Mixing
Projection of a vector in $L^2[0,1]$
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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
asked Jan 10 at 11:32
user118413user118413
1679
$endgroup$
add a comment |
$begingroup$
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
asked Jan 10 at 11:32
user118413user118413
1679
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Jan 10 at 11:32
user118413user118413
1679
$endgroup$
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
asked Jan 10 at 11:32
user118413user118413
1679
asked Jan 10 at 11:32
user118413user118413
1679
edited Jan 10 at 11:46
user118413
asked Jan 10 at 11:32
user118413user118413
1679
asked Jan 10 at 11:32
user118413user118413
1679
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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).
answered Jan 10 at 11:55
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
$begingroup$
thank you very much sir for your kind support
$endgroup$
– user118413
Jan 11 at 5:42
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
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).
answered Jan 10 at 11:55
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
$begingroup$
thank you very much sir for your kind support
$endgroup$
– user118413
Jan 11 at 5:42
add a comment |
$begingroup$
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).
answered Jan 10 at 11:55
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
$begingroup$
thank you very much sir for your kind support
$endgroup$
– user118413
Jan 11 at 5:42
add a comment |
$begingroup$
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).
answered Jan 10 at 11:55
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
$endgroup$
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).
answered Jan 10 at 11:55
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Jan 10 at 11:55
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
answered Jan 10 at 11:55
Kavi Rama MurthyKavi Rama Murthy
57.9k42160
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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