Is the representation of a linear affine space unique?
$begingroup$
Suppose $S subset mathbb R^n$ is a linear affine subspace. I know picking any $s in S$, $S- s =: U$ is a subspace and we can write $S = s + U$. Now consider writing $s = s_{U} + s_{U^{perp}}$ where the subscripts denote the orthogonal projection. Then $S = s_{U^{perp}} + U$. I would like to know whether this representation, i.e., $s_{U^{perp}}$, is a uniquely defined vector.
linear-algebra projection
asked Nov 19 '18 at 18:47
user43210user43210
406
$endgroup$
add a comment |
$begingroup$
Suppose $S subset mathbb R^n$ is a linear affine subspace. I know picking any $s in S$, $S- s =: U$ is a subspace and we can write $S = s + U$. Now consider writing $s = s_{U} + s_{U^{perp}}$ where the subscripts denote the orthogonal projection. Then $S = s_{U^{perp}} + U$. I would like to know whether this representation, i.e., $s_{U^{perp}}$, is a uniquely defined vector.
linear-algebra projection
asked Nov 19 '18 at 18:47
user43210user43210
406
$endgroup$
add a comment |
$begingroup$
Suppose $S subset mathbb R^n$ is a linear affine subspace. I know picking any $s in S$, $S- s =: U$ is a subspace and we can write $S = s + U$. Now consider writing $s = s_{U} + s_{U^{perp}}$ where the subscripts denote the orthogonal projection. Then $S = s_{U^{perp}} + U$. I would like to know whether this representation, i.e., $s_{U^{perp}}$, is a uniquely defined vector.
linear-algebra projection
asked Nov 19 '18 at 18:47
user43210user43210
406
$endgroup$
Suppose $S subset mathbb R^n$ is a linear affine subspace. I know picking any $s in S$, $S- s =: U$ is a subspace and we can write $S = s + U$. Now consider writing $s = s_{U} + s_{U^{perp}}$ where the subscripts denote the orthogonal projection. Then $S = s_{U^{perp}} + U$. I would like to know whether this representation, i.e., $s_{U^{perp}}$, is a uniquely defined vector.
linear-algebra projection
linear-algebra projection
asked Nov 19 '18 at 18:47
user43210user43210
406
asked Nov 19 '18 at 18:47
user43210user43210
406
edited Dec 12 '18 at 17:23
user43210
asked Nov 19 '18 at 18:47
user43210user43210
406
asked Nov 19 '18 at 18:47
user43210user43210
406
asked Nov 19 '18 at 18:47
user43210user43210
406
406
add a comment |
add a comment |
1 Answer
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$begingroup$
Yes, it is.
The $U$ in the representation $S=s+U$ is the same for all $sin S$, because it is defined as $U{s-t:s,tin S}$.
As a consequence you have that $s_U^perp = t_U^perp$ for all $s,tin S$,
because $s-t=(s_U-t_U)+(s_U^perp-t_U^perp) in U$ and $s_U-t_Uin U$ imply $s_U^perp-t_U^perpin U$, so it has to be $0$.
Moreover this vector can be characterized as the projection of $0$ on $S$.
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, it is.
The $U$ in the representation $S=s+U$ is the same for all $sin S$, because it is defined as $U{s-t:s,tin S}$.
As a consequence you have that $s_U^perp = t_U^perp$ for all $s,tin S$,
because $s-t=(s_U-t_U)+(s_U^perp-t_U^perp) in U$ and $s_U-t_Uin U$ imply $s_U^perp-t_U^perpin U$, so it has to be $0$.
Moreover this vector can be characterized as the projection of $0$ on $S$.
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
$endgroup$
add a comment |
$begingroup$
Yes, it is.
The $U$ in the representation $S=s+U$ is the same for all $sin S$, because it is defined as $U{s-t:s,tin S}$.
As a consequence you have that $s_U^perp = t_U^perp$ for all $s,tin S$,
because $s-t=(s_U-t_U)+(s_U^perp-t_U^perp) in U$ and $s_U-t_Uin U$ imply $s_U^perp-t_U^perpin U$, so it has to be $0$.
Moreover this vector can be characterized as the projection of $0$ on $S$.
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
$endgroup$
add a comment |
$begingroup$
Yes, it is.
The $U$ in the representation $S=s+U$ is the same for all $sin S$, because it is defined as $U{s-t:s,tin S}$.
As a consequence you have that $s_U^perp = t_U^perp$ for all $s,tin S$,
because $s-t=(s_U-t_U)+(s_U^perp-t_U^perp) in U$ and $s_U-t_Uin U$ imply $s_U^perp-t_U^perpin U$, so it has to be $0$.
Moreover this vector can be characterized as the projection of $0$ on $S$.
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
$endgroup$
Yes, it is.
The $U$ in the representation $S=s+U$ is the same for all $sin S$, because it is defined as $U{s-t:s,tin S}$.
As a consequence you have that $s_U^perp = t_U^perp$ for all $s,tin S$,
because $s-t=(s_U-t_U)+(s_U^perp-t_U^perp) in U$ and $s_U-t_Uin U$ imply $s_U^perp-t_U^perpin U$, so it has to be $0$.
Moreover this vector can be characterized as the projection of $0$ on $S$.
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
edited Dec 12 '18 at 17:45
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
answered Dec 12 '18 at 17:39
FedericoFederico
5,124514
5,124514
add a comment |
add a comment |
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