$U={ pin P_{4}left( mathbb{R} right) : p''left( 6right) =0}$ Find a basis for $U$












1












$begingroup$



Find a basis for $U$, where
$$U={ pin P_{4}left( mathbb{R} right) : p''left( 6right) =0}.$$




This question is 2.C 5 of Linear Algebra Done right.



I would like to know how this person got the example basis: $${1, x, x^{3}-18x^{2},x^{4}-12x^{3}}.$$



My attempt:



Take $ fleft( xright) = ax^{4}+bx^{3}+cx^{2}+dx + e$



$f''= 12ax^{2}+6bx+2c $



$implies f''left( 6right) = 0 implies c = -6^{3}a-18b$



$implies fleft( xright) = a(x^{4}-6^{3}x^{2})+b(x^{3}-18x^{2})+dx + e$



$implies$ my basis is ${1, x, x^{3}-18x^{2},x^{4}-6^{3}x^{2}}$



Where have I gone wrong?






linear-algebra proof-verification polynomials vector-spaces hamel-basis







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  • 1




    $begingroup$
    You haven't done anything wrong. Both answers are correct. Basis is not unique.
    $endgroup$
    – Kavi Rama Murthy
    Sep 17 '18 at 8:40


















1












$begingroup$



Find a basis for $U$, where
$$U={ pin P_{4}left( mathbb{R} right) : p''left( 6right) =0}.$$




This question is 2.C 5 of Linear Algebra Done right.



I would like to know how this person got the example basis: $${1, x, x^{3}-18x^{2},x^{4}-12x^{3}}.$$



My attempt:



Take $ fleft( xright) = ax^{4}+bx^{3}+cx^{2}+dx + e$



$f''= 12ax^{2}+6bx+2c $



$implies f''left( 6right) = 0 implies c = -6^{3}a-18b$



$implies fleft( xright) = a(x^{4}-6^{3}x^{2})+b(x^{3}-18x^{2})+dx + e$



$implies$ my basis is ${1, x, x^{3}-18x^{2},x^{4}-6^{3}x^{2}}$



Where have I gone wrong?






linear-algebra proof-verification polynomials vector-spaces hamel-basis







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You haven't done anything wrong. Both answers are correct. Basis is not unique.
    $endgroup$
    – Kavi Rama Murthy
    Sep 17 '18 at 8:40
















1












1








1





$begingroup$



Find a basis for $U$, where
$$U={ pin P_{4}left( mathbb{R} right) : p''left( 6right) =0}.$$




This question is 2.C 5 of Linear Algebra Done right.



I would like to know how this person got the example basis: $${1, x, x^{3}-18x^{2},x^{4}-12x^{3}}.$$



My attempt:



Take $ fleft( xright) = ax^{4}+bx^{3}+cx^{2}+dx + e$



$f''= 12ax^{2}+6bx+2c $



$implies f''left( 6right) = 0 implies c = -6^{3}a-18b$



$implies fleft( xright) = a(x^{4}-6^{3}x^{2})+b(x^{3}-18x^{2})+dx + e$



$implies$ my basis is ${1, x, x^{3}-18x^{2},x^{4}-6^{3}x^{2}}$



Where have I gone wrong?






linear-algebra proof-verification polynomials vector-spaces hamel-basis







share|cite|improve this question











$endgroup$





Find a basis for $U$, where
$$U={ pin P_{4}left( mathbb{R} right) : p''left( 6right) =0}.$$




This question is 2.C 5 of Linear Algebra Done right.



I would like to know how this person got the example basis: $${1, x, x^{3}-18x^{2},x^{4}-12x^{3}}.$$



My attempt:



Take $ fleft( xright) = ax^{4}+bx^{3}+cx^{2}+dx + e$



$f''= 12ax^{2}+6bx+2c $



$implies f''left( 6right) = 0 implies c = -6^{3}a-18b$



$implies fleft( xright) = a(x^{4}-6^{3}x^{2})+b(x^{3}-18x^{2})+dx + e$



$implies$ my basis is ${1, x, x^{3}-18x^{2},x^{4}-6^{3}x^{2}}$



Where have I gone wrong?






linear-algebra proof-verification polynomials vector-spaces hamel-basis




linear-algebra proof-verification polynomials vector-spaces hamel-basis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 19:10







user593746

















asked Sep 17 '18 at 8:29









JimSiJimSi

19629




19629








  • 1




    $begingroup$
    You haven't done anything wrong. Both answers are correct. Basis is not unique.
    $endgroup$
    – Kavi Rama Murthy
    Sep 17 '18 at 8:40
















  • 1




    $begingroup$
    You haven't done anything wrong. Both answers are correct. Basis is not unique.
    $endgroup$
    – Kavi Rama Murthy
    Sep 17 '18 at 8:40










1




1




$begingroup$
You haven't done anything wrong. Both answers are correct. Basis is not unique.
$endgroup$
– Kavi Rama Murthy
Sep 17 '18 at 8:40






$begingroup$
You haven't done anything wrong. Both answers are correct. Basis is not unique.
$endgroup$
– Kavi Rama Murthy
Sep 17 '18 at 8:40












