Mixing
Basis for a Subspace of Polynomials of Degree 5
Clash Royale CLAN TAG#URR8PPP
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I have the following question: "Consider the subspace $W$ of $P_5(mathbbR)$ (the set of polynomials of at most degree five) given by $$W=p(1)=p(-1)=0.$$ Find a basis for $W$, and compute its dimension." I have been working on this for a bit, and I know that the standard basis for $P_5(mathbbR)$ won't work, so I came up with $$x^5-x^3,x^4-2x^2+1,x^3-x,x^2-1.$$ However, I'm afraid that the set doesn't span $P_5(mathbbR)$, but I think that it's linearly independent. Once I have all this, I can easily find the dimension. Thank you in advance.
linear-algebra polynomials vector-spaces
edited Jul 18 at 15:50
Arthur
98.8k793175
asked Jul 18 at 15:34
Mr. Frothingslosh
796
add a comment |Â
up vote
6
down vote
favorite
I have the following question: "Consider the subspace $W$ of $P_5(mathbbR)$ (the set of polynomials of at most degree five) given by $$W=p(1)=p(-1)=0.$$ Find a basis for $W$, and compute its dimension." I have been working on this for a bit, and I know that the standard basis for $P_5(mathbbR)$ won't work, so I came up with $$x^5-x^3,x^4-2x^2+1,x^3-x,x^2-1.$$ However, I'm afraid that the set doesn't span $P_5(mathbbR)$, but I think that it's linearly independent. Once I have all this, I can easily find the dimension. Thank you in advance.
linear-algebra polynomials vector-spaces
edited Jul 18 at 15:50
Arthur
98.8k793175
asked Jul 18 at 15:34
Mr. Frothingslosh
796
1
As there are two constraints, two degrees of freedom are consumed and $W$ has dimension $4$.
– Yves Daoust
Jul 18 at 15:43
Yes, that is what I meant. I accidentally typed in the wrong thing. Thank you
– Mr. Frothingslosh
Jul 18 at 15:46
1
@YvesDaoust There are some caveats to that, as restrictions aren't necessarily independent, and some times the resulting subset isn't a vector space. But those are easily ruled out in this case, and it's a good general rule of thumb that each restriction decreases dimension by $1$.
– Arthur
Jul 18 at 15:47
Very interesting. I didn't know that.
– Mr. Frothingslosh
Jul 18 at 15:51
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
I have the following question: "Consider the subspace $W$ of $P_5(mathbbR)$ (the set of polynomials of at most degree five) given by $$W=p(1)=p(-1)=0.$$ Find a basis for $W$, and compute its dimension." I have been working on this for a bit, and I know that the standard basis for $P_5(mathbbR)$ won't work, so I came up with $$x^5-x^3,x^4-2x^2+1,x^3-x,x^2-1.$$ However, I'm afraid that the set doesn't span $P_5(mathbbR)$, but I think that it's linearly independent. Once I have all this, I can easily find the dimension. Thank you in advance.
linear-algebra polynomials vector-spaces
edited Jul 18 at 15:50
Arthur
98.8k793175
asked Jul 18 at 15:34
Mr. Frothingslosh
796
I have the following question: "Consider the subspace $W$ of $P_5(mathbbR)$ (the set of polynomials of at most degree five) given by $$W=p(1)=p(-1)=0.$$ Find a basis for $W$, and compute its dimension." I have been working on this for a bit, and I know that the standard basis for $P_5(mathbbR)$ won't work, so I came up with $$x^5-x^3,x^4-2x^2+1,x^3-x,x^2-1.$$ However, I'm afraid that the set doesn't span $P_5(mathbbR)$, but I think that it's linearly independent. Once I have all this, I can easily find the dimension. Thank you in advance.
linear-algebra polynomials vector-spaces
edited Jul 18 at 15:50
Arthur
98.8k793175
asked Jul 18 at 15:34
Mr. Frothingslosh
796
edited Jul 18 at 15:50
Arthur
98.8k793175
edited Jul 18 at 15:50
Arthur
98.8k793175
edited Jul 18 at 15:50
Arthur
98.8k793175
98.8k793175
asked Jul 18 at 15:34
Mr. Frothingslosh
796
asked Jul 18 at 15:34
Mr. Frothingslosh
796
asked Jul 18 at 15:34
Mr. Frothingslosh
796
796
1
As there are two constraints, two degrees of freedom are consumed and $W$ has dimension $4$.
