Results 1 to 5 of 5

Math Help - Interesting exercise on a particular set of vectors.

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    10

    Interesting exercise on a particular set of vectors.

    Removed by the author
    Last edited by eurialo; October 21st 2010 at 04:22 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Tinyboss's Avatar
    Joined
    Jul 2008
    Posts
    433
    Isn't X just the preimage of a closed set under a continuous map?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    10
    In general, X it is the image of a closed set under a function. If the matrix is invertible, it is also the preimage under some other function, but this is not true if the matrix is not invertible.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member HappyJoe's Avatar
    Joined
    Sep 2010
    From
    Denmark
    Posts
    234
    It is certainly true for non-singular matrices A (as the corresponding linear map has a linear (hence continuous) inverse).

    I'm not sure about the singular matrices. It seems very likely that, in general, singular matrices are closed maps in the sense that they map closed sets to closed sets.

    We do have the isomorphism

    \mathbb{R}/\text{Ker}(A)\simeq \text{Im}(A),

    which gives us a factorization of A (as a linear operator) as the composition of a projection P\colon \mathbb{R}^n\rightarrow\mathbb{R}^n/\text{Ker}(A) composed with the inclusion of \mathbb{R}^n/\text{Ker}(A)\simeq\text{Im}(A) into the codomain \mathbb{R}^m.

    The inclusion is certainly a closed map, and so it suffices to check that projections are also closed maps.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member HappyJoe's Avatar
    Joined
    Sep 2010
    From
    Denmark
    Posts
    234
    Ugh, it is clearly wrong that all singular matrices are closed maps. I must have been too focused on it being true that I didn't care to look for an easy counterexample. You can take the projection (x,y)\mapsto x, for which the closed set given by the graph of f(x) = 1/x maps to an open set of \mathbb{R}.

    For this particular problem, I asked around, and someone came up with this idea:

    The set X is the cone of the unit cube [0,1]^n, i.e. X=C([0,1]^n), where C is the cone operator, that is, C(S) = \{\lambda a | \lambda\geq 0, a\in S\}. For any linear operator A, and any set S, it is true that

    A(C(S)) = C(A(S)).

    Hence A(X) = C(A([0,1]^n)), which is the cone of a compact set (since all linear maps are continuous, even the singular ones), but the cone of a compact set is closed.

    I've also recalled that if Y is a compact topological space, then the projection \pi_X\colon X\times Y\rightarrow X is a closed map - just for bonus info.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Interesting (and funny) exercise
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: October 6th 2010, 05:45 AM
  2. Exercise
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 25th 2009, 03:59 AM
  3. Interesting polynomial exercise
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: June 23rd 2009, 08:19 AM
  4. Help with an exercise
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 10th 2009, 11:35 PM
  5. looking for an interesting problem in vectors
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: July 2nd 2007, 07:44 PM

Search Tags


/mathhelpforum @mathhelpforum