Finding a basis for $W^perp$
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Define an inner product on $mathbbP_4[x]$ over $mathbbR$ as follows:
$$
langle f,g rangle = int_0^1 f(x)g(x) ,mathrmdx
$$
Let $W$ be the subspace of $mathbbP_4[x]$ consisting of $0$ and all polynomials with degree $0$: That is, $W= mathbbR$. Find a basis for $W^perp$.
My attempts:
$mathbbP_4[x]= a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$
Now $p_1 =1$, $p_2 = x$, $p_3 = x^2$, $p_4 = x^3$, $p_5 = x^4$.
Now $p_1= w_1$ so that $| w_1 |^2 = (w_1,w_1) = (1,1)$.
Now $| w_1 |^2 = int_0^1 1 cdot 1 ,mathrmdx = 1$.
Similarly $w_2 = p_2 - fraclangle p_2, w_1 rangle w_1w_1 = (x - frac12)$.
Is this partial answer correct or not?
Please help me; any hints/solution will be appreciated.
Thanks in advance.
linear-algebra inner-product-space
edited Aug 26 at 21:37
Jendrik Stelzner
7,63121037
asked Aug 26 at 16:24
stupid
676111
add a comment |Â
up vote
2
down vote
favorite
Define an inner product on $mathbbP_4[x]$ over $mathbbR$ as follows:
$$
langle f,g rangle = int_0^1 f(x)g(x) ,mathrmdx
$$
Let $W$ be the subspace of $mathbbP_4[x]$ consisting of $0$ and all polynomials with degree $0$: That is, $W= mathbbR$. Find a basis for $W^perp$.
My attempts:
$mathbbP_4[x]= a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$
Now $p_1 =1$, $p_2 = x$, $p_3 = x^2$, $p_4 = x^3$, $p_5 = x^4$.
Now $p_1= w_1$ so that $| w_1 |^2 = (w_1,w_1) = (1,1)$.
Now $| w_1 |^2 = int_0^1 1 cdot 1 ,mathrmdx = 1$.
Similarly $w_2 = p_2 - fraclangle p_2, w_1 rangle w_1w_1 = (x - frac12)$.
Is this partial answer correct or not?
Please help me; any hints/solution will be appreciated.
Thanks in advance.
linear-algebra inner-product-space
edited Aug 26 at 21:37
Jendrik Stelzner
7,63121037
asked Aug 26 at 16:24
stupid
676111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Define an inner product on $mathbbP_4[x]$ over $mathbbR$ as follows:
$$
langle f,g rangle = int_0^1 f(x)g(x) ,mathrmdx
$$
Let $W$ be the subspace of $mathbbP_4[x]$ consisting of $0$ and all polynomials with degree $0$: That is, $W= mathbbR$. Find a basis for $W^perp$.
My attempts:
$mathbbP_4[x]= a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$
Now $p_1 =1$, $p_2 = x$, $p_3 = x^2$, $p_4 = x^3$, $p_5 = x^4$.
Now $p_1= w_1$ so that $| w_1 |^2 = (w_1,w_1) = (1,1)$.
Now $| w_1 |^2 = int_0^1 1 cdot 1 ,mathrmdx = 1$.
Similarly $w_2 = p_2 - fraclangle p_2, w_1 rangle w_1w_1 = (x - frac12)$.
Is this partial answer correct or not?
Please help me; any hints/solution will be appreciated.
Thanks in advance.
linear-algebra inner-product-space
edited Aug 26 at 21:37
Jendrik Stelzner
7,63121037
asked Aug 26 at 16:24
stupid
676111
Define an inner product on $mathbbP_4[x]$ over $mathbbR$ as follows:
$$
langle f,g rangle = int_0^1 f(x)g(x) ,mathrmdx
$$
Let $W$ be the subspace of $mathbbP_4[x]$ consisting of $0$ and all polynomials with degree $0$: That is, $W= mathbbR$. Find a basis for $W^perp$.
My attempts:
$mathbbP_4[x]= a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$
Now $p_1 =1$, $p_2 = x$, $p_3 = x^2$, $p_4 = x^3$, $p_5 = x^4$.
Now $p_1= w_1$ so that $| w_1 |^2 = (w_1,w_1) = (1,1)$.
Now $| w_1 |^2 = int_0^1 1 cdot 1 ,mathrmdx = 1$.
Similarly $w_2 = p_2 - fraclangle p_2, w_1 rangle w_1w_1 = (x - frac12)$.
Is this partial answer correct or not?
