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Determing formula for T inverse

  1. Oct 5, 2007 #1
    Ehm, this is kind of a dumb question (as in, it should be very easy and I'm just having a moment of confusion). If I'm given a linear transformation T from vector space V to vector space W (fininte dim) and I want to compute T-inverse, I know that I can compute the matrix representation of T and invert it. So my question is, once I have inverted the matrix, how do I reconstruct an explicit formula for T-inverse? I know the values of T-inverse on my basis (the coefficients) but it seems that this doesn't quite tell me what formula to use since more than one possibility exists.

    Also, is there a good way to compute T-inverse without the use of the matrix representation? The only way I know how is if the mapping is from R to R, because that's like single variable calculus, but as soon as we have some other space I'm clueless without a matrix rep.

    Ideas? Know what I mean? :shy:
     
  2. jcsd
  3. Oct 5, 2007 #2

    radou

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    Here's a practical example: Let T : R^3 --> R^3 be given with T(x, y, z) = (2x, 4x-y, 2x+3y-z). One should first prove that T is regular, which is pretty much simple. Now, to get to the point, sine T^-1 exists, you have T^-1(x, y, z) = (u, v, w). Multiply with T in order to obtain (x, y, z) = T(u, v, w) = (2u, 4u-v, 2u+3v-w). All you have to do now is solve for u, v and w.
     
  4. Oct 5, 2007 #3
    this is FANTASTIC!!! I get it! haha thanks so much!
     
  5. Oct 7, 2007 #4

    daniel_i_l

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    But both methods are really the same: you could solve this system of 3 equations and 3 unknowns by constructing the coorsponding 3x3 matrix and inverting it.
     
  6. Oct 7, 2007 #5

    mathwonk

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    i understnd completely whaT YOU are asking nd have a perfetly cleaR RESONSE , but it does not quite fit into the margin of this answer section. .........
     
  7. Oct 7, 2007 #6

    mathwonk

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    t inverse is a polynomial in t. just find the characteristic polynomial of t, subtract off the constant term, and then factor out t.
     
  8. Oct 7, 2007 #7

    radou

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    Actually, probably the most important thing about this problem is to be aware of some neat isomorphisms between spaces of matrices and spaces of operators and their consequences.
     
  9. Oct 9, 2007 #8
    the real question is, if I take matrix, invert it, how do I translate my columns back into some explicit formula for T? it's now making sense for polynomials. Still not sure for the more general case, however.
     
  10. Oct 9, 2007 #9

    radou

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    You should know exactly how the matrix representation of some operator T : V --> W is constructed.
     
  11. Oct 9, 2007 #10
    yes, I DO. What I mean, is that given the matrix and NOT the operator, how do I go from the matrix back to the formula for T? Obviously the columns each give the set of coefficients determining our unique vector in the imT. I just am not sure what I need do with my coefficients to make one general formula that applies to all the vectors. Maybe this question doesn't make sense, I don't know.
     
  12. Oct 9, 2007 #11

    matt grime

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    What do you mean by 'a formula for T'? That seems like a very ill-defined concept to me.
     
  13. Oct 11, 2007 #12
    I think I've figured my stuff out... I was somehow confusing myself over nothing. There shouldn't be a "formula" per se, what we are really finding is the coefficents of the vectors in the imT.
     
  14. Oct 11, 2007 #13

    HallsofIvy

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    'a formula for T'

    Given a basis for the vector space T is over, T can be written as a matrix. Is that what you mean by 'a formula for T'?
     
  15. Oct 11, 2007 #14

    radou

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    I think he means something like T(x1, ... , xn) = ... Of course, this only works for operators defined on spaces of ordered n-tuples etc.
     
  16. Oct 12, 2007 #15
    yes, that's what I was getting at. It's okay - I think I've got a firm enough grasp of what I'm doing now to see what's going on. You know how it is, sometimes you look at something TOO hard and then it seems complex when in actuality it's very simple. I tend to do that a lot.
     
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