Need help finding the kernel of a linear transformation $P^2 to mathbb{R}$












5












$begingroup$


The question asks to find the kernel of $S: P^2 to mathbb{R}$ defined by $S(a+bx+cx^2) = a+b+c$.



I know how to find the kernel of a matrix transformation (it's just the null space of the matrix) but I can't conceptualize a transformation from two different types of vector spaces. How would I go about finding a basis ker(S)?






linear-algebra linear-transformations







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    The question asks to find the kernel of $S: P^2 to mathbb{R}$ defined by $S(a+bx+cx^2) = a+b+c$.



    I know how to find the kernel of a matrix transformation (it's just the null space of the matrix) but I can't conceptualize a transformation from two different types of vector spaces. How would I go about finding a basis ker(S)?






    linear-algebra linear-transformations







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      2



      $begingroup$


      The question asks to find the kernel of $S: P^2 to mathbb{R}$ defined by $S(a+bx+cx^2) = a+b+c$.



      I know how to find the kernel of a matrix transformation (it's just the null space of the matrix) but I can't conceptualize a transformation from two different types of vector spaces. How would I go about finding a basis ker(S)?






      linear-algebra linear-transformations







      share|cite|improve this question











      $endgroup$




      The question asks to find the kernel of $S: P^2 to mathbb{R}$ defined by $S(a+bx+cx^2) = a+b+c$.



      I know how to find the kernel of a matrix transformation (it's just the null space of the matrix) but I can't conceptualize a transformation from two different types of vector spaces. How would I go about finding a basis ker(S)?






      linear-algebra linear-transformations




      linear-algebra linear-transformations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '16 at 22:11









      StackTD

      22.7k2050




      22.7k2050










      asked Dec 18 '16 at 21:52









      user5319366user5319366

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      332






















          1 Answer
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          +50







          $begingroup$

          Check the definition. The kernel consists of all the vectors (in your case: second degree polynomials) that are mapped to the zero vector (in your case: the zero polynomial).



          Now since $S left( a+bx+cx^2 right) = a+b+c$, you have:
          $$S left( a+bx+cx^2 right) = 0 iff a+b+c = 0 $$
          So a polynomial of the form $a+bx+cx^2$ is the kernel of $S$ when the sum of its coefficients is equal to $0$.



          For example:





          • $x^2-3x+1$ is not in the kernel because $1-3+1 ne 0$; indeed: $$S(x^2-3x+1) = 1-3+1 = -1$$


          • $x^2-3x+2$ is in the kernel because $1-3+2 = 0$; indeed: $$S(x^2-3x+2) = 1-3+2 = 0$$



          To find a basis of this kernel, notice that you can solve the condition $a+b+c = 0$ for one of the coefficients, e.g. $color{purple}{a=-b-c}$. Vectors in the kernel are thus of the form:
          $$color{purple}{underbrace{-b-c}_{a}}+bx+cx^2 = bleft( color{blue}{-1+x} right) + cleft( color{red}{-1+x^2}right)$$
          Notice that you can always write such a vector as a linear combination of the vectors (polynomials) $color{blue}{-1+x}$ and $color{red}{-1+x^2}$ so these two clearly span the kernel. Verify that they are linearly independent and hence form a basis for the kernel.



          Another way of looking at this kernel: notice that for a polynomial $p(x)=a+bx+cx^2$, you have $p(1)=a+b+c$ so the condition $a+b+c=0$ boils down to $p(1)=0$, i.e. the polynomial has $x=1$ as a root. This allows the kernel to be described as:
          $$mbox{Ker}(S) = left{ p(x) in P^2 ;vert; p(1) = 0right}$$



          You can check that the basis vectors we found above indeed have $x=1$ as a root.



          Thanks to zipirovich for pointing this out in the comments.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $ker(S)={p(x)in P^2;colon; p(1)=0}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root.
            $endgroup$
            – zipirovich
            Dec 18 '16 at 22:19












          • $begingroup$
            @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root.
            $endgroup$
            – StackTD
            Dec 18 '16 at 22:23













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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

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          active

          oldest

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          4





          +50







          $begingroup$

          Check the definition. The kernel consists of all the vectors (in your case: second degree polynomials) that are mapped to the zero vector (in your case: the zero polynomial).



          Now since $S left( a+bx+cx^2 right) = a+b+c$, you have:
          $$S left( a+bx+cx^2 right) = 0 iff a+b+c = 0 $$
          So a polynomial of the form $a+bx+cx^2$ is the kernel of $S$ when the sum of its coefficients is equal to $0$.



