Subspace, Direct Sum, Polynomials, Basis












3












$begingroup$



Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.

a. Find a basis for $U$.

b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.

c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.




If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.



Moreover I am stuck with the part (c).






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, you already have a $W$: it's the subspace generated by $x^2$.
    $endgroup$
    – Bernard
    May 1 '15 at 21:59












  • $begingroup$
    Did I do it correctly?
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:01










  • $begingroup$
    Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
    $endgroup$
    – Bernard
    May 1 '15 at 22:05












  • $begingroup$
    thanks for pointing it out.
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:09
















3












$begingroup$



Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.

a. Find a basis for $U$.

b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.

c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.




If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.



Moreover I am stuck with the part (c).






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Well, you already have a $W$: it's the subspace generated by $x^2$.
    $endgroup$
    – Bernard
    May 1 '15 at 21:59












  • $begingroup$
    Did I do it correctly?
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:01










  • $begingroup$
    Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
    $endgroup$
    – Bernard
    May 1 '15 at 22:05












  • $begingroup$
    thanks for pointing it out.
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:09














3












3








3





$begingroup$



Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.

a. Find a basis for $U$.

b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.

c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.




If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.



Moreover I am stuck with the part (c).






linear-algebra







share|cite|improve this question











$endgroup$





Let $U = {p in mathcal{P}_4(mathbb{F}) ;colon; p''(6) = 0}$.

a. Find a basis for $U$.

b. Extend the basis in part (a) to a basis for $mathcal{P}_4(mathbb{F})$.

c. Find a subspace $W$ of $mathcal{P}_4(mathbb{F})$ such that
$mathcal{P}_4(mathbb{F}) = U oplus W$.




If I take the basis as $1$, $x$, $x^3 -18x^2$, and $x^4-12x^3$.
Now $x^2$ can't be produced by the basis elements so adding $x^2$ to the previous basis I get the basis of $mathcal{P}_4(mathbb{F})$.
Please let me know if I am correct.



Moreover I am stuck with the part (c).






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 May 1 '15 at 21:56









Mike Pierce

11.5k103584




11.5k103584










asked May 1 '15 at 21:48









jyoti prokash royjyoti prokash roy

459510




459510








  • 1




    $begingroup$
    Well, you already have a $W$: it's the subspace generated by $x^2$.
    $endgroup$
    – Bernard
    May 1 '15 at 21:59












  • $begingroup$
    Did I do it correctly?
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:01










  • $begingroup$
    Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
    $endgroup$
    – Bernard
    May 1 '15 at 22:05












  • $begingroup$
    thanks for pointing it out.
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:09














  • 1




    $begingroup$
    Well, you already have a $W$: it's the subspace generated by $x^2$.
    $endgroup$
    – Bernard
    May 1 '15 at 21:59












  • $begingroup$
    Did I do it correctly?
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:01










  • $begingroup$
    Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
    $endgroup$
    – Bernard
    May 1 '15 at 22:05












  • $begingroup$
    thanks for pointing it out.
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:09








1




1




$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59






$begingroup$
Well, you already have a $W$: it's the subspace generated by $x^2$.
$endgroup$
– Bernard
May 1 '15 at 21:59














$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01




$begingroup$
Did I do it correctly?
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:01












$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05






$begingroup$
Roughly speaking, yes. Maybe you should give some details as to why $x^2$ is not in the subspace generated by the first $3$ vectors.
$endgroup$
– Bernard
May 1 '15 at 22:05














$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09




$begingroup$
thanks for pointing it out.
$endgroup$
– jyoti prokash roy
May 1 '15 at 22:09










2 Answers
2






active

oldest

votes


















2












$begingroup$

If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
begin{align}
p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
p''(x)&=2a_2+6a_3x+12a_4x^2
end{align}
so
$$
p''(6)=2a_2+36a_3+432a_4
$$
and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
$$
{1,x,-18x^2+x^3,-216x^2+x^4}
$$
Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.



Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.



So you're correct. I just found a different basis, with a more systematic approach.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thanks a lot,really appreciate your help.
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:09



















0












$begingroup$

Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)



You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
$$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
You have
begin{align*}
p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
end{align*}
and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
So you have
$$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.




The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
    begin{align}
    p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
    p''(x)&=2a_2+6a_3x+12a_4x^2
    end{align}
    so
    $$
    p''(6)=2a_2+36a_3+432a_4
    $$
    and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
    $$
    {1,x,-18x^2+x^3,-216x^2+x^4}
    $$
    Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.



    Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.



