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A question about linear algebra (change of basis of a linear transformation)

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [itex]A \in M_n(F)[/itex] and [itex]v \in F^n[/itex].

    Let [itex]v, Av, A^2v, ... , A^{k-1}v[/itex] be a basis, B, of V.

    Let[itex] T:V \rightarrow V[/itex] be induced by multiplication by A:T(w) = Aw for w in V. Find [itex][T]_B[/itex], the matrix of T with respect to B.


    Thanks in advance



    2. Relevant equations


    [itex][T(w)]_B = [Aw]_B = C^{-1}Aw[/itex]

    3. The attempt at a solution

    Can anybody give me a hint please? I'm trying to do this for an hour but I'm not sure how.

    From here: http://www.khanacademy.org/math/lin...transformation-matrix-with-respect-to-a-basis

    I learned that [itex][T(w)]_B = [Aw]_B = C^{-1}Aw[/itex], where [itex] C= [v| Av| A^2v| ... | A^{k-1}v] [/itex]. But now I don't know what the inverse of C is? :cry:

    Thanks in advance
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 30, 2012 #2

    jbunniii

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    Well, if you write [itex]b_0 = v[/itex], [itex]b_1 = Av[/itex], [itex]b_2 = A^2v[/itex], ..., [itex]b_{n-1} = A^{n-1}v[/itex], then you have [itex]T(b_k) = b_{k+1}[/itex] for [itex]0 \leq k < n-1[/itex]. What does this tell you about the first [itex]n-1[/itex] columns of [itex][T]_B[/itex]?
     
  4. Oct 1, 2012 #3
    They are the same elements as the basis, except for the last one, since it is [itex]A^kv[/itex]...
     
    Last edited: Oct 1, 2012
  5. Oct 1, 2012 #4

    jbunniii

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    So what are the first [itex]n-1[/itex] columns? You should be able to write them using actual numbers.
     
  6. Oct 1, 2012 #5
    The first n-1 columns are: [itex]Av, A^2v, ..., A^{k-1}v[/itex]. Since each element [itex]A^pv[/itex] is transformed into [itex]A^{p+1}v[/itex], right?
     
  7. Oct 1, 2012 #6

    jbunniii

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    Let's take a step back. What is the matrix representation of [itex]b_0 = v[/itex] in terms of the basis [itex]B[/itex]? Your answer should be a column vector containing 1's and 0's.
     
  8. Oct 1, 2012 #7
    I'm not sure I understand why it can be written with numbers. The question tells us what [itex]B[/itex] is, not with numbers but with A's and v's. Do you mean something like this:

    [itex][v]_B[/itex] = [itex] 1.v + 0.Av + ... + 0.A^{k-1}v [/itex]?
     
  9. Oct 1, 2012 #8

    jbunniii

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    That's not what [itex]v_B[/itex] is. It is true that
    [tex]v = 1\cdot v + 0 \cdot Av + \ldots + 0 \cdot A^{k-1}v[/tex]
    And since B is a basis, this is the unique way of writing [itex]v[/itex] as a linear combination of the elements of B. So what is [itex]v_B[/itex]? It is nothing other than the array of coefficients in that linear combination: [itex]v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T[/itex] where by convention this is understood to be a column vector.
     
  10. Oct 1, 2012 #9
    So [itex][T]_B[/itex] is just the kxk identity matrix?
     
  11. Oct 1, 2012 #10

    jbunniii

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    No, because T doesn't map each element of B to itself.
    In fact, T maps [itex]b_0[/itex] to [itex]b_1[/itex]. Therefore the first column of [itex][T]_B[/itex] should be the [itex][b_1]_B = [Av]_B[/itex], which is what?
     
  12. Oct 1, 2012 #11
    But didn't you say that

    [itex]v_B = [\begin{array}{cccccc}1 & 0 & 0 & 0 & \ldots &0\end{array}]^T[/itex]?

    So...

    [itex][Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T[/itex]
    [itex][A^2v]_B = [\begin{array}{cccccc} 0 & 0 & 1 & 0 & \ldots &0\end{array}]^T[/itex]
    ...
    [itex][A^{k-1}v]_B = [\begin{array}{cccccc} 0 & 0 & 0 & 0 & \ldots &1\end{array}]^T[/itex]

    So aren't these the columns of [itex][T]_B[/itex]? So why isn't it the identity matrix?
     
  13. Oct 1, 2012 #12

    jbunniii

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    Well, the first column of [itex][T]_B[/itex] is [itex][Tv]_B = [Av]_B = [\begin{array}{cccccc} 0 & 1 & 0 & 0 & \ldots &0\end{array}]^T[/itex] so already it can't be the identity matrix.
     
  14. Oct 1, 2012 #13
    Oh, I get...thanks. But for the last one, we won't be able to write it with numbers, right? [itex]A^kv[/itex] must be some linear combination of the basis, but we can't know exactly what...right? So we just write...

    [itex][T]_B[/itex] = [\begin{array}{cccccc} c_1 & c_2 & \ldots &c_k\end{array}]^T[/itex]
     
  15. Oct 1, 2012 #14

    jbunniii

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    Offhand I'm not sure about the last column. I'm not sure if there is enough information given to know [itex]T(A^{k-1}v)[/itex] in terms of B. Maybe think about the fact that [itex]v[/itex], [itex]Av[/itex], [itex]A^2v[/itex], ..., [itex]A^{k-1}v[/itex] form a basis. This wouldn't necessarily be true for arbitrary A and v, so this gives you some information about A and v. I'll think about this some more and let you know if I have an idea.
     
  16. Oct 1, 2012 #15

    jbunniii

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    Yeah, I'm pretty sure the last column can be anything. It doesn't even have to be linearly independent of the other columns as there's nothing in the problem that requires T to be invertible.
     
  17. Oct 1, 2012 #16
    Alright, thanks a lot. :biggrin:
     
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