How do I find $[T]_E$ when given this linear transformation $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$?












1














In this question I have been asked to find $[T]_E$.



I was given that $T: R_3[x] rightarrow R_3[x]$ is a linear transformation



defined as: $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$,



$p(x),p(x)',p(x)'' in R_3[x]$ and $E = { 1,x,x^2 }$ is the basis of the transformation.



For solving this question I went to the definition of matrix that represents a transformation that says:



$[T]_E = [[Te_1],[Te_2],[Te_3]]$



Now, my problem is how do I map each vector in E with this transformation (first time seeing transformation with derivatives)



Thank you in advance.






linear-algebra







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  • 1




    The derivative map $D : p(x) mapsto p'(x)$ is a perfectly good linear transformation of $R_3[x]$. Can you compute the matrix representation $[D]_E$ of $D$?
    – Travis
    Apr 21 '16 at 16:58






  • 1




    You're welcome, I hope the suggestion was useful.
    – Travis
    Apr 21 '16 at 17:28
















1














In this question I have been asked to find $[T]_E$.



I was given that $T: R_3[x] rightarrow R_3[x]$ is a linear transformation



defined as: $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$,



$p(x),p(x)',p(x)'' in R_3[x]$ and $E = { 1,x,x^2 }$ is the basis of the transformation.



For solving this question I went to the definition of matrix that represents a transformation that says:



$[T]_E = [[Te_1],[Te_2],[Te_3]]$



Now, my problem is how do I map each vector in E with this transformation (first time seeing transformation with derivatives)



Thank you in advance.






linear-algebra







share|cite|improve this question


















  • 1




    The derivative map $D : p(x) mapsto p'(x)$ is a perfectly good linear transformation of $R_3[x]$. Can you compute the matrix representation $[D]_E$ of $D$?
    – Travis
    Apr 21 '16 at 16:58






  • 1




    You're welcome, I hope the suggestion was useful.
    – Travis
    Apr 21 '16 at 17:28














1












1








1







In this question I have been asked to find $[T]_E$.



I was given that $T: R_3[x] rightarrow R_3[x]$ is a linear transformation



defined as: $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$,



$p(x),p(x)',p(x)'' in R_3[x]$ and $E = { 1,x,x^2 }$ is the basis of the transformation.



For solving this question I went to the definition of matrix that represents a transformation that says:



$[T]_E = [[Te_1],[Te_2],[Te_3]]$



Now, my problem is how do I map each vector in E with this transformation (first time seeing transformation with derivatives)



Thank you in advance.






linear-algebra







share|cite|improve this question













In this question I have been asked to find $[T]_E$.



I was given that $T: R_3[x] rightarrow R_3[x]$ is a linear transformation



defined as: $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$,



$p(x),p(x)',p(x)'' in R_3[x]$ and $E = { 1,x,x^2 }$ is the basis of the transformation.



For solving this question I went to the definition of matrix that represents a transformation that says:



$[T]_E = [[Te_1],[Te_2],[Te_3]]$



Now, my problem is how do I map each vector in E with this transformation (first time seeing transformation with derivatives)



Thank you in advance.






linear-algebra




linear-algebra






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asked Apr 21 '16 at 16:55









LiziPizi

1,3682824




1,3682824








  • 1




    The derivative map $D : p(x) mapsto p'(x)$ is a perfectly good linear transformation of $R_3[x]$. Can you compute the matrix representation $[D]_E$ of $D$?
    – Travis
    Apr 21 '16 at 16:58






  • 1




    You're welcome, I hope the suggestion was useful.
    – Travis
    Apr 21 '16 at 17:28














  • 1




    The derivative map $D : p(x) mapsto p'(x)$ is a perfectly good linear transformation of $R_3[x]$. Can you compute the matrix representation $[D]_E$ of $D$?
    – Travis
    Apr 21 '16 at 16:58






  • 1




    You're welcome, I hope the suggestion was useful.
    – Travis
    Apr 21 '16 at 17:28








1




1




The derivative map $D : p(x) mapsto p'(x)$ is a perfectly good linear transformation of $R_3[x]$. Can you compute the matrix representation $[D]_E$ of $D$?
– Travis
Apr 21 '16 at 16:58




The derivative map $D : p(x) mapsto p'(x)$ is a perfectly good linear transformation of $R_3[x]$. Can you compute the matrix representation $[D]_E$ of $D$?
– Travis
Apr 21 '16 at 16:58