    So you're correct. I just found a different basis, with a more systematic approach.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks a lot,really appreciate your help.
      $endgroup$
      – jyoti prokash roy
      May 1 '15 at 22:09
















    2












    $begingroup$

    If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
    begin{align}
    p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
    p''(x)&=2a_2+6a_3x+12a_4x^2
    end{align}
    so
    $$
    p''(6)=2a_2+36a_3+432a_4
    $$
    and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
    $$
    {1,x,-18x^2+x^3,-216x^2+x^4}
    $$
    Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.



    Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.



    So you're correct. I just found a different basis, with a more systematic approach.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      thanks a lot,really appreciate your help.
      $endgroup$
      – jyoti prokash roy
      May 1 '15 at 22:09














    2












    2








    2





    $begingroup$

    If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
    begin{align}
    p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
    p''(x)&=2a_2+6a_3x+12a_4x^2
    end{align}
    so
    $$
    p''(6)=2a_2+36a_3+432a_4
    $$
    and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
    $$
    {1,x,-18x^2+x^3,-216x^2+x^4}
    $$
    Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.



    Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.



    So you're correct. I just found a different basis, with a more systematic approach.






    share|cite|improve this answer









    $endgroup$



    If $p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$ is a polynomial of degree at most $4$, then
    begin{align}
    p'(x)&=a_1+2a_2x+3a_3x^2+4a_4x^3\
    p''(x)&=2a_2+6a_3x+12a_4x^2
    end{align}
    so
    $$
    p''(6)=2a_2+36a_3+432a_4
    $$
    and $p''(6)=0$ can be written as $a_2=-18a_3-216a_4$. So the free variables are $a_0$, $a_1$, $a_3$ and $a_4$ and a basis is given by
    $$
    {1,x,-18x^2+x^3,-216x^2+x^4}
    $$
    Now you have just to find a polynomial $q$ such that $qnotin U$, for instance $q(x)=x^2$.



    Add this to the above basis; of course, you can also take $W=operatorname{Span}(x^2)$.



    So you're correct. I just found a different basis, with a more systematic approach.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 1 '15 at 22:04









    egregegreg

    183k1486204




    183k1486204












    • $begingroup$
      thanks a lot,really appreciate your help.
      $endgroup$
      – jyoti prokash roy
      May 1 '15 at 22:09


















    • $begingroup$
      thanks a lot,really appreciate your help.
      $endgroup$
      – jyoti prokash roy
      May 1 '15 at 22:09
















    $begingroup$
    thanks a lot,really appreciate your help.
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:09




    $begingroup$
    thanks a lot,really appreciate your help.
    $endgroup$
    – jyoti prokash roy
    May 1 '15 at 22:09











    0












    $begingroup$

    Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)



    You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
    $$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
    This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
    You have
    begin{align*}
    p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
    p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
    p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
    end{align*}
    and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
    So you have
    $$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
    So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.




    The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)



      You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
      $$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
      This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
      You have
      begin{align*}
      p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
      p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
      p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
      end{align*}
      and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
      So you have
      $$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
      So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.




      The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)



        You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
        $$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
        This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
        You have
        begin{align*}
        p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
        p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
        p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
        end{align*}
        and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
        So you have
        $$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
        So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.




        The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.






        share|cite|improve this answer









        $endgroup$



        Let us assume $mathbb F=mathbb R$. (Or $mathbb F$ is any field with characteristic which does not cause any problems - and we'll check later whether there might be some problems for some other fields.)



        You could use the fact that very polynomial in $mathcal P_4$ can be expressed (uniquely) as
        $$p(x)=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4.$$
        This is simply another way of saying that $1$, $x-6$, $(x-6)^2$, $(x-6)^3$, $(x-6)^4$ is a basis. We have tried this basis mainly to simplify a bit checking the values at $x=6$.
        You have
        begin{align*}
        p(x)&=c_0+c_1(x-6)+c_2(x-6)^2+c_3(x-6)^3+c_4(x-6)^4\
        p'(x)&=c_1+2c_2(x-6)+3c_3(x-6)^2+4c_4(x-6)^3\
        p''(x)&=2c_2+6c_3(x-6)+12c_4(x-6)^2
        end{align*}
        and the condition describing $U$ reduces simply to $$p''(6)=c_2=0.$$
        So you have
        $$U={p(x)=c_0+c_1(x-6)+c_3(x-6)^3+c_4(x-6)^4; c_0,c_1,c_2,c_3inmathbb F}.$$
        So we now see that $1$, $x-6$, $(x-6)^3$, $(x-6)^4$ is a basis for $U$. And for the space $W$ we can choose, for example, the span of $(x-6)^2$.




        The only difference is if $mathbb F$ has characteristic $2$, in which case we get $p''(x)=0$ and $U=mathcal P_4$.







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        answered Dec 30 '18 at 9:23









        Martin SleziakMartin Sleziak

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