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Linear Algebra Question

  1. Dec 24, 2008 #1
    I am stuck on a question in my book:
    For a linear transformation T: V->V

    If Rank(T)=Rank(T^2), then prove that Range(T) U Null(T) = {0}.

    I don't know how to get started, I tried initiating a variable x=a_1*x_1 + ...+ a_n*x_n as a linear combination of a basis of V, that is an element of both the range and the null space, but am stuck after that.

    I have the idea that using the dimension theorem (rank nullity theorem), The equation Rank(T^2) + Nullity(T) = Dim(V) will be used in a suitable contradiction.

    I am confused, as I thought as the basis for the range and null space are always disjoint (how the rank nullity theorem is proved), then there's no way Range(T)U Null(T) != {0}.

    This problem made my question my understanding of linear transformations and their ranges and null spaces...
     
  2. jcsd
  3. Dec 24, 2008 #2

    Defennder

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    You sure it's supposed to be R(T) U N(T) = {0} ? Because clearly it doesn't hold for the identity transformation.
     
  4. Dec 24, 2008 #3
    Oh oops, its supposed to be the intersection, not the union of R(T) and N(T)
     
  5. Dec 24, 2008 #4

    HallsofIvy

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    Looks like a candidate for an "indirect" proof. Suppose there were a non-zero vector in both range of T and null set of T: That is, non-zero v such that T(x)= z and T(z)= 0. What can you say about the subspace spanned by z?
     
  6. Dec 24, 2008 #5

    mathwonk

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    assuming the space involved here is finite dimensional, the property that T^2 = T, (which is equivalent to the rank condition), implies T is a projection operator.

    this means that the nulllspace of T does not intersect the image of T except in {0}.
     
  7. Dec 25, 2008 #6
    mathwonk, how are the two conditions equivalent? It seems that if T is, say, a (non-identity) rotation in Rn, then both T and T2 have rank n, but they're not equal.

    Here's roughly how my proof went. I assume V is finite-dimensional.

    First prove that Null(T) = Null(T2); I used the rank-nullity theorem on both T and T2 for this. Suppose v is in both Range(T) and Null(T). Then v = T(w) for some w in V; but v is in Null(T), so T2(w) = T(v) = 0. Thus w is in Null(T2) = Null(T), so v = T(w) = 0.
     
    Last edited: Dec 25, 2008
  8. Dec 25, 2008 #7

    Defennder

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    I'm not understanding you fully. Which part of your proof requires that V is finite dim ?
     
  9. Dec 25, 2008 #8
    The proof that Null(T) = Null(T2) requires that V be finite-dimensional. Here's how I proved it:

    It's clear that Null(T) is a subspace of Null(T2). Suppose V is finite-dimensional. Then by the rank-nullity theorem,

    Rank(T) + dim Null(T) = dim V
    Rank(T2) + dim Null(T2) = dim V

    That, combined with the fact that Rank(T) = Rank(T2), allows me to conclude that the null spaces are finite-dimensional and dim Null(T) = dim Null(T2). Then since Null(T) is a subspace of Null(T2), Null(T) = Null(T2).



    Even if Range(T) = Range(T2), it is not necessarily true that Null(T) = Null(T2) if V is not finite-dimensional. Here's a counterexample: Let V = Rω be the vector space of sequences of real numbers. Define T: V → V by T(v1, v2, v3, ...) = (v2, v4, v6, ...). Then Range(T) = Range(T2) = V, but Null(T) is a proper subset of Null(T2). For instance, (0, 1, 0, 0, ...) is in Null(T2) but not Null(T).
     
    Last edited: Dec 25, 2008
  10. Dec 26, 2008 #9
    Thanks, does the null space and range always disjoint except at 0?

    If v=x1+ ax2 where x1 and x2 are basis for V, then does the basis for the null space and range share x1 and x2, and the subspace the at is formed by their span? I always assumed that they would be disjoint cept at 0.
     
  11. Dec 26, 2008 #10

    Defennder

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    Ah yes. Thanks.
     
  12. Dec 26, 2008 #11
    Under what conditions? If Rank(T) = Rank(T2) and V is finite-dimensional, then the range and null space intersect only at 0, but in general that's not true.
     
  13. Dec 27, 2008 #12
    The proof of the rank nullity theorem starts with a basis for the null space, then extending that to a basis for V, and the symmetric difference between the basis for the null space and the basis for V is the basis for the range. I am thus led to believe that the basis and therefore the space for the range and null space to be disjoint cept at 0.

    I know this is wrong..
     
  14. Dec 27, 2008 #13

    HallsofIvy

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    First, it only makes sense to talk about the "range" and "null space" to be disjoint or having non-trivial intersection if L is a linear transformation from vector space V to itself because if L is from V to U, the null space is in V and the range is in U. The "rank theorem" (that dimension of range+ dimension of null space equals the dimension of U) does not require that.

    But there is a simple counter example to your statement even in that limited situation. Let L: R2 to R2 be given by L(a, b)= (b, 0). The range is the subspace {(x, 0)} of course and the null space is exactly the same thing.
     
    Last edited: Dec 31, 2008
  15. Dec 28, 2008 #14
    Thank you.

    I had misinterpreted the proof of the rank nullity theorem. Going back and reviewing it cleared up my confusion.
     
  16. Dec 30, 2008 #15

    mathwonk

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    thanks for pointing out my error.
     
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