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Linear algebra: Find the matrix of linear transformation

  1. Feb 13, 2016 #1
    1. The problem statement, all variables and given/known data
    Check if [itex]L(p)(x)=(1+4x)p(x)+(x-x^2)p'(x)-(x^2+x^3)p''(x)[/itex] is a linear transformation on [itex]\mathbb{R_2}[x][/itex]. If [itex]L(p)(x)[/itex] is a linear transformation, find it's matrix in standard basis and check if [itex]L(p)(x)[/itex] is invertible. If [itex]L(p)(x)[/itex] is invertible, find the function rule of it's inverse.

    2. Relevant equations
    -Linear transformations

    3. The attempt at a solution

    [itex]L(p)(x)[/itex] is a linear transformation if
    [itex]L(z(x)+w(x))=L(z)(x)+L(w)(x) \forall z,w\in\mathbb{R_2}[x][/itex] and
    [itex]L(\alpha z(x))=\alpha L(z)(x)\forall z\in \mathbb{R_2}[x],\forall\alpha\in\mathbb{R}[/itex]

    Let [itex]p(x)=a+bx+cx^2\Rightarrow L(p)(x)=a+(4a+2b)x+(3b+c)x^2[/itex].
    Let [itex]z(x)=p+qx+rx^2[/itex] and [itex]w(x)=\alpha+\beta x+\gamma x^2[/itex].

    [itex]L(z(x))=p+(4p+2q)+(3q+r)x^2[/itex] and [itex]L(w(x))=\alpha+(4\alpha+2\beta)+(3\beta+\gamma)x^2[/itex] then [itex]L(z(x))+L(w(x))=(p+\alpha)+(4p+2q+4\alpha+2\beta)x+(3q+r+3\beta+\gamma)x^2[/itex]
    [itex]L(z(x)+w(x))=L((p+\alpha)+(q+\beta)x+(r+\gamma)x^2)[/itex].
    From this [itex]\Rightarrow L(z(x)+w(x))=L(z)(x)+L(w)(x)[/itex].

    [itex]L(\alpha z(x))=L(\alpha p+\alpha qx+\alpha rx^2)=\alpha p+(4\alpha p+2\alpha q)x+(3\alpha q+\alpha r)x^2[/itex].
    [itex]\alpha L(z)(x)=\alpha(p+(4p+2q)x+(3q+r)x^2)[/itex].
    From this [itex]\Rightarrow L(\alpha z(x))=\alpha L(z)(x)[/itex] and [itex]L(p)(x)[/itex] is a linear transformation.

    Standard basis for [itex]\mathbb{R_2}[x][/itex] is [itex]\mathcal{B}=\{1,x,x^2\}[/itex].
    [itex][L]_{\mathcal{B}}=
    \begin{bmatrix}
    1 & 0 & 0 \\
    4 & 2 & 0 \\
    0 & 3 & 1 \\
    \end{bmatrix}
    [/itex].

    Is the matrix [itex][L]_{\mathcal{B}}[/itex] correct?
    If not, how to construct it?

    [itex]L(p)(x)[/itex] is invertible because [itex]\det [L]_{\mathcal{B}}\neq 0[/itex].
    [itex][L]^{-1}_{\mathcal{B}}=
    \begin{bmatrix}
    1 & 0 & 0 \\
    -2 & 1/2 & 0 \\
    6 & -3/2 & 1 \\
    \end{bmatrix}
    [/itex]

    Function rule of [itex]L^{-1}(p)(x)[/itex] can be read from [itex][L]^{-1}_{\mathcal{B}}\Rightarrow L^{-1}(p)(x)=a+(-2a+\frac{1}{2}b)x+(6a-\frac{3}{2}b+c)x^2[/itex].
    Is the function rule of [itex]L^{-1}(p)(x)[/itex] correct?
    If not, how to construct it?

    Also, how to check if [itex]L(p)(x)[/itex] is bijective?
     
  2. jcsd
  3. Feb 13, 2016 #2
    Once you proved linearity of ##L##, then it is bijective iff ##\det [L]_{\cal B} \neq 0 ##
    You can check your answer by verifying that ##[L]_{\cal B}[L]_{L({\cal B})}^{-1} = I_3##.

    If a linear transformation is bijective, then its inverse is linear. So when ##[L]_{L({\cal B})}^{-1}## is known, you know the coordinates on ##L({\cal B})## of ##L^{-1}##, and you can re-assemble your function
     
    Last edited: Feb 13, 2016
  4. Feb 13, 2016 #3
    Are matrices of [itex]L[/itex] and [itex]L^{-1}[/itex] correct?
     
