Finding the Matrix of a Linear Transformation With Polynomial Basis












1












$begingroup$


If I have some given linear transformation $T:Vrightarrow V$ such that:



$T(p(x)) = p'(x) + x,p(0) + (x+1),p(1)$,



where $V$ is the space of rational polynomials of degree at most 2 with basis ${1 ,x,x^2}$.



Now I'm trying to figure out what $T$ actually does. $p'(x)$ means derivative of $p(x)$ and $p(0)$ is some rational number when $p(x)$ is evaluated in zero,which is just the constant term of $p(x)$.



What I'm confused by is this $p(1)$. Do I need an extra parameter when applying $T$ to a basis vector ?



For example:
$T(1)=0+x+(x+1),a$



where $a in mathbb{Q}$.






linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $V$ is the space of rational polynomials of degree at most $2$ with basis ${1,x,x^2}$?
    $endgroup$
    – Theo C.
    Nov 29 '18 at 21:56












  • $begingroup$
    @TheoC.Yes I do ,sorry.
    $endgroup$
    – Bustin Jieber
    Nov 29 '18 at 21:58
















1












$begingroup$


If I have some given linear transformation $T:Vrightarrow V$ such that:



$T(p(x)) = p'(x) + x,p(0) + (x+1),p(1)$,



where $V$ is the space of rational polynomials of degree at most 2 with basis ${1 ,x,x^2}$.



Now I'm trying to figure out what $T$ actually does. $p'(x)$ means derivative of $p(x)$ and $p(0)$ is some rational number when $p(x)$ is evaluated in zero,which is just the constant term of $p(x)$.



What I'm confused by is this $p(1)$. Do I need an extra parameter when applying $T$ to a basis vector ?



For example:
$T(1)=0+x+(x+1),a$



where $a in mathbb{Q}$.






linear-transformations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean $V$ is the space of rational polynomials of degree at most $2$ with basis ${1,x,x^2}$?
    $endgroup$
    – Theo C.
    Nov 29 '18 at 21:56












  • $begingroup$
    @TheoC.Yes I do ,sorry.
    $endgroup$
    – Bustin Jieber
    Nov 29 '18 at 21:58














1












1








1





$begingroup$


If I have some given linear transformation $T:Vrightarrow V$ such that:



$T(p(x)) = p'(x) + x,p(0) + (x+1),p(1)$,



where $V$ is the space of rational polynomials of degree at most 2 with basis ${1 ,x,x^2}$.



Now I'm trying to figure out what $T$ actually does. $p'(x)$ means derivative of $p(x)$ and $p(0)$ is some rational number when $p(x)$ is evaluated in zero,which is just the constant term of $p(x)$.



What I'm confused by is this $p(1)$. Do I need an extra parameter when applying $T$ to a basis vector ?



For example:
$T(1)=0+x+(x+1),a$



where $a in mathbb{Q}$.






linear-transformations







share|cite|improve this question











$endgroup$




If I have some given linear transformation $T:Vrightarrow V$ such that:



$T(p(x)) = p'(x) + x,p(0) + (x+1),p(1)$,



where $V$ is the space of rational polynomials of degree at most 2 with basis ${1 ,x,x^2}$.



Now I'm trying to figure out what $T$ actually does. $p'(x)$ means derivative of $p(x)$ and $p(0)$ is some rational number when $p(x)$ is evaluated in zero,which is just the constant term of $p(x)$.



What I'm confused by is this $p(1)$. Do I need an extra parameter when applying $T$ to a basis vector ?



