Subspace of $ P_5$?












1












$begingroup$


Problem:




Let $U$:= {$p$ ∈ $P_5$$(mathbb{R})$ : $p(−1)$ = $p(1)$ = $0$}.Show that $U$ is a subspace of $P_5$$(mathbb{R})$ . Find a basis and determine the dimension of $U$.




Solution:





  1. Proving that $U$ is a subspace:


Let $n$ be the zero polynomial, $n$(-1) =$0$= $n$(1) = $0$ then $n$ ∈ $U$.



Let $f$(1)=$0$ and $g$(1)=$0$ ∈ $U$ and $h$= $f$ +$g$ ,then we obtain that $1$ +$1$=$0$=$h$, so $h$ ∈ $U$



Let further $c$ ∈ $mathbb{R}$ then $f$(1) . $c$= $0$.$c$ = $0$ ∈ $U$, so $P_5$$(mathbb{R})$ is a subspace of $U$



2.Basis and Dimension of $U$



$p$=$a_0$+$a_1$$x$+2$a_2$$x^2$+3$a_3$$x^3$+4$a_4$$x^4$ +5$a_5$$x^5$
Could someone check if my solution of the problem is right and give me a hint how to find the basis of $U$? Is the Basis the answer of the equlation system ?







linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    I don't get why in part $2$ you are multiplying $x^n$ by $n$
    $endgroup$
    – user289143
    Jan 20 at 18:39
















1












$begingroup$


Problem:




Let $U$:= {$p$ ∈ $P_5$$(mathbb{R})$ : $p(−1)$ = $p(1)$ = $0$}.Show that $U$ is a subspace of $P_5$$(mathbb{R})$ . Find a basis and determine the dimension of $U$.




Solution:





  1. Proving that $U$ is a subspace:


Let $n$ be the zero polynomial, $n$(-1) =$0$= $n$(1) = $0$ then $n$ ∈ $U$.



Let $f$(1)=$0$ and $g$(1)=$0$ ∈ $U$ and $h$= $f$ +$g$ ,then we obtain that $1$ +$1$=$0$=$h$, so $h$ ∈ $U$



Let further $c$ ∈ $mathbb{R}$ then $f$(1) . $c$= $0$.$c$ = $0$ ∈ $U$, so $P_5$$(mathbb{R})$ is a subspace of $U$



2.Basis and Dimension of $U$



$p$=$a_0$+$a_1$$x$+2$a_2$$x^2$+3$a_3$$x^3$+4$a_4$$x^4$ +5$a_5$$x^5$
Could someone check if my solution of the problem is right and give me a hint how to find the basis of $U$? Is the Basis the answer of the equlation system ?







linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    I don't get why in part $2$ you are multiplying $x^n$ by $n$
    $endgroup$
    – user289143
    Jan 20 at 18:39














1












1








1





$begingroup$


Problem:




Let $U$:= {$p$ ∈ $P_5$$(mathbb{R})$ : $p(−1)$ = $p(1)$ = $0$}.Show that $U$ is a subspace of $P_5$$(mathbb{R})$ . Find a basis and determine the dimension of $U$.




Solution:





  1. Proving that $U$ is a subspace:


Let $n$ be the zero polynomial, $n$(-1) =$0$= $n$(1) = $0$ then $n$ ∈ $U$.



Let $f$(1)=$0$ and $g$(1)=$0$ ∈ $U$ and $h$= $f$ +$g$ ,then we obtain that $1$ +$1$=$0$=$h$, so $h$ ∈ $U$



Let further $c$ ∈ $mathbb{R}$ then $f$(1) . $c$= $0$.$c$ = $0$ ∈ $U$, so $P_5$$(mathbb{R})$ is a subspace of $U$



2.Basis and Dimension of $U$



$p$=$a_0$+$a_1$$x$+2$a_2$$x^2$+3$a_3$$x^3$+4$a_4$$x^4$ +5$a_5$$x^5$
Could someone check if my solution of the problem is right and give me a hint how to find the basis of $U$? Is the Basis the answer of the equlation system ?







linear-algebra







share|cite|improve this question









$endgroup$




Problem:




Let $U$:= {$p$ ∈ $P_5$$(mathbb{R})$ : $p(−1)$ = $p(1)$ = $0$}.Show that $U$ is a subspace of $P_5$$(mathbb{R})$ . Find a basis and determine the dimension of $U$.




Solution:





  1. Proving that $U$ is a subspace:


Let $n$ be the zero polynomial, $n$(-1) =$0$= $n$(1) = $0$ then $n$ ∈ $U$.