– Yves Daoust
Jul 18 at 15:43
Yes, that is what I meant. I accidentally typed in the wrong thing. Thank you
– Mr. Frothingslosh
Jul 18 at 15:46
1
@YvesDaoust There are some caveats to that, as restrictions aren't necessarily independent, and some times the resulting subset isn't a vector space. But those are easily ruled out in this case, and it's a good general rule of thumb that each restriction decreases dimension by $1$.
– Arthur
Jul 18 at 15:47
Very interesting. I didn't know that.
– Mr. Frothingslosh
Jul 18 at 15:51
add a comment |Â
1
As there are two constraints, two degrees of freedom are consumed and $W$ has dimension $4$.
– Yves Daoust
Jul 18 at 15:43
Yes, that is what I meant. I accidentally typed in the wrong thing. Thank you
– Mr. Frothingslosh
Jul 18 at 15:46
1
@YvesDaoust There are some caveats to that, as restrictions aren't necessarily independent, and some times the resulting subset isn't a vector space. But those are easily ruled out in this case, and it's a good general rule of thumb that each restriction decreases dimension by $1$.
– Arthur
Jul 18 at 15:47
Very interesting. I didn't know that.
– Mr. Frothingslosh
Jul 18 at 15:51
1
1
As there are two constraints, two degrees of freedom are consumed and $W$ has dimension $4$.
– Yves Daoust
Jul 18 at 15:43
As there are two constraints, two degrees of freedom are consumed and $W$ has dimension $4$.
– Yves Daoust
Jul 18 at 15:43
Yes, that is what I meant. I accidentally typed in the wrong thing. Thank you
– Mr. Frothingslosh
Jul 18 at 15:46
Yes, that is what I meant. I accidentally typed in the wrong thing. Thank you
– Mr. Frothingslosh
Jul 18 at 15:46
1
1
@YvesDaoust There are some caveats to that, as restrictions aren't necessarily independent, and some times the resulting subset isn't a vector space. But those are easily ruled out in this case, and it's a good general rule of thumb that each restriction decreases dimension by $1$.
– Arthur
Jul 18 at 15:47
@YvesDaoust There are some caveats to that, as restrictions aren't necessarily independent, and some times the resulting subset isn't a vector space. But those are easily ruled out in this case, and it's a good general rule of thumb that each restriction decreases dimension by $1$.
– Arthur
Jul 18 at 15:47
Very interesting. I didn't know that.
– Mr. Frothingslosh
Jul 18 at 15:51
Very interesting. I didn't know that.
– Mr. Frothingslosh
Jul 18 at 15:51
add a comment |Â
5 Answers
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up vote
12
down vote
accepted
$p(1) = p(-1) = 0$ is equivalent to $p(x)$ being divisible by $x^2-1$. So we can start by writing out a couple of simple multiples of $x^2-1$:
$$
x^2-1, x(x^2-1), x^2(x^2-1), x^3(x^2-1)
$$
These are linearly independent: write $0$ as a linear combination of the above polynomials. Only one of the polynomials has a degree 5 term, so its coefficient must be $0$. Once we know that, there is only one of the polynomials with a degree 4 term, so its coefficient must be $0$, and so on.
Also, together they can build any polynomial which is divisible by $x^2-1$ as a linear combination: if $p(x) = q(x)(x^2-1)$, then express $q(x)$ as a linear combination of $1, x, x^2$ and $x^3$ ($q(x)$ can't have degree higher than $3$, cause then $p(x)$ would have degree more than $5$), and the same coefficients will give you $p(x)$ as a linear combination of the polynomials above.