1




1




You're welcome, I hope the suggestion was useful.
– Travis
Apr 21 '16 at 17:28




You're welcome, I hope the suggestion was useful.
– Travis
Apr 21 '16 at 17:28










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Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$ Compute
$$begin{align}
Te_1 & = T(1)\
& = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\
& = 2\
& = color{#08F}2 cdot 1 + color{#08F}0 cdot x + color{#08F}0 cdot x^2\
& = color{#08F}2 e_1 + color{#08F}0e_2 + color{#08F}0e_3\
[Te_1]_E & = begin{bmatrix}color{#08F}{2\ 0\ 0}end{bmatrix}\\
Te_2 & = T(x)\
& = (1 + 2x^2)x'' + (1 - 2x)x' + x + 0\
& = 1 - x\
& = color{#08F}1 cdot 1 color{#08F}{- 1} cdot x + color{#08F}0 cdot x^2\
& = color{#08F}1 e_1 color{#08F}{- 1}e_2 + color{#08F}0e_3\
[Te_2]_E & = begin{bmatrix}color{#08F}{1\ -1\ 0}end{bmatrix}\\
Te_3 & = T(x^2)\
& = (1 + 2x^2)(x^2)'' + (1 - 2x)(x^2)' + x^2 + 0^2\
& = 2 - 2x + x^2\
& = color{#08F}2 cdot 1 color{#08F}{- 2} cdot x + color{#08F}1 cdot x^2\
& = color{#08F}2 e_1 color{#08F}{- 2}e_2 + color{#08F}1e_3\
[Te_3]_E & = begin{bmatrix}color{#08F}{2\ -2\ 1}end{bmatrix}.end{align}$$
Thus,
$$[T]_E = begin{bmatrix}
2 & 1 & 2\
0 & -1 & -2\
0 & 0 & 1end{bmatrix}.$$






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    Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$ Compute
    $$begin{align}
    Te_1 & = T(1)\
    & = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\
    & = 2\
    & = color{#08F}2 cdot 1 + color{#08F}0 cdot x + color{#08F}0 cdot x^2\
    & = color{#08F}2 e_1 + color{#08F}0e_2 + color{#08F}0e_3\
    [Te_1]_E & = begin{bmatrix}color{#08F}{2\ 0\ 0}end{bmatrix}\\
    Te_2 & = T(x)\
    & = (1 + 2x^2)x'' + (1 - 2x)x' + x + 0\
    & = 1 - x\
    & = color{#08F}1 cdot 1 color{#08F}{- 1} cdot x + color{#08F}0 cdot x^2\
    & = color{#08F}1 e_1 color{#08F}{- 1}e_2 + color{#08F}0e_3\
    [Te_2]_E & = begin{bmatrix}color{#08F}{1\ -1\ 0}end{bmatrix}\\
    Te_3 & = T(x^2)\
    & = (1 + 2x^2)(x^2)'' + (1 - 2x)(x^2)' + x^2 + 0^2\
    & = 2 - 2x + x^2\
    & = color{#08F}2 cdot 1 color{#08F}{- 2} cdot x + color{#08F}1 cdot x^2\
    & = color{#08F}2 e_1 color{#08F}{- 2}e_2 + color{#08F}1e_3\
    [Te_3]_E & = begin{bmatrix}color{#08F}{2\ -2\ 1}end{bmatrix}.end{align}$$
    Thus,
    $$[T]_E = begin{bmatrix}
    2 & 1 & 2\
    0 & -1 & -2\
    0 & 0 & 1end{bmatrix}.$$






    share|cite|improve this answer




























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      Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$ Compute
      $$begin{align}
      Te_1 & = T(1)\
      & = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\
      & = 2\
      & = color{#08F}2 cdot 1 + color{#08F}0 cdot x + color{#08F}0 cdot x^2\
      & = color{#08F}2 e_1 + color{#08F}0e_2 + color{#08F}0e_3\
      [Te_1]_E & = begin{bmatrix}color{#08F}{2\ 0\ 0}end{bmatrix}\\
      Te_2 & = T(x)\
      & = (1 + 2x^2)x'' + (1 - 2x)x' + x + 0\
      & = 1 - x\
      & = color{#08F}1 cdot 1 color{#08F}{- 1} cdot x + color{#08F}0 cdot x^2\
      & = color{#08F}1 e_1 color{#08F}{- 1}e_2 + color{#08F}0e_3\
      [Te_2]_E & = begin{bmatrix}color{#08F}{1\ -1\ 0}end{bmatrix}\\
      Te_3 & = T(x^2)\
      & = (1 + 2x^2)(x^2)'' + (1 - 2x)(x^2)' + x^2 + 0^2\
      & = 2 - 2x + x^2\
      & = color{#08F}2 cdot 1 color{#08F}{- 2} cdot x + color{#08F}1 cdot x^2\
      & = color{#08F}2 e_1 color{#08F}{- 2}e_2 + color{#08F}1e_3\
      [Te_3]_E & = begin{bmatrix}color{#08F}{2\ -2\ 1}end{bmatrix}.end{align}$$
      Thus,
      $$[T]_E = begin{bmatrix}
      2 & 1 & 2\
      0 & -1 & -2\
      0 & 0 & 1end{bmatrix}.$$






