x_U$ is the orthogonal projection of $x$ onto $U$, but not in general.
For instance, if $n = 2$, with $U$ being the span of $(1, 0)$ and $V$ being the span of $(1, 1)$, then the projection is not orthogonal. Given $x = (0, 1)$, we actually have $x_U = (-1, 0)$ and $x_V = (1, 1)$.
answered Jan 4 at 13:51
ArthurArthur
111k7105186
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28
add a comment |
Technically, $x_U$ is just some vector in $U$ which, together with $x_V$ happens to make up $x$. You don't need to interpret it as anything special to do the problem.
But yes, it is a projection of $x$ onto $U$. But not necessarily orthogonal. It is the projection "along" $V$. So if $U$ and $V$ are orthogonal, then yes, $x_U$ is the orthogonal projection of $x$ onto $U$, but not in general.
For instance, if $n = 2$, with $U$ being the span of $(1, 0)$ and $V$ being the span of $(1, 1)$, then the projection is not orthogonal. Given $x = (0, 1)$, we actually have $x_U = (-1, 0)$ and $x_V = (1, 1)$.
answered Jan 4 at 13:51
ArthurArthur
111k7105186
Technically, $x_U$ is just some vector in $U$ which, together with $x_V$ happens to make up $x$. You don't need to interpret it as anything special to do the problem.
But yes, it is a projection of $x$ onto $U$. But not necessarily orthogonal. It is the projection "along" $V$. So if $U$ and $V$ are orthogonal, then yes, $x_U$ is the orthogonal projection of $x$ onto $U$, but not in general.
For instance, if $n = 2$, with $U$ being the span of $(1, 0)$ and $V$ being the span of $(1, 1)$, then the projection is not orthogonal. Given $x = (0, 1)$, we actually have $x_U = (-1, 0)$ and $x_V = (1, 1)$.
answered Jan 4 at 13:51
ArthurArthur
111k7105186
answered Jan 4 at 13:51
ArthurArthur
111k7105186
answered Jan 4 at 13:51
ArthurArthur
111k7105186
answered Jan 4 at 13:51
ArthurArthur
111k7105186
111k7105186
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28
add a comment |
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28
add a comment |
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There is the recent post "How many orthogonal projectors on a given subspace are there?", and the answer is "one". This is explained there. The problem you are going to solve deals with Direct sum decompositions $Uoplus V$ of $,mathbb C^n$.
– Hanno
Jan 4 at 15:30
Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:31
This holds for all $x$ if $U$ is 1-dimensional and $uin U$ is a unit vector.
– Hanno
Jan 5 at 9:36
oh should then be $x_u =( <$x,$u>/ $u times $u) * $u
– Kai
Jan 5 at 9:38