answered Jul 18 at 15:38
Arthur
98.8k793175
It was a typo. Thank you
– Mr. Frothingslosh
Jul 18 at 15:49
add a comment |Â
up vote
6
down vote
The general polynomial of degree $5$ is
$$p(x)=ax^5+bx_4+cx_3+dx^2+ex+f$$
and from the condition $p(1)=p(-1)=0$ we obtain
$a+b+c+d+e+f=0$
$-a+b-c+d-e+f=0$
from which, by adding and subtracting the two equations, we obtain
$2b+2d+2f=0 implies f=-b-d$
$2a+2c+2e=0 implies e=-a-c$
and finally
$$p(x)=ax^5+bx_4+cx_3+dx^2+(-a-c)x+(-b-d)\=a(x^5-x)+b(x^4-1)+c(x^3-x)+d(x^2-1)$$
therefore a basis is given by:
$$x^5-x,x^4-1,x^3-x,x^2-1$$
answered Jul 18 at 15:44
gimusi
65.4k73584
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up vote
2
down vote
Your answer is correct and span $W$.
One way to see it is the following:
- It is linearly independent as the degree are distincts.
- Consider:
beginalign
phi:&P_5(mathbbR) to mathbbR^2\
&p mapsto (p(-1),p(1))
endalign
then $W=ker(phi)$. But as $phi((x+1)/2)=(1,0)$ and $phi((x-1)/2)=(0,1)$ you have $dim(phi(P_5(mathbbR)))=2$ so:
$$dim(W)=dim(ker(phi))=dim(P_5(mathbbR))-dim(phi(P_5(mathbbR))=6-2=4$$
and $4$ linearly independent vectors of a subspace of dimension $4$ is indeed a basis.
answered Jul 18 at 15:42
Delta-u
4,742518
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up vote
1
down vote
The polynomials in $W$ are all divisible by $x^2-1$, and you can consider the set of cubic polynomials
$$fracw(x)x^2-1:win W,$$ which is in fact $P_3$. It is easy to show that the $p_3(x)(x^2-1)$ are linearly independent and span $W$.
answered Jul 18 at 15:49
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
Written in linear algebra terms, given the basis $e_i = x^i$ and a polynomial $p$ expressed as
$$beginbmatrix
p_0 \
p_1 \
p_2 \
p_3\
p_4\
p_5
endbmatrix$$
evaluating $p$ at 1 is the same as $[1,1,1,1,1,1]p$, and evaluating it at -1 is $[1,-1,1,-1,1,-1]p$
We have that both are equal to zero, so
$$beginbmatrix
1&1&1&1&1&1 \
1&-1&1&-1&1&-1
endbmatrixp =beginbmatrix
0 \
0
endbmatrix$$
So we can take the augmented matrix
$$beginbmatrix
1&1&1&1&1&1 &|0 \
1&-1&1&-1&1&-1&|0
endbmatrix$$
And reduce it to row-echelon form
$$beginbmatrix
1&0&1&0&1&0 &|0 \
0&1&0&1&0&1&|0
endbmatrix$$
This tells you that among even $i$, you can pick two $p_i$ to be free and the third will be fixed, and similarly for odd $i$. For instance, we can have $p_i$ free for $i<4$, and then $p_5=-(p_3+p_1)$ and $p_4=-(p_2+p_0)$. If we successively set one free variable to 1 and the other free variables to zero, this gives
$$1-x^4$$
$$x-x^5$$
$$x^2-x^4$$
$$x^3-x^5$$
Another basis, that could be written more succinctly, would be $b_i = e_i+2-e_imod2$, which gives
$$b_0 = x^2-1$$
$$b_1=x^3-x$$
$$b_2=x^4-1$$
$$b_3 = x^5-x$$
answered Jul 18 at 21:52
Acccumulation
4,4732314
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
12
down vote
accepted
$p(1) = p(-1) = 0$ is equivalent to $p(x)$ being divisible by $x^2-1$. So we can start by writing out a couple of simple multiples of $x^2-1$:
$$
x^2-1, x(x^2-1), x^2(x^2-1), x^3(x^2-1)
$$
These are linearly independent: write $0$ as a linear combination of the above polynomials. Only one of the polynomials has a degree 5 term, so its coefficient must be $0$. Once we know that, there is only one of the polynomials with a degree 4 term, so its coefficient must be $0$, and so on.
