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Linear algebra, basis, linear transformation and matrix representation

  1. Sep 3, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    In a given basis [itex]\{ e_i \}[/itex] of a vector space, a linear transformation and a given vector of this vector space are respectively determined by [itex]\begin{pmatrix} 2 & 1 & 0 \\ 1 & 2 & 0\\ 0&0&5\\ \end{pmatrix}[/itex] and [itex]\begin{pmatrix} 1 \\ 2 \\3 \end{pmatrix}[/itex].
    Find the matrix representations of the transformation and the vector in a new basis such that the old one is represented by [itex]e_1 =\begin{pmatrix} 1 \\ 1 \\0 \end{pmatrix}[/itex], [itex]e_2=\begin{pmatrix} 1 \\ -1 \\0 \end{pmatrix}[/itex] and [itex]e_3=\begin{pmatrix} 0 \\ 0 \\1 \end{pmatrix}[/itex].

    2. Relevant equations
    Not even sure.


    3. The attempt at a solution
    I suspect it has to do with eigenvalues since the problem in the assignments is right after an eigenvalue problem. I would not have thought of this otherwise.
    It reminds me of 2 similar matrices, say A and B, then A=P^(-1)BP but nothing more than this.
    I've found that the given matrix has 3 different eigenvalues, thus 3 eigenvectors (hence non 0 vectors) that are linearly independent (since the 3 eigenvalues are all different) and thus the matrix is similar to a triangular one with the eigenvalues 1, 3 and 5 on the main diagonal.
    So now I know that there exist an invertible matrix P such that A=P^(-1)BP. But I'm really stuck at seeing how this can help me.
    Any tip's welcome.
     
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  3. Sep 3, 2011 #2

    micromass

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    First you'll need to find the change-of-basis matrix. That is, a matrix A such that if x are coordinates in the old basis, then Ax are coordinates in the new basis.

    You know already what [itex]Ae_1,~Ae_2,~Ae_3[/itex] are. So can you use this to find the matrix A?
     
  4. Sep 4, 2011 #3

    fluidistic

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    Thanks for your help. I'm confused, I didn't know I knew what [itex]Ae_i[/itex] are.
    Are these the column vectors of the first matrix of my first post?
     
  5. Sep 4, 2011 #4

    micromass

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    Oh, no, sorry. I was thinking of the [itex]e_i[/itex]as the original basis.

    So, if I rephrase it, you know what the image of (1,0,0), (0,1,0) and (0,0,1) under A are?
     
  6. Sep 4, 2011 #5

    fluidistic

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    Ah ok. :)
    Hmm no either. [itex]\{ e_i \}[/itex] isn't necessarily the canonical basis as far as I know...
     
  7. Sep 4, 2011 #6

    micromass

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    The coordinates of any basis is always (1,0,0), (0,1,0), (0,0,1). Regardless of whether it is the canonical basis.

    By definition, the coordinates of a vector v with respect to [itex]\{e_1,e_2,e_3\}[/itex] are [itex](\alpha_1,\alpha_2,\alpha_3)[/itex] such that

    [itex]v=\alpha_1e_1+\alpha_2e_2+\alpha_3e_3[/itex].

    From this, we see that the coordinates of [itex]e_1[/itex] are always (1,0,0).
     
  8. Sep 4, 2011 #7

    I like Serena

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    Hi fluidistic! :smile:

    First let's define our symbols.
    Let's call your linear transformation matrix A, and your vector v.

    What you need is a matrix B that defines a basis transformation.
    B is defined by the fact that for instance the unit vector (1,0,0) is transformed to your e1. That is: B (1,0,0) = e1.

    Can you deduce what the matrix B is?

    If you apply this B to your vector v, you'll find the represention of the v in the alternative basis.
     
  9. Sep 4, 2011 #8

    fluidistic

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    Hi!
    Is [itex]B= \begin{pmatrix} 1 & 1 & 0 \\ 1& -1 & 0\\ 0& 0 & 1 \end{pmatrix}[/itex]? It seems to work for me. Its columns vectors are the one given in the problem statement.
     
  10. Sep 4, 2011 #9

    I like Serena

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    Yep, that's it! :smile:
    It's simply the given unit vectors as columns of the matrix B.

    Calculate Bv and you have your representation of v in the basis defined by B.

    For your matrix A, you need B-1AB.
    That is, first you transform a vector to the alternate basis (wrt which A has been defined), then you apply A, and then you transform the result back to your original basis.
     
  11. Sep 4, 2011 #10

    fluidistic

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    I reached [itex]v'= \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix}[/itex]. And A' is a diagonal matrix with the eigenvalues of A as entries... (3, 1 and 5).
     
  12. Sep 4, 2011 #11

    I like Serena

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    That's it! :smile:

    I'm surprised myself that A' is diagonal, but I see that it is.

    In retrospect it is clear that it is, since the given basis consists of exactly the eigenvectors of A.
    Note that A is diagonized by a base transition consisting of its eigenvectors.

    I suspect this problem was intended to demonstrate this feature of eigenvectors.
     
  13. Sep 4, 2011 #12

    fluidistic

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    Yes it is. In my first post I saw that A was diagonalizable.
    So [itex]A=P^{-1}A'P[/itex]. I didn't know that [itex]P^{-1}=B^{-1}[/itex].
    I had calculated the eigenvalues of A, so that I think I had calculated A'. (Up to the order of the eigenvalue on the main diagonal, but I don't think it's relevant).
    Is there a way I could have found P=B which is faster than the one I've done?
    In other words, knowing A' from start, is there a quick way to solve the problem?
     
  14. Sep 4, 2011 #13

    I like Serena

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    Well, this problem is actually not about eigenvectors.
    It's about a base transformation.
    You can't assume that the new basis consists of the eigenvectors!
     
  15. Sep 4, 2011 #14

    fluidistic

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    Ok. I think I understand. :smile:
    Thanks guys!
     
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