          For example:





          • $x^2-3x+1$ is not in the kernel because $1-3+1 ne 0$; indeed: $$S(x^2-3x+1) = 1-3+1 = -1$$


          • $x^2-3x+2$ is in the kernel because $1-3+2 = 0$; indeed: $$S(x^2-3x+2) = 1-3+2 = 0$$



          To find a basis of this kernel, notice that you can solve the condition $a+b+c = 0$ for one of the coefficients, e.g. $color{purple}{a=-b-c}$. Vectors in the kernel are thus of the form:
          $$color{purple}{underbrace{-b-c}_{a}}+bx+cx^2 = bleft( color{blue}{-1+x} right) + cleft( color{red}{-1+x^2}right)$$
          Notice that you can always write such a vector as a linear combination of the vectors (polynomials) $color{blue}{-1+x}$ and $color{red}{-1+x^2}$ so these two clearly span the kernel. Verify that they are linearly independent and hence form a basis for the kernel.



          Another way of looking at this kernel: notice that for a polynomial $p(x)=a+bx+cx^2$, you have $p(1)=a+b+c$ so the condition $a+b+c=0$ boils down to $p(1)=0$, i.e. the polynomial has $x=1$ as a root. This allows the kernel to be described as:
          $$mbox{Ker}(S) = left{ p(x) in P^2 ;vert; p(1) = 0right}$$



          You can check that the basis vectors we found above indeed have $x=1$ as a root.



          Thanks to zipirovich for pointing this out in the comments.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $ker(S)={p(x)in P^2;colon; p(1)=0}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root.
            $endgroup$
            – zipirovich
            Dec 18 '16 at 22:19












          • $begingroup$
            @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root.
            $endgroup$
            – StackTD
            Dec 18 '16 at 22:23


















          4





          +50







          $begingroup$

          Check the definition. The kernel consists of all the vectors (in your case: second degree polynomials) that are mapped to the zero vector (in your case: the zero polynomial).



          Now since $S left( a+bx+cx^2 right) = a+b+c$, you have:
          $$S left( a+bx+cx^2 right) = 0 iff a+b+c = 0 $$
          So a polynomial of the form $a+bx+cx^2$ is the kernel of $S$ when the sum of its coefficients is equal to $0$.



          For example:





          • $x^2-3x+1$ is not in the kernel because $1-3+1 ne 0$; indeed: $$S(x^2-3x+1) = 1-3+1 = -1$$


          • $x^2-3x+2$ is in the kernel because $1-3+2 = 0$; indeed: $$S(x^2-3x+2) = 1-3+2 = 0$$



          To find a basis of this kernel, notice that you can solve the condition $a+b+c = 0$ for one of the coefficients, e.g. $color{purple}{a=-b-c}$. Vectors in the kernel are thus of the form:
          $$color{purple}{underbrace{-b-c}_{a}}+bx+cx^2 = bleft( color{blue}{-1+x} right) + cleft( color{red}{-1+x^2}right)$$
          Notice that you can always write such a vector as a linear combination of the vectors (polynomials) $color{blue}{-1+x}$ and $color{red}{-1+x^2}$ so these two clearly span the kernel. Verify that they are linearly independent and hence form a basis for the kernel.



          Another way of looking at this kernel: notice that for a polynomial $p(x)=a+bx+cx^2$, you have $p(1)=a+b+c$ so the condition $a+b+c=0$ boils down to $p(1)=0$, i.e. the polynomial has $x=1$ as a root. This allows the kernel to be described as:
          $$mbox{Ker}(S) = left{ p(x) in P^2 ;vert; p(1) = 0right}$$



          You can check that the basis vectors we found above indeed have $x=1$ as a root.



          Thanks to zipirovich for pointing this out in the comments.






          share|cite|improve this answer











          $endgroup$









          • 3




            $begingroup$
            The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $ker(S)={p(x)in P^2;colon; p(1)=0}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root.
            $endgroup$
            – zipirovich
            Dec 18 '16 at 22:19












          • $begingroup$
            @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root.
            $endgroup$
            – StackTD
            Dec 18 '16 at 22:23
















          4





          +50







          4





          +50



          4




          +50



          $begingroup$

          Check the definition. The kernel consists of all the vectors (in your case: second degree polynomials) that are mapped to the zero vector (in your case: the zero polynomial).



          Now since $S left( a+bx+cx^2 right) = a+b+c$, you have:
          $$S left( a+bx+cx^2 right) = 0 iff a+b+c = 0 $$
          So a polynomial of the form $a+bx+cx^2$ is the kernel of $S$ when the sum of its coefficients is equal to $0$.