Also, together they can build any polynomial which is divisible by $x^2-1$ as a linear combination: if $p(x) = q(x)(x^2-1)$, then express $q(x)$ as a linear combination of $1, x, x^2$ and $x^3$ ($q(x)$ can't have degree higher than $3$, cause then $p(x)$ would have degree more than $5$), and the same coefficients will give you $p(x)$ as a linear combination of the polynomials above.
answered Jul 18 at 15:38
Arthur
98.8k793175
It was a typo. Thank you
– Mr. Frothingslosh
Jul 18 at 15:49
add a comment |Â
up vote
12
down vote
accepted
$p(1) = p(-1) = 0$ is equivalent to $p(x)$ being divisible by $x^2-1$. So we can start by writing out a couple of simple multiples of $x^2-1$:
$$
x^2-1, x(x^2-1), x^2(x^2-1), x^3(x^2-1)
$$
These are linearly independent: write $0$ as a linear combination of the above polynomials. Only one of the polynomials has a degree 5 term, so its coefficient must be $0$. Once we know that, there is only one of the polynomials with a degree 4 term, so its coefficient must be $0$, and so on.
Also, together they can build any polynomial which is divisible by $x^2-1$ as a linear combination: if $p(x) = q(x)(x^2-1)$, then express $q(x)$ as a linear combination of $1, x, x^2$ and $x^3$ ($q(x)$ can't have degree higher than $3$, cause then $p(x)$ would have degree more than $5$), and the same coefficients will give you $p(x)$ as a linear combination of the polynomials above.
answered Jul 18 at 15:38
Arthur
98.8k793175
It was a typo. Thank you
– Mr. Frothingslosh
Jul 18 at 15:49
add a comment |Â
up vote
12
down vote
accepted
up vote
12
down vote
accepted
$p(1) = p(-1) = 0$ is equivalent to $p(x)$ being divisible by $x^2-1$. So we can start by writing out a couple of simple multiples of $x^2-1$:
$$
x^2-1, x(x^2-1), x^2(x^2-1), x^3(x^2-1)
$$
These are linearly independent: write $0$ as a linear combination of the above polynomials. Only one of the polynomials has a degree 5 term, so its coefficient must be $0$. Once we know that, there is only one of the polynomials with a degree 4 term, so its coefficient must be $0$, and so on.
Also, together they can build any polynomial which is divisible by $x^2-1$ as a linear combination: if $p(x) = q(x)(x^2-1)$, then express $q(x)$ as a linear combination of $1, x, x^2$ and $x^3$ ($q(x)$ can't have degree higher than $3$, cause then $p(x)$ would have degree more than $5$), and the same coefficients will give you $p(x)$ as a linear combination of the polynomials above.
answered Jul 18 at 15:38
Arthur
98.8k793175
$p(1) = p(-1) = 0$ is equivalent to $p(x)$ being divisible by $x^2-1$. So we can start by writing out a couple of simple multiples of $x^2-1$:
$$
x^2-1, x(x^2-1), x^2(x^2-1), x^3(x^2-1)
$$
These are linearly independent: write $0$ as a linear combination of the above polynomials. Only one of the polynomials has a degree 5 term, so its coefficient must be $0$. Once we know that, there is only one of the polynomials with a degree 4 term, so its coefficient must be $0$, and so on.
Also, together they can build any polynomial which is divisible by $x^2-1$ as a linear combination: if $p(x) = q(x)(x^2-1)$, then express $q(x)$ as a linear combination of $1, x, x^2$ and $x^3$ ($q(x)$ can't have degree higher than $3$, cause then $p(x)$ would have degree more than $5$), and the same coefficients will give you $p(x)$ as a linear combination of the polynomials above.
