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 15 hours ago
amWhy
192k28224439
asked 18 hours ago
Kai
164
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 15 hours ago
amWhy
192k28224439
asked 18 hours ago
Kai
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
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 15 hours ago
amWhy
192k28224439
asked 18 hours ago
Kai
164
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 15 hours ago
amWhy
192k28224439
asked 18 hours ago
Kai
164
edited 15 hours ago
amWhy
192k28224439
asked 18 hours ago
Kai
164
edited 15 hours ago
amWhy
192k28224439
edited 15 hours ago
amWhy
192k28224439
edited 15 hours ago
amWhy
192k28224439
192k28224439
asked 18 hours ago
Kai
164
asked 18 hours ago
Kai
164
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
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
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
add a comment |
1 Answer
1
active
oldest
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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 $x_V = (1, 1)$.
answered 17 hours ago
Arthur
111k7105186
add a comment |
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1 Answer
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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 17 hours ago
Arthur
111k7105186
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 17 hours ago
Arthur
111k7105186
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 17 hours ago
Arthur
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 17 hours ago
Arthur
111k7105186
answered 17 hours ago
Arthur
111k7105186
answered 17 hours ago
Arthur
111k7105186
answered 17 hours ago
Arthur
111k7105186
111k7105186
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
16 hours ago