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







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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
    16 hours ago
















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
    16 hours ago














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 15 hours ago









amWhy

192k28224439




192k28224439










asked 18 hours ago









Kai

164




164












  • 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
    16 hours ago


















  • 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
    16 hours ago
















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
16 hours ago




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
16 hours ago










1 Answer
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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 $x_V = (1, 1)$.






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    1 Answer
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    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


























      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
























        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 17 hours ago









        Arthur

        111k7105186




        111k7105186






























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