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Matrix of linear transformation

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Let [itex]A:\mathbb R_2[x]\rightarrow \mathbb R_2[x][/itex] is a linear transformation defined as [itex](A(p))(x)=p'(x+1)[/itex] where [itex]\mathbb R_2[x][/itex] is the space of polynomials of the second order. Find all [itex]a,b,c\in\mathbb R[/itex] such that the matrix [itex]\begin{bmatrix}
    a & 1 & 0 \\
    b & 0 & 1 \\
    c & 0 & 0 \\
    \end{bmatrix}[/itex] is the matrix of linear transformation [itex]A[/itex] with respect to some arbitrary basis of [itex]\mathbb R_2[x][/itex].

    2. Relevant equations
    -Polynomial vector space
    -Basis

    3. The attempt at a solution
    If we choose the standard basis, [tex]\mathcal B=\{1,x,x^2\}\Rightarrow p(x)=\alpha+\beta x+\gamma x^2,\alpha,\beta,\gamma\in\mathbb R\Rightarrow (A(p))(x)=\beta+(\beta+2\gamma)x+2\gamma x^2\Rightarrow[/tex]

    [itex]A(1)=0x^2+0x+1,A(x)=0x^2+1x+1,A(x^2)=2x^2+0x+0[/itex]

    Setting [itex]A(1),A(x),A(x^2)[/itex] as column vectors gives the matrix [itex]\begin{bmatrix}
    0 & 0 & 2 \\
    0 & 1 & 0 \\
    1 & 1 & 0 \\
    \end{bmatrix}[/itex] that is not of the form of given matrix [itex]\begin{bmatrix}
    a & 1 & 0 \\
    b & 0 & 1 \\
    c & 0 & 0 \\
    \end{bmatrix}[/itex].

    This means that we can't choose the standard basis to get matrix of [itex]A[/itex] that will be of the form [itex]\begin{bmatrix}
    a & 1 & 0 \\
    b & 0 & 1 \\
    c & 0 & 0 \\
    \end{bmatrix}[/itex].

    Question: Do we have to guess a proper basis? If not, then how to find one?
     
  2. jcsd
  3. Mar 27, 2016 #2

    micromass

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    What is the relation between the matrix of a linear map in one basis and the matrix of the same linear map in another basis?
     
  4. Mar 27, 2016 #3

    micromass

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    Alternatively, do you know a property of linear maps which is independent of the basis?
     
  5. Mar 27, 2016 #4
    Let [itex]B=\{b_1,b_2,b_3\}[/itex] is a standard basis, and [itex]B'=\{{b'}_1,{b'}_2,{b'}_3\}[/itex].
    Matrix of changing basis from [itex]B[/itex] to [itex]B'[/itex] is defined as [itex]S=[[b_1]_{B'},...,[{b'}_n]_{B'}][/itex].

    Finding [itex][b_1]_{B'},...,[{b'}_n]_{B'}[/itex] gives matrix [itex]S[/itex].

    This should be the reversed process, right? We know vectors [itex][b_1]_{B'},...,[{b'}_n]_{B'}[/itex] in some basis,
    and we need to find [itex]b_1,...,b_n[/itex] from basis [itex]B[/itex].
     
  6. Mar 27, 2016 #5

    micromass

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    Right, but that is not my point. Let's say I give you two matrices ##A## and ##B## which are matrices of the same linear map but with different bases. Do you know anything about how ##A## and ##B## are related? Does similarity tell you something?
     
  7. Mar 27, 2016 #6
    If [itex]B[/itex] is the matrix after transition to new basis, then [itex]B=A^{-1}[/itex]?
     
  8. Mar 27, 2016 #7
    @micromass The only condition is [itex]c\neq 0[/itex] since from [itex]b_1=1,b_2=x,b_3=x^2[/itex] follows [itex]{b'}_1=x,{b'}_2=x^2,{b'}_3=\frac{1-ax-bx^2}{c}[/itex].
    Is this correct?
     