answered Jul 18 at 15:38
Arthur
98.8k793175
edited Jul 18 at 15:51
answered Jul 18 at 15:38
Arthur
98.8k793175
answered Jul 18 at 15:38
Arthur
98.8k793175
answered Jul 18 at 15:38
Arthur
98.8k793175
98.8k793175
It was a typo. Thank you
– Mr. Frothingslosh
Jul 18 at 15:49
add a comment |Â
It was a typo. Thank you
– Mr. Frothingslosh
Jul 18 at 15:49
It was a typo. Thank you
– Mr. Frothingslosh
Jul 18 at 15:49
It was a typo. Thank you
– Mr. Frothingslosh
Jul 18 at 15:49
add a comment |Â
up vote
6
down vote
The general polynomial of degree $5$ is
$$p(x)=ax^5+bx_4+cx_3+dx^2+ex+f$$
and from the condition $p(1)=p(-1)=0$ we obtain
$a+b+c+d+e+f=0$
$-a+b-c+d-e+f=0$
from which, by adding and subtracting the two equations, we obtain
$2b+2d+2f=0 implies f=-b-d$
$2a+2c+2e=0 implies e=-a-c$
and finally
$$p(x)=ax^5+bx_4+cx_3+dx^2+(-a-c)x+(-b-d)\=a(x^5-x)+b(x^4-1)+c(x^3-x)+d(x^2-1)$$
therefore a basis is given by:
$$x^5-x,x^4-1,x^3-x,x^2-1$$
answered Jul 18 at 15:44
gimusi
65.4k73584
add a comment |Â
up vote
6
down vote
The general polynomial of degree $5$ is
$$p(x)=ax^5+bx_4+cx_3+dx^2+ex+f$$
and from the condition $p(1)=p(-1)=0$ we obtain
$a+b+c+d+e+f=0$
$-a+b-c+d-e+f=0$
from which, by adding and subtracting the two equations, we obtain
$2b+2d+2f=0 implies f=-b-d$
$2a+2c+2e=0 implies e=-a-c$
and finally
$$p(x)=ax^5+bx_4+cx_3+dx^2+(-a-c)x+(-b-d)\=a(x^5-x)+b(x^4-1)+c(x^3-x)+d(x^2-1)$$
therefore a basis is given by:
$$x^5-x,x^4-1,x^3-x,x^2-1$$
answered Jul 18 at 15:44
gimusi
65.4k73584
add a comment |Â
up vote
6
down vote
up vote
6
down vote
The general polynomial of degree $5$ is
$$p(x)=ax^5+bx_4+cx_3+dx^2+ex+f$$
and from the condition $p(1)=p(-1)=0$ we obtain
$a+b+c+d+e+f=0$
$-a+b-c+d-e+f=0$
from which, by adding and subtracting the two equations, we obtain
$2b+2d+2f=0 implies f=-b-d$
$2a+2c+2e=0 implies e=-a-c$
and finally
$$p(x)=ax^5+bx_4+cx_3+dx^2+(-a-c)x+(-b-d)\=a(x^5-x)+b(x^4-1)+c(x^3-x)+d(x^2-1)$$
therefore a basis is given by:
$$x^5-x,x^4-1,x^3-x,x^2-1$$
answered Jul 18 at 15:44
gimusi
65.4k73584
The general polynomial of degree $5$ is
$$p(x)=ax^5+bx_4+cx_3+dx^2+ex+f$$
and from the condition $p(1)=p(-1)=0$ we obtain
$a+b+c+d+e+f=0$
$-a+b-c+d-e+f=0$
from which, by adding and subtracting the two equations, we obtain
$2b+2d+2f=0 implies f=-b-d$
$2a+2c+2e=0 implies e=-a-c$
and finally
$$p(x)=ax^5+bx_4+cx_3+dx^2+(-a-c)x+(-b-d)\=a(x^5-x)+b(x^4-1)+c(x^3-x)+d(x^2-1)$$
therefore a basis is given by:
$$x^5-x,x^4-1,x^3-x,x^2-1$$
answered Jul 18 at 15:44
gimusi
65.4k73584
answered Jul 18 at 15:44
gimusi
65.4k73584
answered Jul 18 at 15:44
gimusi
65.4k73584
answered Jul 18 at 15:44
gimusi
65.4k73584
65.4k73584
add a comment |Â
add a comment |Â
up vote
2
down vote
Your answer is correct and span $W$.
One way to see it is the following:
- It is linearly independent as the degree are distincts.