Let $f$(1)=$0$ and $g$(1)=$0$ ∈ $U$ and $h$= $f$ +$g$ ,then we obtain that $1$ +$1$=$0$=$h$, so $h$ ∈ $U$



Let further $c$ ∈ $mathbb{R}$ then $f$(1) . $c$= $0$.$c$ = $0$ ∈ $U$, so $P_5$$(mathbb{R})$ is a subspace of $U$



2.Basis and Dimension of $U$



$p$=$a_0$+$a_1$$x$+2$a_2$$x^2$+3$a_3$$x^3$+4$a_4$$x^4$ +5$a_5$$x^5$
Could someone check if my solution of the problem is right and give me a hint how to find the basis of $U$? Is the Basis the answer of the equlation system ?







linear-algebra




linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 20 at 18:31









KaiKai

446




446












  • $begingroup$
    I don't get why in part $2$ you are multiplying $x^n$ by $n$
    $endgroup$
    – user289143
    Jan 20 at 18:39


















  • $begingroup$
    I don't get why in part $2$ you are multiplying $x^n$ by $n$
    $endgroup$
    – user289143
    Jan 20 at 18:39
















$begingroup$
I don't get why in part $2$ you are multiplying $x^n$ by $n$
$endgroup$
– user289143
Jan 20 at 18:39




$begingroup$
I don't get why in part $2$ you are multiplying $x^n$ by $n$
$endgroup$
– user289143
Jan 20 at 18:39










1 Answer
1






active

oldest

votes


















2












$begingroup$

Clearly the zero polynomial belongs to $U$.

Now if $f,g in U$, we have $(f+g)(1)=f(1)+g(1)=0$ and $(f+g)(-1)=f(-1)+g(-1)=0$, hence $f+g in U$.
Now if $f in U$ and $c in mathbb{R}$, $(cf)(1)=c cdot f(1)= c cdot 0=0$ and $(cf)(-1)=c cdot f(-1)= c cdot 0=0$, hence $cf in U$.

Therefore $U$ is a subspace of $P_5(mathbb{R})$




For part $2)$ You can write every polynomial in $U$ as $p(x)=(x-1)(x+1)q(x)$ for some $q(x)$ with $mathrm{deg} q(x) leq 3$. Hence a basis for $U$ is given by $(x-1)(x+1), x(x-1)(x+1), x^2(x-1)(x+1), x^3(x-1)(x+1)$ and $mathrm{dim} U =4$, i.e. if $p(x) in U$ we have $p(x)=a_0(x-1)(x+1)+a_1x(x-1)(x+1)+a_2x^2(x-1)(x+1)+a_3x^3(x-1)(x+1)=(x-1)(x+1)(a_0+a_1x+a_2x^2+a_3x^3)$ where $a_0, a_1, a_2, a_3$ are uniquely determined






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You forgot to multiply the basis elements by $(x-1)(x+1)$..
    $endgroup$
    – Berci
    Jan 20 at 18:55










  • $begingroup$
    Yes sorry, I was giving a basis for the $q(x)$
    $endgroup$
    – user289143
    Jan 20 at 18:57










  • $begingroup$
    @289143 So the Basis is then made of the solution of all four equalations?
    $endgroup$
    – Kai
    Jan 20 at 18:58






  • 1




    $begingroup$
    No, the basis is given by these $4$ polynomials: you can write every $p(x) in U$ as a linear combination of these polynomials in a uniquely way
    $endgroup$
    – user289143
    Jan 20 at 19:00








  • 1




    $begingroup$
    Exactly, because if $p(alpha)=0$ I know that $p(x)=(x-alpha)q(x)$ with $q(x) neq 0$
    $endgroup$
    – user289143
    Jan 20 at 19:55











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1 Answer
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1 Answer
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active

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active

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active

oldest

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2












$begingroup$

Clearly the zero polynomial belongs to $U$.

Now if $f,g in U$, we have $(f+g)(1)=f(1)+g(1)=0$ and $(f+g)(-1)=f(-1)+g(-1)=0$, hence $f+g in U$.
Now if $f in U$ and $c in mathbb{R}$, $(cf)(1)=c cdot f(1)= c cdot 0=0$ and $(cf)(-1)=c cdot f(-1)= c cdot 0=0$, hence $cf in U$.

Therefore $U$ is a subspace of $P_5(mathbb{R})$




For part $2)$ You can write every polynomial in $U$ as $p(x)=(x-1)(x+1)q(x)$ for some $q(x)$ with $mathrm{deg} q(x) leq 3$. Hence a basis for $U$ is given by $(x-1)(x+1), x(x-1)(x+1), x^2(x-1)(x+1), x^3(x-1)(x+1)$ and $mathrm{dim} U =4$, i.e. if $p(x) in U$ we have $p(x)=a_0(x-1)(x+1)+a_1x(x-1)(x+1)+a_2x^2(x-1)(x+1)+a_3x^3(x-1)(x+1)=(x-1)(x+1)(a_0+a_1x+a_2x^2+a_3x^3)$ where $a_0, a_1, a_2, a_3$ are uniquely determined






