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Transition Matrix and Ordered Bases

  1. Apr 14, 2013 #1
    Let B and C be ordered bases for ℝn. Let P be the matrix whose columns are the vectors in B and let Q be the matrix whose columns are the vectors in C. Prove that the transition matrix from B to C equals Q-1P.


    I am stuck. Here is what I have.

    I know that if B is the standard basis in ℝn, then the transition matrix from B to C is given by [1st vector in C 2nd vector in C ........... nth vector in C]-1.

    Also, if C is a standard basis in ℝn, then the transition matrix from B to C is given by [1st vector in B 2 vector in B ......... nth vector in B].

    Since I konw what the transition matrix is from B to C given different standard bases, I am having a difficult time relating this to teh columns of each.
     
  2. jcsd
  3. Apr 14, 2013 #2

    Fredrik

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    You're going to need a notation for the basis vectors. I suggest
    \begin{align} B&=\{e_1,\dots,e_n\}\\
    C &=\{f_1,\dots,f_n\}
    \end{align} How do you define the transition matrix from B to C? Is it the M defined by
    $$f_i=\sum_j M_{ij} e_j$$ or the M defined by
    $$f_i=Me_i=\sum_j (Me_i)_j e_j=\sum_j M_{ji} e_j?$$ (The latter M is the transpose of the former). You want to prove that (with one of these choices of M), we have ##M=Q^{-1}P##. This is equivalent to ##QM=P##, which is equivalent to ##P_{ij}=(QM)_{ij}=##what? Use the definition of matrix multiplication to rewrite ##(QM)_{ij}##. Then you can start thinking about rows and columns.
     
  4. Apr 15, 2013 #3
    Im confused with the notation of the matrix. How do you rewrite the rows and columns
     
  5. Apr 15, 2013 #4

    Fredrik

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    I'm not sure if you're asking about what I did or about the problem.

    One notation that can be useful is to denote the number on row i, column j of a matrix A by ##A^i_j## instead of ##A_{ij}##. Then you can just denote the ith row by ##A^i##.

    So for example, we have ##P_i=e_i## for all i.
     
  6. Apr 16, 2013 #5

    HallsofIvy

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    If we have basis [itex]\{u_1, u_2, \cdot\cdot\cdot, u_n\}[/itex] for vector space U, then we can represent a vector [itex]u= a_1u_1+ a_2u_2+ \cdot\cdot\cdot+ a_nu_n[/itex] as the array [itex]\left<a_1, a_2, \cdot\cdot\cdot, a_n\right>[/itex].
    In particular the basis vectors themselves are very easy:
    [itex]u_1= \left< 1, 0, \cdot\cdot\cdot, 0\right>[/itex]
    [itex]u_2= \left<0, 1, \cdot\cdot\cdot, 0\right>[/itex]
    ... [itex]u_n= \left<0, 0, \cdot\cdot\cdot, 1\right>[/itex]
    Now, look at what when you multiply each of those, written as a column by a matrix:
    Multiplying [itex]u_1[/itex] gives just the first column, multiplying [itex]u_2[/itex] gives the second column, etc.
    Example:
    [tex]\begin{bmatrix}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix}= \begin{bmatrix}a_{12} \\ a_{22} \\ a_{32}\end{bmatrix}[/tex]
    which will then be the coefficients of the expansion of Au in whatever basis we are using for the range space. That is, to represent linear transformation A from U to V, using a given ordered basis for each, apply A to each basis vector for U in turn, writing the result as a linear combination of the basis vectors for V. The coefficients of that linear combination will be the columns of the matrix representation.
     
    Last edited by a moderator: Apr 16, 2013
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