Projection of a vector in $L^2[0,1]$












1












$begingroup$


Let $V$ be a closed subspace of $L^2[0,1]$ and let $f,g in L^2[0,1] $ be given by $f(x)=x$ and $g(x)=x^2$. If $V^{perp}= span (f)$ and $Pg$ is the orthogonal projection of $g$ on $V$, then $(g-Pg)(x)$, $x in [0,1]$ is



(A)$frac{3x}{4}$ (B) $frac{x}{4}$ (C) $frac{3x^2}{4}$ (D) $frac{x^2}{4}$



The value is actually the projection of $x^2$ on $V^{perp}$. But I dont know how to solve this. Can someone help me please?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't see how $f$ comes into play. Is the definition of $Pg$ right?
    $endgroup$
    – Paul
    Jan 10 at 11:38












  • $begingroup$
    $f=x$ and $Pg$ is the projection of $g$ on $V$
    $endgroup$
    – user118413
    Jan 10 at 11:46












  • $begingroup$
    Which scalar product is used here? The standart $langle a,brangle = intlimits^1_0 a b dx$?
    $endgroup$
    – denklo
    Jan 10 at 11:51












  • $begingroup$
    Yes, the standard one
    $endgroup$
    – user118413
    Jan 10 at 11:53
















1












$begingroup$


Let $V$ be a closed subspace of $L^2[0,1]$ and let $f,g in L^2[0,1] $ be given by $f(x)=x$ and $g(x)=x^2$. If $V^{perp}= span (f)$ and $Pg$ is the orthogonal projection of $g$ on $V$, then $(g-Pg)(x)$, $x in [0,1]$ is



(A)$frac{3x}{4}$ (B) $frac{x}{4}$ (C) $frac{3x^2}{4}$ (D) $frac{x^2}{4}$



The value is actually the projection of $x^2$ on $V^{perp}$. But I dont know how to solve this. Can someone help me please?






linear-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    I don't see how $f$ comes into play. Is the definition of $Pg$ right?
    $endgroup$
    – Paul
    Jan 10 at 11:38












  • $begingroup$
    $f=x$ and $Pg$ is the projection of $g$ on $V$
    $endgroup$
    – user118413
    Jan 10 at 11:46












  • $begingroup$
    Which scalar product is used here? The standart $langle a,brangle = intlimits^1_0 a b dx$?
    $endgroup$
    – denklo
    Jan 10 at 11:51












  • $begingroup$
    Yes, the standard one
    $endgroup$
    – user118413
    Jan 10 at 11:53














1












1








1





$begingroup$


Let $V$ be a closed subspace of $L^2[0,1]$ and let $f,g in L^2[0,1] $ be given by $f(x)=x$ and $g(x)=x^2$. If $V^{perp}= span (f)$ and $Pg$ is the orthogonal projection of $g$ on $V$, then $(g-Pg)(x)$, $x in [0,1]$ is



(A)$frac{3x}{4}$ (B) $frac{x}{4}$ (C) $frac{3x^2}{4}$ (D) $frac{x^2}{4}$



The value is actually the projection of $x^2$ on $V^{perp}$. But I dont know how to solve this. Can someone help me please?






linear-algebra







share|cite|improve this question











$endgroup$




Let $V$ be a closed subspace of $L^2[0,1]$ and let $f,g in L^2[0,1] $ be given by $f(x)=x$ and $g(x)=x^2$. If $V^{perp}= span (f)$ and $Pg$ is the orthogonal projection of $g$ on $V$, then $(g-Pg)(x)$, $x in [0,1]$ is



(A)$frac{3x}{4}$ (B) $frac{x}{4}$ (C) $frac{3x^2}{4}$ (D) $frac{x^2}{4}$



The value is actually the projection of $x^2$ on $V^{perp}$. But I dont know how to solve this. Can someone help me please?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 11:46







user118413

















asked Jan 10 at 11:32









user118413user118413

1679




1679












  • $begingroup$
    I don't see how $f$ comes into play. Is the definition of $Pg$ right?
    $endgroup$
    – Paul
    Jan 10 at 11:38












  • $begingroup$
    $f=x$ and $Pg$ is the projection of $g$ on $V$
    $endgroup$
    – user118413
    Jan 10 at 11:46