2 Answers
2






active

oldest

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3












$begingroup$

There is nothing wrong with both bases. Observe that your basis element $x^4-6^3x^2$ is a linear combination of the "other" basis:
$$x^4-6^3x^2=(x^4-12x^3)+12(x^3-18x^2),.$$
And vice versa, the basis element $x^4-12x^3$ of the "other" basis is obtained via a linear combination of your basis:
$$x^4-12x^3=(x^4-6^3x^2)-12(x^3-18x^2),.$$
As Kavi Rama Murthy said, a vector space does not have a unique basis.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Alternatively, write



    $$f(x) = a_0 + a_1(x-6) + a_2(x-6)^2 + a_3(x-6)^3 + a_4(x-6)^4$$



    so $f in U$ if and only if $$0 = f''(6) = Big(2a_2 + 6a_3(x-6) + 12a_4(x-6)^4Big)Bigg|_{x = 6} = 2a_2$$



    meaning $a_2 = 0$. Therefore, another basis for $U$ is $$Big{1, x-6 , (x-6)^3, (x-6)^4Big}$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      There is nothing wrong with both bases. Observe that your basis element $x^4-6^3x^2$ is a linear combination of the "other" basis:
      $$x^4-6^3x^2=(x^4-12x^3)+12(x^3-18x^2),.$$
      And vice versa, the basis element $x^4-12x^3$ of the "other" basis is obtained via a linear combination of your basis:
      $$x^4-12x^3=(x^4-6^3x^2)-12(x^3-18x^2),.$$
      As Kavi Rama Murthy said, a vector space does not have a unique basis.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        There is nothing wrong with both bases. Observe that your basis element $x^4-6^3x^2$ is a linear combination of the "other" basis:
        $$x^4-6^3x^2=(x^4-12x^3)+12(x^3-18x^2),.$$
        And vice versa, the basis element $x^4-12x^3$ of the "other" basis is obtained via a linear combination of your basis:
        $$x^4-12x^3=(x^4-6^3x^2)-12(x^3-18x^2),.$$
        As Kavi Rama Murthy said, a vector space does not have a unique basis.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          There is nothing wrong with both bases. Observe that your basis element $x^4-6^3x^2$ is a linear combination of the "other" basis:
          $$x^4-6^3x^2=(x^4-12x^3)+12(x^3-18x^2),.$$
          And vice versa, the basis element $x^4-12x^3$ of the "other" basis is obtained via a linear combination of your basis:
          $$x^4-12x^3=(x^4-6^3x^2)-12(x^3-18x^2),.$$
          As Kavi Rama Murthy said, a vector space does not have a unique basis.






          share|cite|improve this answer









          $endgroup$



          There is nothing wrong with both bases. Observe that your basis element $x^4-6^3x^2$ is a linear combination of the "other" basis:
          $$x^4-6^3x^2=(x^4-12x^3)+12(x^3-18x^2),.$$
          And vice versa, the basis element $x^4-12x^3$ of the "other" basis is obtained via a linear combination of your basis:
          $$x^4-12x^3=(x^4-6^3x^2)-12(x^3-18x^2),.$$
          As Kavi Rama Murthy said, a vector space does not have a unique basis.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 17 '18 at 8:47







          user593746






























              1












              $begingroup$

              Alternatively, write



              $$f(x) = a_0 + a_1(x-6) + a_2(x-6)^2 + a_3(x-6)^3 + a_4(x-6)^4$$



              so $f in U$ if and only if $$0 = f''(6) = Big(2a_2 + 6a_3(x-6) + 12a_4(x-6)^4Big)Bigg|_{x = 6} = 2a_2$$



              meaning $a_2 = 0$. Therefore, another basis for $U$ is $$Big{1, x-6 , (x-6)^3, (x-6)^4Big}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Alternatively, write



                $$f(x) = a_0 + a_1(x-6) + a_2(x-6)^2 + a_3(x-6)^3 + a_4(x-6)^4$$



                so $f in U$ if and only if $$0 = f''(6) = Big(2a_2 + 6a_3(x-6) + 12a_4(x-6)^4Big)Bigg|_{x = 6} = 2a_2$$



                meaning $a_2 = 0$. Therefore, another basis for $U$ is $$Big{1, x-6 , (x-6)^3, (x-6)^4Big}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Alternatively, write



                  $$f(x) = a_0 + a_1(x-6) + a_2(x-6)^2 + a_3(x-6)^3 + a_4(x-6)^4$$



                  so $f in U$ if and only if $$0 = f''(6) = Big(2a_2 + 6a_3(x-6) + 12a_4(x-6)^4Big)Bigg|_{x = 6} = 2a_2$$



                  meaning $a_2 = 0$. Therefore, another basis for $U$ is $$Big{1, x-6 , (x-6)^3, (x-6)^4Big}$$






                  share|cite|improve this answer









                  $endgroup$



                  Alternatively, write



                  $$f(x) = a_0 + a_1(x-6) + a_2(x-6)^2 + a_3(x-6)^3 + a_4(x-6)^4$$



                  so $f in U$ if and only if $$0 = f''(6) = Big(2a_2 + 6a_3(x-6) + 12a_4(x-6)^4Big)Bigg|_{x = 6} = 2a_2$$



                  meaning $a_2 = 0$. Therefore, another basis for $U$ is $$Big{1, x-6 , (x-6)^3, (x-6)^4Big}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 17 '18 at 9:28









                  mechanodroidmechanodroid

                  27.3k62446




                  27.3k62446






























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