General Projections












0














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







share|cite|improve this question
























  • 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


















0














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







share|cite|improve this question
























  • 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
















0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 15:43









amWhy

192k28225439




192k28225439










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




















  • 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












1 Answer
1






active

oldest

votes


















2














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 $x_V = (1, 1)$.






share|cite|improve this answer





















  • Is then $x_u = < $x , $u > $u ?
    – Kai
    Jan 5 at 9:28













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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 $x_V = (1, 1)$.






share|cite|improve this answer





















  • Is then $x_u = < $x , $u > $u ?
    – Kai
    Jan 5 at 9:28


















2














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 $x_V = (1, 1)$.






share|cite|improve this answer





















  • Is then $x_u = < $x , $u > $u ?
    – Kai
    Jan 5 at 9:28
















2












2








2






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 $x_V = (1, 1)$.






share|cite|improve this answer












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 $x_V = (1, 1)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 13:51









ArthurArthur

111k7105186




111k7105186












  • 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






Is then $x_u = < $x , $u > $u ?
– Kai
Jan 5 at 9:28




















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