Results 1 to 4 of 4

Math Help - Apparently, matrices of linear transformation are my weakness

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    13

    Apparently, matrices of linear transformation are my weakness

    I have another eigenvector problem, where the transformation is given by T(f(x))= xf'(x) + f(2)x + f(3) . I haven't slightest clue how to represent this transformation with a matrix, and as such am completely stumped.
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    Quote Originally Posted by kaelbu View Post
    I have another eigenvector problem, where the transformation is given by T(f(x))= xf'(x) + f(2)x + f(3) . I haven't slightest clue how to represent this transformation with a matrix, and as such am completely stumped.
    Thanks!
    You haven't said what the vector space is- what kinds of functions f(x) can be. The most general would be "the space of all differentiable functions" but that is an infinite dimensional vector space and T cannot be represented by a matrix.

    It is impossible to represent T as a matrix until we know what the vector space is! Please state the entire problem.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2010
    Posts
    13
    Oh! Sorry! V= P2(R)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,707
    Thanks
    1470
    So V is the set of polynomials of degree 2 or less over the real numbers.

    In order to represent a linear transformation as a matrix you have to first have a basis and the standard basis for this space is \{1, x, x^2\}. Apply the linear transformation to each vector in the basis in turn and write the result in terms of the basis. The coefficients in that linear combination form one column of the basis.

    Here, T(f(x))= xf'+ f(2)x+ f(3) so T(1)= x(0)+ 1x+ 1= 1+ x. Since that is 1(1+ 1(x)+ 0(x^2), the first column of the matrix is \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}.

    T(x)= x(1)+ 2x+ 3= 3x+ 3= 3(1)+ 3(x)+ 0(x^2) and the second column of the matrix is \begin{bmatrix}3 \\ 3 \\ 0\end{bmatrix}.

    Finally, T(x^2)= x(2x)+ 4x+ 9= 9(1)+ 4(x)+ 2(x^2) so the third column of the matrix is \begin{bmatrix}9 \\ 4\\ 2\end{bmatrix}.

    The matrix representation of this linear transformation is \begin{bmatrix}1 & 3 & 9 \\ 1 & 3 & 4 \\ 0 & 0 & 2\end{bmatrix}.

    But you shouldn't have to write a linear transformation as a matrix to find eigenvalues- eigenvalue problems show up in many fields of mathematics, often in "function spaces" of infinite dimension where the transformations cannot be written as matrices.

    Here, any "vector" is of the form f(x)= ax^2+ bx+ c and so f'= 2ax+ b, f(2)= 4a+ 2b+ c, and f(3)= 9a+ 3b+ c.

    T(f)= (2ax+ b)x+ (4a+ 2b+ c)x+ (9a+ 3b+ c) = 2ax^2+ (4a+ 3b+ c)x+ (9a+ 3b+ c)= \lambda f = a\lambda x^2+ b\lambda x+ c\lambda
    so, comparing "like" coefficients, we must have a\lambda= 2a, b\lambda= 4a+ 3b+ c, and c\lambda= 9a+ 3b+ c.

    The first equation, a\lambda= 2a is the same as \lambda= 2 so that either a= 0 or the eigenvalue is 2.

    Set a= 0 to find other eigenvalues.
    Last edited by HallsofIvy; May 14th 2010 at 02:18 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 30th 2011, 03:36 PM
  2. Linear Transformation and Matrices #2.
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 16th 2011, 04:21 PM
  3. Linear Transformation and Matrices
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 16th 2011, 04:15 PM
  4. Transformation matrices
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 4th 2010, 01:37 AM
  5. Matrices for linear transformation
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 16th 2009, 06:59 PM

Search Tags


/mathhelpforum @mathhelpforum