Projection Formula












1














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







share|cite|improve this question




















  • 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
















1














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







share|cite|improve this question




















  • 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














1












1








1







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Matt Samuel

37.5k63565




37.5k63565










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














  • 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










1 Answer
1






active

oldest

votes


















1














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.






share|cite|improve this answer





















  • 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













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1 Answer
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1














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.






share|cite|improve this answer





















  • 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


















1














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.






share|cite|improve this answer





















  • 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
















1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • 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




















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