Equality of orthogonal projection norms. [closed]












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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







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closed as off-topic by Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos Nov 27 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

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    0














    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







    share|cite|improve this question













    closed as off-topic by Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos Nov 27 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.
















      0












      0








      0







      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







      share|cite|improve this question













      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






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 27 at 10:19









      Alfred F.

      697




      697




      closed as off-topic by Kavi Rama Murthy, Saad, Christopher, Arnaud D., José Carlos Santos Nov 27 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 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.






















          1 Answer
          1






          active

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          2














          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 ?






          share|cite|improve this answer





















          • Perfect, thank you.
            – Alfred F.
            Nov 27 at 10:42


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          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 ?






          share|cite|improve this answer





















          • Perfect, thank you.
            – Alfred F.
            Nov 27 at 10:42
















          2














          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 ?






          share|cite|improve this answer





















          • Perfect, thank you.
            – Alfred F.
            Nov 27 at 10:42














          2












          2








          2






          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 ?






          share|cite|improve this answer












          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 ?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 at 10:30









          Fred

          44.2k1845




          44.2k1845












          • Perfect, thank you.
            – Alfred F.
            Nov 27 at 10:42


















          • Perfect, thank you.
            – Alfred F.
            Nov 27 at 10:42
















          Perfect, thank you.
          – Alfred F.
          Nov 27 at 10:42




          Perfect, thank you.
          – Alfred F.
          Nov 27 at 10:42



          -

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