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You forgot to multiply the basis elements by $(x-1)(x+1)$..
    $endgroup$
    – Berci
    Jan 20 at 18:55










  • $begingroup$
    Yes sorry, I was giving a basis for the $q(x)$
    $endgroup$
    – user289143
    Jan 20 at 18:57










  • $begingroup$
    @289143 So the Basis is then made of the solution of all four equalations?
    $endgroup$
    – Kai
    Jan 20 at 18:58






  • 1




    $begingroup$
    No, the basis is given by these $4$ polynomials: you can write every $p(x) in U$ as a linear combination of these polynomials in a uniquely way
    $endgroup$
    – user289143
    Jan 20 at 19:00








  • 1




    $begingroup$
    Exactly, because if $p(alpha)=0$ I know that $p(x)=(x-alpha)q(x)$ with $q(x) neq 0$
    $endgroup$
    – user289143
    Jan 20 at 19:55
















2












$begingroup$

Clearly the zero polynomial belongs to $U$.

Now if $f,g in U$, we have $(f+g)(1)=f(1)+g(1)=0$ and $(f+g)(-1)=f(-1)+g(-1)=0$, hence $f+g in U$.
Now if $f in U$ and $c in mathbb{R}$, $(cf)(1)=c cdot f(1)= c cdot 0=0$ and $(cf)(-1)=c cdot f(-1)= c cdot 0=0$, hence $cf in U$.

Therefore $U$ is a subspace of $P_5(mathbb{R})$




For part $2)$ You can write every polynomial in $U$ as $p(x)=(x-1)(x+1)q(x)$ for some $q(x)$ with $mathrm{deg} q(x) leq 3$. Hence a basis for $U$ is given by $(x-1)(x+1), x(x-1)(x+1), x^2(x-1)(x+1), x^3(x-1)(x+1)$ and $mathrm{dim} U =4$, i.e. if $p(x) in U$ we have $p(x)=a_0(x-1)(x+1)+a_1x(x-1)(x+1)+a_2x^2(x-1)(x+1)+a_3x^3(x-1)(x+1)=(x-1)(x+1)(a_0+a_1x+a_2x^2+a_3x^3)$ where $a_0, a_1, a_2, a_3$ are uniquely determined






share|cite|improve this answer











$endgroup$













  • $begingroup$
    You forgot to multiply the basis elements by $(x-1)(x+1)$..
    $endgroup$
    – Berci
    Jan 20 at 18:55










  • $begingroup$
    Yes sorry, I was giving a basis for the $q(x)$
    $endgroup$
    – user289143
    Jan 20 at 18:57










  • $begingroup$
    @289143 So the Basis is then made of the solution of all four equalations?
    $endgroup$
    – Kai
    Jan 20 at 18:58






  • 1




    $begingroup$
    No, the basis is given by these $4$ polynomials: you can write every $p(x) in U$ as a linear combination of these polynomials in a uniquely way
    $endgroup$
    – user289143
    Jan 20 at 19:00








  • 1




    $begingroup$
    Exactly, because if $p(alpha)=0$ I know that $p(x)=(x-alpha)q(x)$ with $q(x) neq 0$
    $endgroup$
    – user289143
    Jan 20 at 19:55














2












2








2





$begingroup$

Clearly the zero polynomial belongs to $U$.

Now if $f,g in U$, we have $(f+g)(1)=f(1)+g(1)=0$ and $(f+g)(-1)=f(-1)+g(-1)=0$, hence $f+g in U$.
Now if $f in U$ and $c in mathbb{R}$, $(cf)(1)=c cdot f(1)= c cdot 0=0$ and $(cf)(-1)=c cdot f(-1)= c cdot 0=0$, hence $cf in U$.

Therefore $U$ is a subspace of $P_5(mathbb{R})$




For part $2)$ You can write every polynomial in $U$ as $p(x)=(x-1)(x+1)q(x)$ for some $q(x)$ with $mathrm{deg} q(x) leq 3$. Hence a basis for $U$ is given by $(x-1)(x+1), x(x-1)(x+1), x^2(x-1)(x+1), x^3(x-1)(x+1)$ and $mathrm{dim} U =4$, i.e. if $p(x) in U$ we have $p(x)=a_0(x-1)(x+1)+a_1x(x-1)(x+1)+a_2x^2(x-1)(x+1)+a_3x^3(x-1)(x+1)=(x-1)(x+1)(a_0+a_1x+a_2x^2+a_3x^3)$ where $a_0, a_1, a_2, a_3$ are uniquely determined






share|cite|improve this answer











$endgroup$



Clearly the zero polynomial belongs to $U$.

