how to prove a spanning set of polynomial












3












$begingroup$


I am struggling so much understanding this concept of subspace and span.



The question is, Given that



$P2:W={(x+1)(ax+b)| a,b in R}$
show that
${x^2+x, x^2+2x+1}$
is a spanning set of $W$.



I don't know if I got this concept right, but I've tried to do things by letting
$p(x)=x^2+x$
$q(x)=x^2+2x+1$
then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.



but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?






linear-algebra vector-spaces







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    3












    $begingroup$


    I am struggling so much understanding this concept of subspace and span.



    The question is, Given that



    $P2:W={(x+1)(ax+b)| a,b in R}$
    show that
    ${x^2+x, x^2+2x+1}$
    is a spanning set of $W$.



    I don't know if I got this concept right, but I've tried to do things by letting
    $p(x)=x^2+x$
    $q(x)=x^2+2x+1$
    then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.



    but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?






    linear-algebra vector-spaces







    share|cite|improve this question









    New contributor




    nokdoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      3












      3








      3


      1



      $begingroup$


      I am struggling so much understanding this concept of subspace and span.



      The question is, Given that



      $P2:W={(x+1)(ax+b)| a,b in R}$
      show that
      ${x^2+x, x^2+2x+1}$
      is a spanning set of $W$.



      I don't know if I got this concept right, but I've tried to do things by letting
      $p(x)=x^2+x$
      $q(x)=x^2+2x+1$
      then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.



      but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?






      linear-algebra vector-spaces







      share|cite|improve this question









      New contributor




      nokdoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I am struggling so much understanding this concept of subspace and span.



      The question is, Given that



      $P2:W={(x+1)(ax+b)| a,b in R}$
      show that
      ${x^2+x, x^2+2x+1}$
      is a spanning set of $W$.



      I don't know if I got this concept right, but I've tried to do things by letting
      $p(x)=x^2+x$
      $q(x)=x^2+2x+1$
      then multiplying them a coefficient $alpha$ and $beta$ each and adding to fit in with $W$.



      but then I got an answer saying $a=alpha + beta$, $a+b = alpha+ 2beta, b=beta$, which means... no solution? I am guessing? so this does not span $W$. Am I right?






      linear-algebra vector-spaces




      linear-algebra vector-spaces






      share|cite|improve this question









      New contributor




      nokdoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      nokdoo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      Thomas Shelby

      2,685421




      2,685421






      New contributor




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      asked 4 hours ago









      nokdoonokdoo

      161




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          3 Answers
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          $begingroup$

          $(x+1)(ax+b)=ax^2+(b+a)x+b.$



          Claim: There exists $alpha, betain Bbb R $ such that
          $$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
          $$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
          Comparing coefficients, we get
          $$a=alpha+betatag2$$
          $$b+a=alpha+2betatag3$$
          $$b=beta tag4$$
          Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
          In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.



            On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              Hint: First, $W$ is a $2$-dimensional vector space (easy to see).



              Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)



              Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.



              Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
              .






              share|cite|improve this answer











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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                $(x+1)(ax+b)=ax^2+(b+a)x+b.$



                Claim: There exists $alpha, betain Bbb R $ such that
                $$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
                $$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
                Comparing coefficients, we get
                $$a=alpha+betatag2$$
                $$b+a=alpha+2betatag3$$
                $$b=beta tag4$$
                Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
                In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  $(x+1)(ax+b)=ax^2+(b+a)x+b.$



                  Claim: There exists $alpha, betain Bbb R $ such that
                  $$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
                  $$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
                  Comparing coefficients, we get
                  $$a=alpha+betatag2$$
                  $$b+a=alpha+2betatag3$$
                  $$b=beta tag4$$
                  Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
                  In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    $(x+1)(ax+b)=ax^2+(b+a)x+b.$



                    Claim: There exists $alpha, betain Bbb R $ such that
                    $$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
                    $$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
                    Comparing coefficients, we get
                    $$a=alpha+betatag2$$
                    $$b+a=alpha+2betatag3$$
                    $$b=beta tag4$$
                    Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
                    In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.






                    share|cite|improve this answer









                    $endgroup$



                    $(x+1)(ax+b)=ax^2+(b+a)x+b.$



                    Claim: There exists $alpha, betain Bbb R $ such that
                    $$ax^2+(b+a)x+b=alpha(x^2+x)+beta( x^2+2x+1)tag1$$
                    $$ax^2+(b+a)x+b=(alpha+beta)x^2+(alpha+2beta)x+beta$$
                    Comparing coefficients, we get
                    $$a=alpha+betatag2$$
                    $$b+a=alpha+2betatag3$$
                    $$b=beta tag4$$
                    Substituting $(4) $ in $(1) $ gives $alpha=a-b$ (you can verify the solution using $(3)$).
                    In other words, $forall a,binBbb R $, there exists $alpha=a-b, beta =b $ such that $(1) $ holds. Hence ${x^2+x,x^2+2x+1} $ is indeed a spanning set.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    Thomas ShelbyThomas Shelby

                    2,685421




                    2,685421























                        2












                        $begingroup$

                        No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.



                        On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.






                        share|cite|improve this answer











                        $endgroup$


















                          2












                          $begingroup$

                          No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.



                          On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.






                          share|cite|improve this answer











                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.



                            On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.






                            share|cite|improve this answer











                            $endgroup$



                            No, you are wrong. Note that$$x^2+x=(x+1)xtext{ and that }x^2+2x+1=(x+1)(x+1).$$So, both polynomails $x^2+x$ and $x^2+2x+1$ belong indeed to $W$ and therefore they span a subspace of $W$.



                            On the other hand, if $(x+1)(ax+b)in W$, thenbegin{align}(x+1)(ax+b)&=(a-b)(x+1)x+b(x+1)(x+1)\&=(a-b)(x^2+x)+b(x^2+2x+1)end{align}which belongs, of course, to $operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$. So, $W=operatorname{span}bigl({x^2+x,x^2+2x+1}bigr)$.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 3 hours ago

























                            answered 4 hours ago









                            José Carlos SantosJosé Carlos Santos

                            157k22126227




                            157k22126227























                                1












                                $begingroup$

                                Hint: First, $W$ is a $2$-dimensional vector space (easy to see).



                                Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)



                                Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.



                                Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
                                .






                                share|cite|improve this answer











                                $endgroup$


















                                  1












                                  $begingroup$

                                  Hint: First, $W$ is a $2$-dimensional vector space (easy to see).



                                  Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)



                                  Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.



                                  Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
                                  .






                                  share|cite|improve this answer











                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    Hint: First, $W$ is a $2$-dimensional vector space (easy to see).



                                    Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)



                                    Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.



                                    Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
                                    .






                                    share|cite|improve this answer











                                    $endgroup$



                                    Hint: First, $W$ is a $2$-dimensional vector space (easy to see).



                                    Now, ${x^2+x,x^2+2x+1}$ is linearly independent (easy to see)



                                    Let $a=1,b=0$. We get $x^2+x$. Now let $a=1,b=1$. We get $x^2+2x+1$.



                                    Thus $W=operatorname{span} {x^2+x,x^2+2x+1}$.
                                    .







                                    share|cite|improve this answer














                                    share|cite|improve this answer



                                    share|cite|improve this answer








                                    edited 3 hours ago

























                                    answered 3 hours ago









                                    Chris CusterChris Custer

                                    11.8k3825




                                    11.8k3825






















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