Please help me; any hints/solution will be appreciated.
Thanks in advance.
linear-algebra inner-product-space
edited Aug 26 at 21:37
Jendrik Stelzner
7,63121037
asked Aug 26 at 16:24
stupid
676111
edited Aug 26 at 21:37
Jendrik Stelzner
7,63121037
edited Aug 26 at 21:37
Jendrik Stelzner
7,63121037
edited Aug 26 at 21:37
Jendrik Stelzner
7,63121037
7,63121037
asked Aug 26 at 16:24
stupid
676111
asked Aug 26 at 16:24
stupid
676111
asked Aug 26 at 16:24
stupid
676111
676111
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $f(x)=kin W$ and $gin W^perp$ then
$$langle f,grangle = kint_0^1 g(x) dx=0$$
that is
$$a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$$
then
$$g(x)=a_1(x-1/2)+a_2(x^2-1/3)+a_3(x^3-1/4)+a_4(x^4-1/5)$$
therefore a basis for $W^perp$ is given by
$$x-1/2, x^2-1/3, x^3-1/4, x^4-1/5$$
answered Aug 26 at 16:44
gimusi
70.4k73786
thanks @gimusi but im not getting how u write $a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$?? how it come
– stupid
Aug 26 at 16:56
1
from the condition that $int_0^1 g(x) dx=0$, I’ve skipped all the intermediate steps but they are trivial
– gimusi
Aug 26 at 16:58
okk im getting now .... but liitle more doubts $a_0$ is not in g(x) ..then how its come on
– stupid
Aug 26 at 17:04
1
@stupid $$g(x)=a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$$ then use the condition found for the coefficients that is $$a_0=-frac12a_1-frac13a_2-frac14a_3-frac15a_4$$.
– gimusi
Aug 26 at 17:07
1
I didn’t give all details since you can find out easily by yourself! You are welcome, Bye
– gimusi
Aug 26 at 17:10
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $f(x)=kin W$ and $gin W^perp$ then
$$langle f,grangle = kint_0^1 g(x) dx=0$$
that is
$$a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$$
then
$$g(x)=a_1(x-1/2)+a_2(x^2-1/3)+a_3(x^3-1/4)+a_4(x^4-1/5)$$
therefore a basis for $W^perp$ is given by
$$x-1/2, x^2-1/3, x^3-1/4, x^4-1/5$$
answered Aug 26 at 16:44
gimusi
70.4k73786
thanks @gimusi but im not getting how u write $a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$?? how it come
– stupid
Aug 26 at 16:56
1
from the condition that $int_0^1 g(x) dx=0$, I’ve skipped all the intermediate steps but they are trivial
– gimusi
Aug 26 at 16:58
okk im getting now .... but liitle more doubts $a_0$ is not in g(x) ..then how its come on
– stupid
Aug 26 at 17:04
1
@stupid $$g(x)=a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$$ then use the condition found for the coefficients that is $$a_0=-frac12a_1-frac13a_2-frac14a_3-frac15a_4$$.
– gimusi
Aug 26 at 17:07
1
I didn’t give all details since you can find out easily by yourself! You are welcome, Bye
– gimusi
Aug 26 at 17:10
 |Â
show 1 more comment
up vote
2
down vote
accepted
Let $f(x)=kin W$ and $gin W^perp$ then
$$langle f,grangle = kint_0^1 g(x) dx=0$$
that is
$$a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$$
then
$$g(x)=a_1(x-1/2)+a_2(x^2-1/3)+a_3(x^3-1/4)+a_4(x^4-1/5)$$
therefore a basis for $W^perp$ is given by
$$x-1/2, x^2-1/3, x^3-1/4, x^4-1/5$$
answered Aug 26 at 16:44
gimusi
70.4k73786
thanks @gimusi but im not getting how u write $a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$?? how it come
– stupid
Aug 26 at 16:56
1
from the condition that $int_0^1 g(x) dx=0$, I’ve skipped all the intermediate steps but they are trivial
– gimusi
Aug 26 at 16:58
okk im getting now .... but liitle more doubts $a_0$ is not in g(x) ..then how its come on
– stupid
Aug 26 at 17:04
1
@stupid $$g(x)=a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$$ then use the condition found for the coefficients that is $$a_0=-frac12a_1-frac13a_2-frac14a_3-frac15a_4$$.