  • $begingroup$
    Which scalar product is used here? The standart $langle a,brangle = intlimits^1_0 a b dx$?
    $endgroup$
    – denklo
    Jan 10 at 11:51












  • $begingroup$
    Yes, the standard one
    $endgroup$
    – user118413
    Jan 10 at 11:53


















  • $begingroup$
    I don't see how $f$ comes into play. Is the definition of $Pg$ right?
    $endgroup$
    – Paul
    Jan 10 at 11:38












  • $begingroup$
    $f=x$ and $Pg$ is the projection of $g$ on $V$
    $endgroup$
    – user118413
    Jan 10 at 11:46












  • $begingroup$
    Which scalar product is used here? The standart $langle a,brangle = intlimits^1_0 a b dx$?
    $endgroup$
    – denklo
    Jan 10 at 11:51












  • $begingroup$
    Yes, the standard one
    $endgroup$
    – user118413
    Jan 10 at 11:53
















$begingroup$
I don't see how $f$ comes into play. Is the definition of $Pg$ right?
$endgroup$
– Paul
Jan 10 at 11:38






$begingroup$
I don't see how $f$ comes into play. Is the definition of $Pg$ right?
$endgroup$
– Paul
Jan 10 at 11:38














$begingroup$
$f=x$ and $Pg$ is the projection of $g$ on $V$
$endgroup$
– user118413
Jan 10 at 11:46






$begingroup$
$f=x$ and $Pg$ is the projection of $g$ on $V$
$endgroup$
– user118413
Jan 10 at 11:46














$begingroup$
Which scalar product is used here? The standart $langle a,brangle = intlimits^1_0 a b dx$?
$endgroup$
– denklo
Jan 10 at 11:51






$begingroup$
Which scalar product is used here? The standart $langle a,brangle = intlimits^1_0 a b dx$?
$endgroup$
– denklo
Jan 10 at 11:51














$begingroup$
Yes, the standard one
$endgroup$
– user118413
Jan 10 at 11:53




$begingroup$
Yes, the standard one
$endgroup$
– user118413
Jan 10 at 11:53










1 Answer
1






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oldest

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4












$begingroup$

If $P$ is projection on $M$ then $I-P$ is the projection on $M^{perp}$. Hence you are just asked to find the projection on span of $f$ which is $frac {langle f, g rangle} {|f|^{2}} f$. The correct answer is (A).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much sir for your kind support
    $endgroup$
    – user118413
    Jan 11 at 5:42











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

If $P$ is projection on $M$ then $I-P$ is the projection on $M^{perp}$. Hence you are just asked to find the projection on span of $f$ which is $frac {langle f, g rangle} {|f|^{2}} f$. The correct answer is (A).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much sir for your kind support
    $endgroup$
    – user118413
    Jan 11 at 5:42
















4












$begingroup$

If $P$ is projection on $M$ then $I-P$ is the projection on $M^{perp}$. Hence you are just asked to find the projection on span of $f$ which is $frac {langle f, g rangle} {|f|^{2}} f$. The correct answer is (A).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you very much sir for your kind support
    $endgroup$
    – user118413
    Jan 11 at 5:42














4












4








4





$begingroup$

If $P$ is projection on $M$ then $I-P$ is the projection on $M^{perp}$. Hence you are just asked to find the projection on span of $f$ which is $frac {langle f, g rangle} {|f|^{2}} f$. The correct answer is (A).






share|cite|improve this answer









$endgroup$



If $P$ is projection on $M$ then $I-P$ is the projection on $M^{perp}$. Hence you are just asked to find the projection on span of $f$ which is $frac {langle f, g rangle} {|f|^{2}} f$. The correct answer is (A).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 11:55









Kavi Rama MurthyKavi Rama Murthy

57.9k42160




57.9k42160












  • $begingroup$
    thank you very much sir for your kind support
    $endgroup$
    – user118413
    Jan 11 at 5:42


















  • $begingroup$
    thank you very much sir for your kind support
    $endgroup$
    – user118413
    Jan 11 at 5:42
















$begingroup$
thank you very much sir for your kind support
$endgroup$
– user118413
Jan 11 at 5:42




$begingroup$
thank you very much sir for your kind support
$endgroup$
– user118413
Jan 11 at 5:42


















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