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Linear transformation and change of basis

  1. May 23, 2014 #1
    1. The problem statement, all variables and given/known data

    Let B = {(1, -2),(2, -3)} and S be the standard basis of R2
    and [-8,-4;9,4]
    be a linear transformation expressed in terms of the standard basis?




    3. The attempt at a solution

    1) What is the change of basis matrix PSB ?
    1,2
    -2,-3

    2)What is the change of basis matrix PBS ?
    -3,-2
    2,1

    What is the linear transform expressed in terms of the basis B?
    i spent 4 hours going through the proof for this part but unable to bridge a connection. Something is lacking in my understanding.
    Give me a step by step guidance on this one.
     
  2. jcsd
  3. May 23, 2014 #2
  4. May 23, 2014 #3
    Khan is good but his proof and subscript can at times be sloppy. I enjoy Mathispower4u (discovered it before Physicforum did) but it seems there are no videos on change on basis on Mathispower4u.

    But anyway, I have a much rigorous proof on hand that I wish to understand fully.

    The transformation machine that maps the vector v with respect to the standard basis, s, is given as [f(v)]s.
    [f(v)]s = Ass [v]s
    Ass = [ [f(e1)]s [f(e2)]s...[f(en)]s ] (I don't get this part)

    There's more to come but I hope to understand this part first before proceeding...
     
    Last edited: May 23, 2014
  5. May 23, 2014 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The basic point is this: if we have basis {u, v} then every vector can be written in the form au+ bv and represented as "[itex]\begin{bmatrix}a \\ b\end{bmatrix}[/itex]". In particular, u= 1u+ 0v and is represented as [itex]\begin{bmatrix}1 \\ 0\end{bmatrix}[/itex] while v= 0u+ 1v and is represented as [itex]\begin{bmatrix}0 \\ 1 \end{bmatrix}[/itex].


    Now suppose linear transformation, A, is represented by the matrix
    [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}[/tex]
    Then Au is
    [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}a \\ c \end{bmatrix}[/tex]
    and Av is
    [tex]\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}b \\ d\end{bmatrix}[/tex]

    That is, applying the linear transformation to the basis vectors, and then writing the result as a linear combination of the basis vectors, the coefficents give the columns.

    In this particular case,
    [tex]\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}1 \\ -2\end{bmatrix}= \begin{bmatrix}0 \\ 1\end{bmatrix}[/tex]
    and we can write [itex]\begin{bmatrix}0 \\ 1\end{bmatrix}= -2\begin{bmatrix}1 \\ -2\end{bmatrix}+ 1\begin{bmatrix}2 \\ -3\end{bmatrix}[/itex]

    so the first column of the matrix representation is [itex]\begin{bmatrix}-2 \\ 1\end{bmatrix}[/itex]

    Similarly,
    [tex]\begin{bmatrix}-8 & -4\\ 9 & 4\end{bmatrix}\begin{bmatrix}2 \\ -3\end{bmatrix}= \begin{bmatrix}-4 \\ 6\end{bmatrix}[/tex]
    and [itex]\begin{bmatrix}-4 \\ 6\end{bmatrix}= 0\begin{bmatrix}1 \\-2\end{bmatrix}- 2\begin{bmatrix}2 \\ -3\end{bmatrix}[/itex]
    so the second column is [itex]\begin{bmatrix}0 \\ -2\end{bmatrix}[/itex]

    The matrix representation in this (ordered) basis is
    [tex]\begin{bmatrix}-2 & 0 \\1 & -2\end{bmatrix}[/tex]
     
  6. May 23, 2014 #5

    Would the outcome be any different if I first went down the route of

    [-8;-9] = α1[1;0] + α2[0;1]
    [-4;4] = β1[1;0] + β1[0;1]

    that is, I express each column of the matrix A as a linear combination of the vectors in S.
     
  7. May 23, 2014 #6
    Ok I got this. Basically, all I had to find was [T(x)s]B.
     
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