For example:
$T(1)=0+x+(x+1),a$



where $a in mathbb{Q}$.






linear-transformations




linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 21:58







Bustin Jieber

















asked Nov 29 '18 at 21:52









Bustin JieberBustin Jieber

83




83












  • $begingroup$
    Do you mean $V$ is the space of rational polynomials of degree at most $2$ with basis ${1,x,x^2}$?
    $endgroup$
    – Theo C.
    Nov 29 '18 at 21:56












  • $begingroup$
    @TheoC.Yes I do ,sorry.
    $endgroup$
    – Bustin Jieber
    Nov 29 '18 at 21:58


















  • $begingroup$
    Do you mean $V$ is the space of rational polynomials of degree at most $2$ with basis ${1,x,x^2}$?
    $endgroup$
    – Theo C.
    Nov 29 '18 at 21:56












  • $begingroup$
    @TheoC.Yes I do ,sorry.
    $endgroup$
    – Bustin Jieber
    Nov 29 '18 at 21:58
















$begingroup$
Do you mean $V$ is the space of rational polynomials of degree at most $2$ with basis ${1,x,x^2}$?
$endgroup$
– Theo C.
Nov 29 '18 at 21:56






$begingroup$
Do you mean $V$ is the space of rational polynomials of degree at most $2$ with basis ${1,x,x^2}$?
$endgroup$
– Theo C.
Nov 29 '18 at 21:56














$begingroup$
@TheoC.Yes I do ,sorry.
$endgroup$
– Bustin Jieber
Nov 29 '18 at 21:58




$begingroup$
@TheoC.Yes I do ,sorry.
$endgroup$
– Bustin Jieber
Nov 29 '18 at 21:58










2 Answers
2






active

oldest

votes


















1












$begingroup$

We'll consider how $T$ maps the given basis ${1,x,x^2}$ in order to find the matrix of this transformation in said basis:



$$T(1) = 0+1*x+(x+1)*1 = 2x+1$$
$$T(x) = 1 + 0*x + (x+1)*1 = x+2$$
$$T(x^2) = 2x + 0*x + (x+1)*1 =3x+1 $$



So, we have
$$T = begin{bmatrix} 1&2&1\2&1&3\0&0&0end{bmatrix}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ohh I see, thank you. Just to be sure ,If it was $p'(0)$ instead of $p'(x)$ for, let's say, $T(x^2)$ ,transformation would look like $T(x^2) = 2*0 + 0*x+(x+1)*1$ ?
    $endgroup$
    – Bustin Jieber
    Nov 29 '18 at 22:11










  • $begingroup$
    That looks right to me!
    $endgroup$
    – Theo C.
    Nov 29 '18 at 23:03



















0












$begingroup$

You have $p (x):=1$, the constant polynomial. Then $T (1)=0+x1+(x+1)1=2x+1$.






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









    1












    $begingroup$

    We'll consider how $T$ maps the given basis ${1,x,x^2}$ in order to find the matrix of this transformation in said basis:



    $$T(1) = 0+1*x+(x+1)*1 = 2x+1$$
    $$T(x) = 1 + 0*x + (x+1)*1 = x+2$$
    $$T(x^2) = 2x + 0*x + (x+1)*1 =3x+1 $$



    So, we have
    $$T = begin{bmatrix} 1&2&1\2&1&3\0&0&0end{bmatrix}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ohh I see, thank you. Just to be sure ,If it was $p'(0)$ instead of $p'(x)$ for, let's say, $T(x^2)$ ,transformation would look like $T(x^2) = 2*0 + 0*x+(x+1)*1$ ?
      $endgroup$
      – Bustin Jieber
      Nov 29 '18 at 22:11










    • $begingroup$
      That looks right to me!
      $endgroup$
      – Theo C.
      Nov 29 '18 at 23:03
















    1












    $begingroup$

    We'll consider how $T$ maps the given basis ${1,x,x^2}$ in order to find the matrix of this transformation in said basis:



    $$T(1) = 0+1*x+(x+1)*1 = 2x+1$$
    $$T(x) = 1 + 0*x + (x+1)*1 = x+2$$
    $$T(x^2) = 2x + 0*x + (x+1)*1 =3x+1 $$



    So, we have
    $$T = begin{bmatrix} 1&2&1\2&1&3\0&0&0end{bmatrix}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ohh I see, thank you. Just to be sure ,If it was $p'(0)$ instead of $p'(x)$ for, let's say, $T(x^2)$ ,transformation would look like $T(x^2) = 2*0 + 0*x+(x+1)*1$ ?
      $endgroup$
      – Bustin Jieber
      Nov 29 '18 at 22:11