- Consider:
beginalign
phi:&P_5(mathbbR) to mathbbR^2\
&p mapsto (p(-1),p(1))
endalign
then $W=ker(phi)$. But as $phi((x+1)/2)=(1,0)$ and $phi((x-1)/2)=(0,1)$ you have $dim(phi(P_5(mathbbR)))=2$ so:
$$dim(W)=dim(ker(phi))=dim(P_5(mathbbR))-dim(phi(P_5(mathbbR))=6-2=4$$
and $4$ linearly independent vectors of a subspace of dimension $4$ is indeed a basis.
answered Jul 18 at 15:42
Delta-u
4,742518
add a comment |Â
up vote
2
down vote
Your answer is correct and span $W$.
One way to see it is the following:
- It is linearly independent as the degree are distincts.
- Consider:
beginalign
phi:&P_5(mathbbR) to mathbbR^2\
&p mapsto (p(-1),p(1))
endalign
then $W=ker(phi)$. But as $phi((x+1)/2)=(1,0)$ and $phi((x-1)/2)=(0,1)$ you have $dim(phi(P_5(mathbbR)))=2$ so:
$$dim(W)=dim(ker(phi))=dim(P_5(mathbbR))-dim(phi(P_5(mathbbR))=6-2=4$$
and $4$ linearly independent vectors of a subspace of dimension $4$ is indeed a basis.
answered Jul 18 at 15:42
Delta-u
4,742518
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Your answer is correct and span $W$.
One way to see it is the following:
- It is linearly independent as the degree are distincts.
- Consider:
beginalign
phi:&P_5(mathbbR) to mathbbR^2\
&p mapsto (p(-1),p(1))
endalign
then $W=ker(phi)$. But as $phi((x+1)/2)=(1,0)$ and $phi((x-1)/2)=(0,1)$ you have $dim(phi(P_5(mathbbR)))=2$ so:
$$dim(W)=dim(ker(phi))=dim(P_5(mathbbR))-dim(phi(P_5(mathbbR))=6-2=4$$
and $4$ linearly independent vectors of a subspace of dimension $4$ is indeed a basis.
answered Jul 18 at 15:42
Delta-u
4,742518
Your answer is correct and span $W$.
One way to see it is the following:
- It is linearly independent as the degree are distincts.
- Consider:
beginalign
phi:&P_5(mathbbR) to mathbbR^2\
&p mapsto (p(-1),p(1))
endalign
then $W=ker(phi)$. But as $phi((x+1)/2)=(1,0)$ and $phi((x-1)/2)=(0,1)$ you have $dim(phi(P_5(mathbbR)))=2$ so:
$$dim(W)=dim(ker(phi))=dim(P_5(mathbbR))-dim(phi(P_5(mathbbR))=6-2=4$$
and $4$ linearly independent vectors of a subspace of dimension $4$ is indeed a basis.
answered Jul 18 at 15:42
Delta-u
4,742518
answered Jul 18 at 15:42
Delta-u
4,742518
answered Jul 18 at 15:42
Delta-u
4,742518
answered Jul 18 at 15:42
Delta-u
4,742518
4,742518
add a comment |Â
add a comment |Â
up vote
1
down vote
The polynomials in $W$ are all divisible by $x^2-1$, and you can consider the set of cubic polynomials
$$fracw(x)x^2-1:win W,$$ which is in fact $P_3$. It is easy to show that the $p_3(x)(x^2-1)$ are linearly independent and span $W$.
answered Jul 18 at 15:49
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
The polynomials in $W$ are all divisible by $x^2-1$, and you can consider the set of cubic polynomials
$$fracw(x)x^2-1:win W,$$ which is in fact $P_3$. It is easy to show that the $p_3(x)(x^2-1)$ are linearly independent and span $W$.
answered Jul 18 at 15:49
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The polynomials in $W$ are all divisible by $x^2-1$, and you can consider the set of cubic polynomials
$$fracw(x)x^2-1:win W,$$ which is in fact $P_3$. It is easy to show that the $p_3(x)(x^2-1)$ are linearly independent and span $W$.