  5. Feb 13, 2016 #4
    ##L## is correct, and ##L^{-1}## is correct if ##LL^{-1} = I_3##
     
  6. Feb 13, 2016 #5
    Could you explain the method for finding the matrix of linear transformation that involves polynomials?
     
  7. Feb 13, 2016 #6
    First of all, your results are correct, I didn't tell you anything because you could check the answer by yourself.

    Yes, you do exactly as usual. If you take the standard basis of ## \mathbb{R}_2[X]##, which is ##{\cal B} = (1,X,X^2)##, you are required to find a decomposition on ##{\cal B}## of ##L(1), L(X), L(X^2)## and assemble their coordinates column-wise. This is the matrix of your transformation
     
  8. Feb 13, 2016 #7
    If your decomposition gives
    ## L(1) = a + b X + c X^2##, ## L(X) = d + e X + f X^2##, ## L(X^2) = g + h X + i X^2##,

    ##[L]_{\cal B} = \begin{pmatrix} a & d & g \\ b & e & h \\ c & f & i \end{pmatrix}##
     
  9. Feb 13, 2016 #8
    I don't understand what are the polynomials of [itex]L(1),L(x),L(x^2)[/itex] if [itex]L(p)(x)=a+x(4a+2b)+(3b+c)x^2[/itex].
    Could you explain how to evaluate [itex]L(1),L(x),L(x^2)[/itex]?
     
  10. Feb 13, 2016 #9
    Is your transformation linear with respect to polynomial ##p## ?
     
  11. Feb 13, 2016 #10
    Yes, and the basis is standard for [itex]\mathbb{R_2}[x][/itex].
     
  12. Feb 13, 2016 #11
    Really ? Do we have ##L(\lambda p + q) = \lambda L(p) + L(q) ## ?
     
  13. Feb 13, 2016 #12
    You are going nowhere with the last transformation because it is not even linear. It doesn't make any sense to look for its matrix representation in a given basis.
    If you want to exercise, take the original transformation in post #1, and tell us what is the matrix of ##L## in basis ##{\cal B} = ( 3, X-1, X^2 + 1) ##
     
  14. Feb 13, 2016 #13
    In basis [itex]\mathcal{B}=\{3,x-1,x^2+1\}\Rightarrow L(3)=12x+3,L(x-1)=3x^2-2x-1,L(x^2+1)=x^2+4x+1[/itex].
    [itex][L]_{\mathcal{B}}=
    \begin{bmatrix}
    0 & 3 & 1 \\
    12 & -2 & 4 \\
    3 & -1 & 1 \\
    \end{bmatrix}
    [/itex].

    But for this new basis I get that [itex]\det \neq 0[/itex] and in standard basis (correction from first post), [itex]\det = 0[/itex].
    What is wrong here?
     
  15. Feb 13, 2016 #14
    No this is wrong, you didn't decompose on ##(3,X-1,X^2 + 1) ## but on ##(1,X,X^2)##
     
  16. Feb 13, 2016 #15
    Matrix in standard basis [itex]\mathcal{B}=\{1,x,x^2\}[/itex] is [itex]
    \begin{bmatrix}
    4 & 3 & 1 \\
    1 & 2 & 0 \\
    0 & 0 & 0 \\
    \end{bmatrix}
    [/itex] and in [itex]\mathcal{B}=\{3,x-1,x^2+1\}[/itex] is
    [itex]
    \begin{bmatrix}
    0 & 3 & 1 \\
    12 & -2 & 4 \\
    3 & -1 & 1 \\
    \end{bmatrix}
    [/itex].

    What is wrong here?
     
  17. Feb 13, 2016 #16
    The matrix ##[L]_{\cal B} ## is the column wise assembly of the coordinates of ##L({\cal B}) = ( L(e_1), L(e_2), L(e_3) ) ## expressed in ##{ \cal B} ##

    You should find :

    ##L(e_1) = a e_1 + b e_2 + c e_3 ##
    ##L(e_2 ) = d e_1 + e e_2 + f e_3##
    ##L(e_3) = g e_1 + h e_2 + i e_3 ##
     
  18. Feb 13, 2016 #17
    So in standard basis the matrix is [itex]
    \begin{bmatrix}
    1 & 0 & 0 \\
    4 & 2 & 0 \\
    0 & 3 & 1 \\
    \end{bmatrix}
    [/itex] and in the new basis is [itex]
    \begin{bmatrix}
    5 & -2 & 4/3 \\
    12 & -2 & 4 \\
    0 & 3 & 1 \\
    \end{bmatrix}
    [/itex].
     
    Last edited: Feb 13, 2016
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