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Some questions about reducible matrices and operators

  1. Apr 3, 2007 #1
    Hello ,

    This is regarding the equality of nullity between A and PAP(inverse).

    If my understanding is correct then the thing should be according to the diagram below

    V---------------->R1(isomorphic to V)
    | |
    | |
    \/ \/
    W----------------->R2(isomorphic to W)

    So A : V->W
    P: V->R1
    P: W->R2

    This second thing I am not convinced since the size of W is less than V because of the nullity in this case is assumed to be greater than 1.

    If my understanding is correct what one does is that sincce R1 is isomorphic to V so

    R1---->R2 is R1->V->W->R2. Which means take bases in R1 and apply P(inverse) which is correct since inverse is defined for the isomorphic transformation and then applying A u land up all the vectors from V which u had got from R1 to 0 in W. Then u apply P to all the vectors and claim that none of them go to 0 in R2 other than the 0s of W. What is confusing is that how do u define the same linear transformation in both the two vector spaces of two different dimensions in general and if possible how do u prove that only the 0s in W go to 0s in R2 and no body else.
    Last edited: Apr 3, 2007
  2. jcsd
  3. Apr 4, 2007 #2

    matt grime

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    How can P be both a map from V to R_1, and W to R_2? It can't and it isn't.

    If P is invertible, then it is square, as well. So even if we assume A is a map from V to V, that is V is isomorphic to W, then it is not correct to assert "W has smaller size [dimension?] than V".

    By equality of nullity do you mean the dimension of the kernel/null space? That is simple - show that they are isomorphic vector spaces. (P is invertible).
  4. Apr 5, 2007 #3
    Actually after I wrote down the query I happened to refer again to Kunze Huffman and found that this is a standard theorem regarding transformation of linear operator from one basis to another.

    Then I realized that the point which was not clear was that if Tv is a vector in basis B then how could with respect to B' I could write P(Tv) where P is the matrix of transformation from B to B'. What is unclear is that when u are doing this u are actually trying to premultiply a vector which is already in the space W, but according to theorem 8 on pg 53 of the book it says that

    Suppose I is an n x n invertible matrix over F. Let V be an n dimensional vector space over F and let B be an ordered basis of V . Then there exists unique ordered basis B' of V such that [alpha] in basis B= P[alpha ] in basis B'.

    So how is it here the vector space in which the vector is going to reside and the basis are completely different. Am I missing something very obvious. My thinking is that probably even if the space W has smaller dimension than V it is extended by adding 0s to it to equate V and then trying to apply the above technique. Still I am highly confused of applying a matrix NxN which would transform V -> V on something in W.

    Sorry but I do not know how to use the subscripts here for clarity but hopefully I have been able to make my doubt across.
    Last edited: Apr 5, 2007
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