answered Jul 18 at 15:49
Yves Daoust
111k665204
The polynomials in $W$ are all divisible by $x^2-1$, and you can consider the set of cubic polynomials
$$fracw(x)x^2-1:win W,$$ which is in fact $P_3$. It is easy to show that the $p_3(x)(x^2-1)$ are linearly independent and span $W$.
answered Jul 18 at 15:49
Yves Daoust
111k665204
edited Jul 18 at 16:04
answered Jul 18 at 15:49
Yves Daoust
111k665204
answered Jul 18 at 15:49
Yves Daoust
111k665204
answered Jul 18 at 15:49
Yves Daoust
111k665204
111k665204
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Written in linear algebra terms, given the basis $e_i = x^i$ and a polynomial $p$ expressed as
$$beginbmatrix
p_0 \
p_1 \
p_2 \
p_3\
p_4\
p_5
endbmatrix$$
evaluating $p$ at 1 is the same as $[1,1,1,1,1,1]p$, and evaluating it at -1 is $[1,-1,1,-1,1,-1]p$
We have that both are equal to zero, so
$$beginbmatrix
1&1&1&1&1&1 \
1&-1&1&-1&1&-1
endbmatrixp =beginbmatrix
0 \
0
endbmatrix$$
So we can take the augmented matrix
$$beginbmatrix
1&1&1&1&1&1 &|0 \
1&-1&1&-1&1&-1&|0
endbmatrix$$
And reduce it to row-echelon form
$$beginbmatrix
1&0&1&0&1&0 &|0 \
0&1&0&1&0&1&|0
endbmatrix$$
This tells you that among even $i$, you can pick two $p_i$ to be free and the third will be fixed, and similarly for odd $i$. For instance, we can have $p_i$ free for $i<4$, and then $p_5=-(p_3+p_1)$ and $p_4=-(p_2+p_0)$. If we successively set one free variable to 1 and the other free variables to zero, this gives
$$1-x^4$$
$$x-x^5$$
$$x^2-x^4$$
$$x^3-x^5$$
Another basis, that could be written more succinctly, would be $b_i = e_i+2-e_imod2$, which gives
$$b_0 = x^2-1$$
$$b_1=x^3-x$$
$$b_2=x^4-1$$
$$b_3 = x^5-x$$
answered Jul 18 at 21:52
Acccumulation
4,4732314
add a comment |Â
up vote
1
down vote
Written in linear algebra terms, given the basis $e_i = x^i$ and a polynomial $p$ expressed as
$$beginbmatrix
p_0 \
p_1 \
p_2 \
p_3\
p_4\
p_5
endbmatrix$$
evaluating $p$ at 1 is the same as $[1,1,1,1,1,1]p$, and evaluating it at -1 is $[1,-1,1,-1,1,-1]p$
We have that both are equal to zero, so
$$beginbmatrix
1&1&1&1&1&1 \
1&-1&1&-1&1&-1
endbmatrixp =beginbmatrix
0 \
0
endbmatrix$$
So we can take the augmented matrix
$$beginbmatrix
1&1&1&1&1&1 &|0 \
1&-1&1&-1&1&-1&|0
endbmatrix$$
And reduce it to row-echelon form
$$beginbmatrix
1&0&1&0&1&0 &|0 \
0&1&0&1&0&1&|0
endbmatrix$$
This tells you that among even $i$, you can pick two $p_i$ to be free and the third will be fixed, and similarly for odd $i$. For instance, we can have $p_i$ free for $i<4$, and then $p_5=-(p_3+p_1)$ and $p_4=-(p_2+p_0)$. If we successively set one free variable to 1 and the other free variables to zero, this gives
$$1-x^4$$
$$x-x^5$$
$$x^2-x^4$$
$$x^3-x^5$$
Another basis, that could be written more succinctly, would be $b_i = e_i+2-e_imod2$, which gives
$$b_0 = x^2-1$$
$$b_1=x^3-x$$
$$b_2=x^4-1$$
$$b_3 = x^5-x$$
answered Jul 18 at 21:52
Acccumulation
4,4732314
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Written in linear algebra terms, given the basis $e_i = x^i$ and a polynomial $p$ expressed as
$$beginbmatrix