– gimusi
Aug 26 at 17:07
1
I didn’t give all details since you can find out easily by yourself! You are welcome, Bye
– gimusi
Aug 26 at 17:10
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $f(x)=kin W$ and $gin W^perp$ then
$$langle f,grangle = kint_0^1 g(x) dx=0$$
that is
$$a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$$
then
$$g(x)=a_1(x-1/2)+a_2(x^2-1/3)+a_3(x^3-1/4)+a_4(x^4-1/5)$$
therefore a basis for $W^perp$ is given by
$$x-1/2, x^2-1/3, x^3-1/4, x^4-1/5$$
answered Aug 26 at 16:44
gimusi
70.4k73786
Let $f(x)=kin W$ and $gin W^perp$ then
$$langle f,grangle = kint_0^1 g(x) dx=0$$
that is
$$a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$$
then
$$g(x)=a_1(x-1/2)+a_2(x^2-1/3)+a_3(x^3-1/4)+a_4(x^4-1/5)$$
therefore a basis for $W^perp$ is given by
$$x-1/2, x^2-1/3, x^3-1/4, x^4-1/5$$
answered Aug 26 at 16:44
gimusi
70.4k73786
answered Aug 26 at 16:44
gimusi
70.4k73786
answered Aug 26 at 16:44
gimusi
70.4k73786
answered Aug 26 at 16:44
gimusi
70.4k73786
70.4k73786
thanks @gimusi but im not getting how u write $a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$?? how it come
– stupid
Aug 26 at 16:56
1
from the condition that $int_0^1 g(x) dx=0$, I’ve skipped all the intermediate steps but they are trivial
– gimusi
Aug 26 at 16:58
okk im getting now .... but liitle more doubts $a_0$ is not in g(x) ..then how its come on
– stupid
Aug 26 at 17:04
1
@stupid $$g(x)=a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$$ then use the condition found for the coefficients that is $$a_0=-frac12a_1-frac13a_2-frac14a_3-frac15a_4$$.
– gimusi
Aug 26 at 17:07
1
I didn’t give all details since you can find out easily by yourself! You are welcome, Bye
– gimusi
Aug 26 at 17:10
 |Â
show 1 more comment
thanks @gimusi but im not getting how u write $a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$?? how it come
– stupid
Aug 26 at 16:56
1
from the condition that $int_0^1 g(x) dx=0$, I’ve skipped all the intermediate steps but they are trivial
– gimusi
Aug 26 at 16:58
okk im getting now .... but liitle more doubts $a_0$ is not in g(x) ..then how its come on
– stupid
Aug 26 at 17:04
1
@stupid $$g(x)=a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$$ then use the condition found for the coefficients that is $$a_0=-frac12a_1-frac13a_2-frac14a_3-frac15a_4$$.
– gimusi
Aug 26 at 17:07
1
I didn’t give all details since you can find out easily by yourself! You are welcome, Bye
– gimusi
Aug 26 at 17:10
thanks @gimusi but im not getting how u write $a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$?? how it come
– stupid
Aug 26 at 16:56
thanks @gimusi but im not getting how u write $a_0+frac12a_1+frac13a_2+frac14a_3+frac15a_4=0$?? how it come
– stupid
Aug 26 at 16:56
1
1
from the condition that $int_0^1 g(x) dx=0$, I’ve skipped all the intermediate steps but they are trivial
– gimusi
Aug 26 at 16:58
from the condition that $int_0^1 g(x) dx=0$, I’ve skipped all the intermediate steps but they are trivial
– gimusi
Aug 26 at 16:58
okk im getting now .... but liitle more doubts $a_0$ is not in g(x) ..then how its come on
– stupid
Aug 26 at 17:04
okk im getting now .... but liitle more doubts $a_0$ is not in g(x) ..then how its come on
– stupid
Aug 26 at 17:04
1
1
@stupid $$g(x)=a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$$ then use the condition found for the coefficients that is $$a_0=-frac12a_1-frac13a_2-frac14a_3-frac15a_4$$.
– gimusi
Aug 26 at 17:07
@stupid $$g(x)=a_0 + a_1x + a_2x^2 +a_3x^3 + a_4x^4$$ then use the condition found for the coefficients that is $$a_0=-frac12a_1-frac13a_2-frac14a_3-frac15a_4$$.
– gimusi
Aug 26 at 17:07
1
1
I didn’t give all details since you can find out easily by yourself! You are welcome, Bye
– gimusi
Aug 26 at 17:10
I didn’t give all details since you can find out easily by yourself! You are welcome, Bye
– gimusi
Aug 26 at 17:10
 |Â
show 1 more comment
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