      share|cite|improve this answer


























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        0








        0






        Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$ Compute
        $$begin{align}
        Te_1 & = T(1)\
        & = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\
        & = 2\
        & = color{#08F}2 cdot 1 + color{#08F}0 cdot x + color{#08F}0 cdot x^2\
        & = color{#08F}2 e_1 + color{#08F}0e_2 + color{#08F}0e_3\
        [Te_1]_E & = begin{bmatrix}color{#08F}{2\ 0\ 0}end{bmatrix}\\
        Te_2 & = T(x)\
        & = (1 + 2x^2)x'' + (1 - 2x)x' + x + 0\
        & = 1 - x\
        & = color{#08F}1 cdot 1 color{#08F}{- 1} cdot x + color{#08F}0 cdot x^2\
        & = color{#08F}1 e_1 color{#08F}{- 1}e_2 + color{#08F}0e_3\
        [Te_2]_E & = begin{bmatrix}color{#08F}{1\ -1\ 0}end{bmatrix}\\
        Te_3 & = T(x^2)\
        & = (1 + 2x^2)(x^2)'' + (1 - 2x)(x^2)' + x^2 + 0^2\
        & = 2 - 2x + x^2\
        & = color{#08F}2 cdot 1 color{#08F}{- 2} cdot x + color{#08F}1 cdot x^2\
        & = color{#08F}2 e_1 color{#08F}{- 2}e_2 + color{#08F}1e_3\
        [Te_3]_E & = begin{bmatrix}color{#08F}{2\ -2\ 1}end{bmatrix}.end{align}$$
        Thus,
        $$[T]_E = begin{bmatrix}
        2 & 1 & 2\
        0 & -1 & -2\
        0 & 0 & 1end{bmatrix}.$$






        share|cite|improve this answer














        Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$ Compute
        $$begin{align}
        Te_1 & = T(1)\
        & = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\
        & = 2\
        & = color{#08F}2 cdot 1 + color{#08F}0 cdot x + color{#08F}0 cdot x^2\
        & = color{#08F}2 e_1 + color{#08F}0e_2 + color{#08F}0e_3\
        [Te_1]_E & = begin{bmatrix}color{#08F}{2\ 0\ 0}end{bmatrix}\\
        Te_2 & = T(x)\
        & = (1 + 2x^2)x'' + (1 - 2x)x' + x + 0\
        & = 1 - x\
        & = color{#08F}1 cdot 1 color{#08F}{- 1} cdot x + color{#08F}0 cdot x^2\
        & = color{#08F}1 e_1 color{#08F}{- 1}e_2 + color{#08F}0e_3\
        [Te_2]_E & = begin{bmatrix}color{#08F}{1\ -1\ 0}end{bmatrix}\\
        Te_3 & = T(x^2)\
        & = (1 + 2x^2)(x^2)'' + (1 - 2x)(x^2)' + x^2 + 0^2\
        & = 2 - 2x + x^2\
        & = color{#08F}2 cdot 1 color{#08F}{- 2} cdot x + color{#08F}1 cdot x^2\
        & = color{#08F}2 e_1 color{#08F}{- 2}e_2 + color{#08F}1e_3\
        [Te_3]_E & = begin{bmatrix}color{#08F}{2\ -2\ 1}end{bmatrix}.end{align}$$
        Thus,
        $$[T]_E = begin{bmatrix}
        2 & 1 & 2\
        0 & -1 & -2\
        0 & 0 & 1end{bmatrix}.$$







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        edited Nov 26 at 14:44

























        answered Nov 25 at 18:50









        Maurice P

        1,3651732




        1,3651732






























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