p_0 \
p_1 \
p_2 \
p_3\
p_4\
p_5
endbmatrix$$
evaluating $p$ at 1 is the same as $[1,1,1,1,1,1]p$, and evaluating it at -1 is $[1,-1,1,-1,1,-1]p$
We have that both are equal to zero, so
$$beginbmatrix
1&1&1&1&1&1 \
1&-1&1&-1&1&-1
endbmatrixp =beginbmatrix
0 \
0
endbmatrix$$
So we can take the augmented matrix
$$beginbmatrix
1&1&1&1&1&1 &|0 \
1&-1&1&-1&1&-1&|0
endbmatrix$$
And reduce it to row-echelon form
$$beginbmatrix
1&0&1&0&1&0 &|0 \
0&1&0&1&0&1&|0
endbmatrix$$
This tells you that among even $i$, you can pick two $p_i$ to be free and the third will be fixed, and similarly for odd $i$. For instance, we can have $p_i$ free for $i<4$, and then $p_5=-(p_3+p_1)$ and $p_4=-(p_2+p_0)$. If we successively set one free variable to 1 and the other free variables to zero, this gives
$$1-x^4$$
$$x-x^5$$
$$x^2-x^4$$
$$x^3-x^5$$
Another basis, that could be written more succinctly, would be $b_i = e_i+2-e_imod2$, which gives
$$b_0 = x^2-1$$
$$b_1=x^3-x$$
$$b_2=x^4-1$$
$$b_3 = x^5-x$$
answered Jul 18 at 21:52
Acccumulation
4,4732314
Written in linear algebra terms, given the basis $e_i = x^i$ and a polynomial $p$ expressed as
$$beginbmatrix
p_0 \
p_1 \
p_2 \
p_3\
p_4\
p_5
endbmatrix$$
evaluating $p$ at 1 is the same as $[1,1,1,1,1,1]p$, and evaluating it at -1 is $[1,-1,1,-1,1,-1]p$
We have that both are equal to zero, so
$$beginbmatrix
1&1&1&1&1&1 \
1&-1&1&-1&1&-1
endbmatrixp =beginbmatrix
0 \
0
endbmatrix$$
So we can take the augmented matrix
$$beginbmatrix
1&1&1&1&1&1 &|0 \
1&-1&1&-1&1&-1&|0
endbmatrix$$
And reduce it to row-echelon form
$$beginbmatrix
1&0&1&0&1&0 &|0 \
0&1&0&1&0&1&|0
endbmatrix$$
This tells you that among even $i$, you can pick two $p_i$ to be free and the third will be fixed, and similarly for odd $i$. For instance, we can have $p_i$ free for $i<4$, and then $p_5=-(p_3+p_1)$ and $p_4=-(p_2+p_0)$. If we successively set one free variable to 1 and the other free variables to zero, this gives
$$1-x^4$$
$$x-x^5$$
$$x^2-x^4$$
$$x^3-x^5$$
Another basis, that could be written more succinctly, would be $b_i = e_i+2-e_imod2$, which gives
$$b_0 = x^2-1$$
$$b_1=x^3-x$$
$$b_2=x^4-1$$
$$b_3 = x^5-x$$
answered Jul 18 at 21:52
Acccumulation
4,4732314
answered Jul 18 at 21:52
Acccumulation
4,4732314
answered Jul 18 at 21:52
Acccumulation
4,4732314
answered Jul 18 at 21:52
Acccumulation
4,4732314
4,4732314
add a comment |Â
add a comment |Â
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1
As there are two constraints, two degrees of freedom are consumed and $W$ has dimension $4$.
– Yves Daoust
Jul 18 at 15:43
Yes, that is what I meant. I accidentally typed in the wrong thing. Thank you
– Mr. Frothingslosh
Jul 18 at 15:46
1
@YvesDaoust There are some caveats to that, as restrictions aren't necessarily independent, and some times the resulting subset isn't a vector space. But those are easily ruled out in this case, and it's a good general rule of thumb that each restriction decreases dimension by $1$.
– Arthur
Jul 18 at 15:47
Very interesting. I didn't know that.
– Mr. Frothingslosh
Jul 18 at 15:51