Now if $f,g in U$, we have $(f+g)(1)=f(1)+g(1)=0$ and $(f+g)(-1)=f(-1)+g(-1)=0$, hence $f+g in U$.
Now if $f in U$ and $c in mathbb{R}$, $(cf)(1)=c cdot f(1)= c cdot 0=0$ and $(cf)(-1)=c cdot f(-1)= c cdot 0=0$, hence $cf in U$.

Therefore $U$ is a subspace of $P_5(mathbb{R})$




For part $2)$ You can write every polynomial in $U$ as $p(x)=(x-1)(x+1)q(x)$ for some $q(x)$ with $mathrm{deg} q(x) leq 3$. Hence a basis for $U$ is given by $(x-1)(x+1), x(x-1)(x+1), x^2(x-1)(x+1), x^3(x-1)(x+1)$ and $mathrm{dim} U =4$, i.e. if $p(x) in U$ we have $p(x)=a_0(x-1)(x+1)+a_1x(x-1)(x+1)+a_2x^2(x-1)(x+1)+a_3x^3(x-1)(x+1)=(x-1)(x+1)(a_0+a_1x+a_2x^2+a_3x^3)$ where $a_0, a_1, a_2, a_3$ are uniquely determined







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 20 at 19:02

























answered Jan 20 at 18:37









user289143user289143

903313




903313












  • $begingroup$
    You forgot to multiply the basis elements by $(x-1)(x+1)$..
    $endgroup$
    – Berci
    Jan 20 at 18:55










  • $begingroup$
    Yes sorry, I was giving a basis for the $q(x)$
    $endgroup$
    – user289143
    Jan 20 at 18:57










  • $begingroup$
    @289143 So the Basis is then made of the solution of all four equalations?
    $endgroup$
    – Kai
    Jan 20 at 18:58






  • 1




    $begingroup$
    No, the basis is given by these $4$ polynomials: you can write every $p(x) in U$ as a linear combination of these polynomials in a uniquely way
    $endgroup$
    – user289143
    Jan 20 at 19:00








  • 1




    $begingroup$
    Exactly, because if $p(alpha)=0$ I know that $p(x)=(x-alpha)q(x)$ with $q(x) neq 0$
    $endgroup$
    – user289143
    Jan 20 at 19:55


















  • $begingroup$
    You forgot to multiply the basis elements by $(x-1)(x+1)$..
    $endgroup$
    – Berci
    Jan 20 at 18:55










  • $begingroup$
    Yes sorry, I was giving a basis for the $q(x)$
    $endgroup$
    – user289143
    Jan 20 at 18:57










  • $begingroup$
    @289143 So the Basis is then made of the solution of all four equalations?
    $endgroup$
    – Kai
    Jan 20 at 18:58






  • 1




    $begingroup$
    No, the basis is given by these $4$ polynomials: you can write every $p(x) in U$ as a linear combination of these polynomials in a uniquely way
    $endgroup$
    – user289143
    Jan 20 at 19:00








  • 1




    $begingroup$
    Exactly, because if $p(alpha)=0$ I know that $p(x)=(x-alpha)q(x)$ with $q(x) neq 0$
    $endgroup$
    – user289143
    Jan 20 at 19:55
















$begingroup$
You forgot to multiply the basis elements by $(x-1)(x+1)$..
$endgroup$
– Berci
Jan 20 at 18:55




$begingroup$
You forgot to multiply the basis elements by $(x-1)(x+1)$..
$endgroup$
– Berci
Jan 20 at 18:55












$begingroup$
Yes sorry, I was giving a basis for the $q(x)$
$endgroup$
– user289143
Jan 20 at 18:57




$begingroup$
Yes sorry, I was giving a basis for the $q(x)$
$endgroup$
– user289143
Jan 20 at 18:57












$begingroup$
@289143 So the Basis is then made of the solution of all four equalations?
$endgroup$
– Kai
Jan 20 at 18:58




$begingroup$
@289143 So the Basis is then made of the solution of all four equalations?
$endgroup$
– Kai
Jan 20 at 18:58




1




1




$begingroup$
No, the basis is given by these $4$ polynomials: you can write every $p(x) in U$ as a linear combination of these polynomials in a uniquely way
$endgroup$
– user289143
Jan 20 at 19:00






$begingroup$
No, the basis is given by these $4$ polynomials: you can write every $p(x) in U$ as a linear combination of these polynomials in a uniquely way
$endgroup$
– user289143
Jan 20 at 19:00






1




1




$begingroup$
Exactly, because if $p(alpha)=0$ I know that $p(x)=(x-alpha)q(x)$ with $q(x) neq 0$
$endgroup$
– user289143
Jan 20 at 19:55




$begingroup$
Exactly, because if $p(alpha)=0$ I know that $p(x)=(x-alpha)q(x)$ with $q(x) neq 0$
$endgroup$
– user289143
Jan 20 at 19:55


















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