Equality of orthogonal projection norms. [closed]
Let $F,G$ be two sub vector spaces of $mathbb{R}^n$, and denote $P_F,P_G$ the orthogonal projection matrices on $F$ and $G$ respectively. Suppose that for all $v in mathbb{R}^n$ we have:
begin{equation*}
||P_Fv||^2 = ||P_Gv||^2
end{equation*}
where $||cdot||$ is the euclidean norm. Is it true that one of the following assertion holds?
i) $F subset G$
ii) $G subset F$
iii) F=G
linear-algebra orthogonality projection-matrices
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
closed as off-topic by Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos Nov 27 '18 at 13:34
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- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos
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Let $F,G$ be two sub vector spaces of $mathbb{R}^n$, and denote $P_F,P_G$ the orthogonal projection matrices on $F$ and $G$ respectively. Suppose that for all $v in mathbb{R}^n$ we have:
begin{equation*}
||P_Fv||^2 = ||P_Gv||^2
end{equation*}
where $||cdot||$ is the euclidean norm. Is it true that one of the following assertion holds?
i) $F subset G$
ii) $G subset F$
iii) F=G
linear-algebra orthogonality projection-matrices
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
closed as off-topic by Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos Nov 27 '18 at 13:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let $F,G$ be two sub vector spaces of $mathbb{R}^n$, and denote $P_F,P_G$ the orthogonal projection matrices on $F$ and $G$ respectively. Suppose that for all $v in mathbb{R}^n$ we have:
begin{equation*}
||P_Fv||^2 = ||P_Gv||^2
end{equation*}
where $||cdot||$ is the euclidean norm. Is it true that one of the following assertion holds?
i) $F subset G$
ii) $G subset F$
iii) F=G
linear-algebra orthogonality projection-matrices
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
Let $F,G$ be two sub vector spaces of $mathbb{R}^n$, and denote $P_F,P_G$ the orthogonal projection matrices on $F$ and $G$ respectively. Suppose that for all $v in mathbb{R}^n$ we have:
begin{equation*}
||P_Fv||^2 = ||P_Gv||^2
end{equation*}
where $||cdot||$ is the euclidean norm. Is it true that one of the following assertion holds?
i) $F subset G$
ii) $G subset F$
iii) F=G
linear-algebra orthogonality projection-matrices
linear-algebra orthogonality projection-matrices
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
asked Nov 27 '18 at 10:19
Alfred F.Alfred F.
697
697
closed as off-topic by Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos Nov 27 '18 at 13:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos Nov 27 '18 at 13:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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From $||P_Fv||^2 = ||P_Gv||^2$ for all $v in mathbb R^n$ we get
$v in F^{perp} iff P_Fv=0 iff P_Gv=0 iff v in G^{perp} .$
Hence $F^{perp}=G^{perp}$.
Can you proceed ?
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
Perfect, thank you.
– Alfred F.
Nov 27 '18 at 10:42
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
From $||P_Fv||^2 = ||P_Gv||^2$ for all $v in mathbb R^n$ we get
$v in F^{perp} iff P_Fv=0 iff P_Gv=0 iff v in G^{perp} .$
Hence $F^{perp}=G^{perp}$.
Can you proceed ?
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
Perfect, thank you.
– Alfred F.
Nov 27 '18 at 10:42
add a comment |
From $||P_Fv||^2 = ||P_Gv||^2$ for all $v in mathbb R^n$ we get
$v in F^{perp} iff P_Fv=0 iff P_Gv=0 iff v in G^{perp} .$
Hence $F^{perp}=G^{perp}$.
Can you proceed ?
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
Perfect, thank you.
– Alfred F.
Nov 27 '18 at 10:42
add a comment |
From $||P_Fv||^2 = ||P_Gv||^2$ for all $v in mathbb R^n$ we get
$v in F^{perp} iff P_Fv=0 iff P_Gv=0 iff v in G^{perp} .$
Hence $F^{perp}=G^{perp}$.
Can you proceed ?
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
From $||P_Fv||^2 = ||P_Gv||^2$ for all $v in mathbb R^n$ we get
$v in F^{perp} iff P_Fv=0 iff P_Gv=0 iff v in G^{perp} .$
Hence $F^{perp}=G^{perp}$.
Can you proceed ?
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
answered Nov 27 '18 at 10:30
FredFred
44.2k1845
44.2k1845
Perfect, thank you.
– Alfred F.
Nov 27 '18 at 10:42
add a comment |
Perfect, thank you.
– Alfred F.
Nov 27 '18 at 10:42
Perfect, thank you.
– Alfred F.
Nov 27 '18 at 10:42
Perfect, thank you.
– Alfred F.
Nov 27 '18 at 10:42
add a comment |