  9. Mar 28, 2016 #8
    ATTEMPT EDITED:

    Let [itex]p(x)=a+bx+cx^2[/itex] be a polynomial in standard basis [itex]\mathcal B=\{1,x,x^2\}[/itex] of [itex]\mathbb R^2[x][/itex].
    Then, linear transformation [itex]A[/itex] is defined as

    [itex](A(p))(x)=p'(x+1)=(2+b)+2cx+0x^2[/itex]

    From the given matrix, [itex]\begin{bmatrix}
    a & 1 & 0 \\
    b & 0 & 1 \\
    c & 0 & 0 \\
    \end{bmatrix}[/itex]

    we have

    [tex][b_1]_{B'}= \begin{bmatrix}
    a \\
    b \\
    c \\
    \end{bmatrix},[b_2]_{B'}= \begin{bmatrix}
    1 \\
    0 \\
    0 \\
    \end{bmatrix},[b_3]_{B'}= \begin{bmatrix}
    0 \\
    1 \\
    0 \\
    \end{bmatrix}[/tex]

    where [itex]\mathcal {B'}[/itex] is some basis different from standard basis [itex]\mathcal B[/itex].

    Now, we have

    [tex]b_1=a\cdot {b'}_1+b\cdot {b'}_2+c\cdot {b'}_3[/tex]

    [tex]b_2=1\cdot {b'}_1+0\cdot {b'}_2+0\cdot {b'}_3[/tex]

    [tex]b_3=0\cdot {b'}_1+1\cdot {b'}_2+0\cdot {b'}_3[/tex]

    where [itex]b_1,b_2,b_3[/itex] form standard basis [itex]\mathcal B[/itex], and
    [itex]{b'}_1,{b'}_2,{b'}_3[/itex] form new basis [itex]\mathcal{B'}[/itex].

    From above equations,

    [tex]{b'}_1=x[/tex]

    [tex]{b'}_2=x^2[/tex]

    [tex]{b'}_3=\frac{1-ax-bx^2}{c},c\neq 0[/tex]

    and we have [itex]\mathcal{B'}=\left\{x,x^2,\frac{1-ax-bx^2}{c}\right\}[/itex].

    Now we need to find the matrix of linear transformation [itex]A[/itex] with respect to basis [itex]\mathcal{B'}[/itex].

    [tex]\begin{bmatrix}
    a & 1 & 0\\
    b & 0 & 1\\
    c & 0 & 0\\
    \end{bmatrix} \begin{bmatrix}
    2+b \\
    2c \\
    0 \\
    \end{bmatrix} =\begin{bmatrix}
    2a+ab+2c \\
    2b+b^2 \\
    2c+bc \\
    \end{bmatrix}[/tex]

    Linear transformation [itex]A[/itex] in basis [itex]\mathcal B'[/itex] is given by

    [tex]((A(p))(x))'=(2a+ab+2c)x+b(2+b)x^2+c(2+b)\frac{1-ax-bx^2}{c}[/tex]

    [tex]=(2a+ab+2c)x+b(2+b)x^2+(2+b)(1-ax-bx^2)[/tex]

    Conclusion: For all [itex]a,b[/itex] and for all [itex]c\neq 0[/itex], the matrix given by [itex]\begin{bmatrix}
    a & 1 & 0 \\
    b & 0 & 1 \\
    c & 0 & 0 \\
    \end{bmatrix}[/itex] is the matrix of linear transformation [itex]A[/itex] in some basis [itex]\mathcal B'[/itex].

    Is this correct?
     
  10. Mar 28, 2016 #9

    Mark44

    Staff: Mentor

    @gruba, this is a pretty strong hint:
     
  11. Mar 28, 2016 #10
    So, is it correct that one is the inverse of the other?
     
  12. Mar 28, 2016 #11

    Mark44

    Staff: Mentor

    No.
     
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