    • $begingroup$
      That looks right to me!
      $endgroup$
      – Theo C.
      Nov 29 '18 at 23:03














    1












    1








    1





    $begingroup$

    We'll consider how $T$ maps the given basis ${1,x,x^2}$ in order to find the matrix of this transformation in said basis:



    $$T(1) = 0+1*x+(x+1)*1 = 2x+1$$
    $$T(x) = 1 + 0*x + (x+1)*1 = x+2$$
    $$T(x^2) = 2x + 0*x + (x+1)*1 =3x+1 $$



    So, we have
    $$T = begin{bmatrix} 1&2&1\2&1&3\0&0&0end{bmatrix}$$






    share|cite|improve this answer









    $endgroup$



    We'll consider how $T$ maps the given basis ${1,x,x^2}$ in order to find the matrix of this transformation in said basis:



    $$T(1) = 0+1*x+(x+1)*1 = 2x+1$$
    $$T(x) = 1 + 0*x + (x+1)*1 = x+2$$
    $$T(x^2) = 2x + 0*x + (x+1)*1 =3x+1 $$



    So, we have
    $$T = begin{bmatrix} 1&2&1\2&1&3\0&0&0end{bmatrix}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 '18 at 22:03









    Theo C.Theo C.

    25928




    25928












    • $begingroup$
      Ohh I see, thank you. Just to be sure ,If it was $p'(0)$ instead of $p'(x)$ for, let's say, $T(x^2)$ ,transformation would look like $T(x^2) = 2*0 + 0*x+(x+1)*1$ ?
      $endgroup$
      – Bustin Jieber
      Nov 29 '18 at 22:11










    • $begingroup$
      That looks right to me!
      $endgroup$
      – Theo C.
      Nov 29 '18 at 23:03


















    • $begingroup$
      Ohh I see, thank you. Just to be sure ,If it was $p'(0)$ instead of $p'(x)$ for, let's say, $T(x^2)$ ,transformation would look like $T(x^2) = 2*0 + 0*x+(x+1)*1$ ?
      $endgroup$
      – Bustin Jieber
      Nov 29 '18 at 22:11










    • $begingroup$
      That looks right to me!
      $endgroup$
      – Theo C.
      Nov 29 '18 at 23:03
















    $begingroup$
    Ohh I see, thank you. Just to be sure ,If it was $p'(0)$ instead of $p'(x)$ for, let's say, $T(x^2)$ ,transformation would look like $T(x^2) = 2*0 + 0*x+(x+1)*1$ ?
    $endgroup$
    – Bustin Jieber
    Nov 29 '18 at 22:11




    $begingroup$
    Ohh I see, thank you. Just to be sure ,If it was $p'(0)$ instead of $p'(x)$ for, let's say, $T(x^2)$ ,transformation would look like $T(x^2) = 2*0 + 0*x+(x+1)*1$ ?
    $endgroup$
    – Bustin Jieber
    Nov 29 '18 at 22:11












    $begingroup$
    That looks right to me!
    $endgroup$
    – Theo C.
    Nov 29 '18 at 23:03




    $begingroup$
    That looks right to me!
    $endgroup$
    – Theo C.
    Nov 29 '18 at 23:03











    0












    $begingroup$

    You have $p (x):=1$, the constant polynomial. Then $T (1)=0+x1+(x+1)1=2x+1$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      You have $p (x):=1$, the constant polynomial. Then $T (1)=0+x1+(x+1)1=2x+1$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        You have $p (x):=1$, the constant polynomial. Then $T (1)=0+x1+(x+1)1=2x+1$.






        share|cite|improve this answer









        $endgroup$



        You have $p (x):=1$, the constant polynomial. Then $T (1)=0+x1+(x+1)1=2x+1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 21:57









        AnyADAnyAD

        2,088812




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