          For example:





          • $x^2-3x+1$ is not in the kernel because $1-3+1 ne 0$; indeed: $$S(x^2-3x+1) = 1-3+1 = -1$$


          • $x^2-3x+2$ is in the kernel because $1-3+2 = 0$; indeed: $$S(x^2-3x+2) = 1-3+2 = 0$$



          To find a basis of this kernel, notice that you can solve the condition $a+b+c = 0$ for one of the coefficients, e.g. $color{purple}{a=-b-c}$. Vectors in the kernel are thus of the form:
          $$color{purple}{underbrace{-b-c}_{a}}+bx+cx^2 = bleft( color{blue}{-1+x} right) + cleft( color{red}{-1+x^2}right)$$
          Notice that you can always write such a vector as a linear combination of the vectors (polynomials) $color{blue}{-1+x}$ and $color{red}{-1+x^2}$ so these two clearly span the kernel. Verify that they are linearly independent and hence form a basis for the kernel.



          Another way of looking at this kernel: notice that for a polynomial $p(x)=a+bx+cx^2$, you have $p(1)=a+b+c$ so the condition $a+b+c=0$ boils down to $p(1)=0$, i.e. the polynomial has $x=1$ as a root. This allows the kernel to be described as:
          $$mbox{Ker}(S) = left{ p(x) in P^2 ;vert; p(1) = 0right}$$



          You can check that the basis vectors we found above indeed have $x=1$ as a root.



          Thanks to zipirovich for pointing this out in the comments.






          share|cite|improve this answer











          $endgroup$



          Check the definition. The kernel consists of all the vectors (in your case: second degree polynomials) that are mapped to the zero vector (in your case: the zero polynomial).



          Now since $S left( a+bx+cx^2 right) = a+b+c$, you have:
          $$S left( a+bx+cx^2 right) = 0 iff a+b+c = 0 $$
          So a polynomial of the form $a+bx+cx^2$ is the kernel of $S$ when the sum of its coefficients is equal to $0$.



          For example:





          • $x^2-3x+1$ is not in the kernel because $1-3+1 ne 0$; indeed: $$S(x^2-3x+1) = 1-3+1 = -1$$


          • $x^2-3x+2$ is in the kernel because $1-3+2 = 0$; indeed: $$S(x^2-3x+2) = 1-3+2 = 0$$



          To find a basis of this kernel, notice that you can solve the condition $a+b+c = 0$ for one of the coefficients, e.g. $color{purple}{a=-b-c}$. Vectors in the kernel are thus of the form:
          $$color{purple}{underbrace{-b-c}_{a}}+bx+cx^2 = bleft( color{blue}{-1+x} right) + cleft( color{red}{-1+x^2}right)$$
          Notice that you can always write such a vector as a linear combination of the vectors (polynomials) $color{blue}{-1+x}$ and $color{red}{-1+x^2}$ so these two clearly span the kernel. Verify that they are linearly independent and hence form a basis for the kernel.



          Another way of looking at this kernel: notice that for a polynomial $p(x)=a+bx+cx^2$, you have $p(1)=a+b+c$ so the condition $a+b+c=0$ boils down to $p(1)=0$, i.e. the polynomial has $x=1$ as a root. This allows the kernel to be described as:
          $$mbox{Ker}(S) = left{ p(x) in P^2 ;vert; p(1) = 0right}$$



          You can check that the basis vectors we found above indeed have $x=1$ as a root.



          Thanks to zipirovich for pointing this out in the comments.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 9:18

























          answered Dec 18 '16 at 22:04









          StackTDStackTD

          22.7k2050




          22.7k2050








          • 3




            $begingroup$
            The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $ker(S)={p(x)in P^2;colon; p(1)=0}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root.
            $endgroup$
            – zipirovich
            Dec 18 '16 at 22:19












          • $begingroup$
            @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root.
            $endgroup$
            – StackTD
            Dec 18 '16 at 22:23
















          • 3




            $begingroup$
            The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $ker(S)={p(x)in P^2;colon; p(1)=0}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root.
            $endgroup$
            – zipirovich
            Dec 18 '16 at 22:19












          • $begingroup$
            @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root.
            $endgroup$
            – StackTD
            Dec 18 '16 at 22:23










          3




          3




          $begingroup$
          The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $ker(S)={p(x)in P^2;colon; p(1)=0}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root.
          $endgroup$
          – zipirovich
          Dec 18 '16 at 22:19






          $begingroup$
          The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $ker(S)={p(x)in P^2;colon; p(1)=0}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root.
          $endgroup$
          – zipirovich
          Dec 18 '16 at 22:19














          $begingroup$
          @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root.
          $endgroup$
          – StackTD
          Dec 18 '16 at 22:23






          $begingroup$
          @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root.
          $endgroup$
          – StackTD
          Dec 18 '16 at 22:23




















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