Results 1 to 7 of 7

Math Help - Help with a (seemingly?) simple proof.

  1. #1
    Member
    Joined
    Mar 2009
    Posts
    168

    Help with a (seemingly?) simple proof.

    I've been brushing up on some Linear Algebra, and came across this puzzler in a proof book. I'm stuck, and looking for someone who has a proof for this. Here's the statement:

    Let V be a vector space of dimension n, with a basis B = {u(1),...,u(n)}.
    Let W be a subspace of V with dimension k with 1 <= k <= n.
    There exists a subset of B which is a base for W. Prove this is true, or provide a counterexample.

    Any help on this would be appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by paupsers View Post
    I've been brushing up on some Linear Algebra, and came across this puzzler in a proof book. I'm stuck, and looking for someone who has a proof for this. Here's the statement:

    Let V be a vector space of dimension n, with a basis B = {u(1),...,u(n)}.
    Let W be a subspace of V with dimension k with 1 <= k <= n.
    There exists a subset of B which is a base for W. Prove this is true, or provide a counterexample.

    Any help on this would be appreciated!
    This is not true. Consider V=\mathbb{R}^3 with basis \{ \bold{i},\bold{j},\bold{k}\}. The space spammed by \bold{i},\bold{j},\bold{k} is the x-axis,y-axis, z-axis respectively. The space spammed by \{\bold{i},\bold{j}\},\{ \bold{i},\bold{k}\},\{\bold{j},\bold{k}\} is the xy-plane, xz-plane, yz-plane respectively. Therefore, any subspace that is distinct from these will not be spammed by a subset of this basis. For example, the plane x+y+z=0 is a subspace but it one one of those coordinate planes and so it cannot be spammed by a subset of the basis.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    isn't it spanned? or is that a joke?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Mar 2009
    Posts
    168
    "Therefore, any subspace that is distinct from these will not be spammed by a subset of this basis."

    Can you (or someone) please elaborate on this part? I'm trying to understand your logic here...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Sampras's Avatar
    Joined
    May 2009
    Posts
    301
    Quote Originally Posted by paupsers View Post
    "Therefore, any subspace that is distinct from these will not be spammed by a subset of this basis."

    Can you (or someone) please elaborate on this part? I'm trying to understand your logic here...
    We considered a basis  B = \{\bold{i}, \bold{j}, \bold{k} \} of  \mathbb{R}^{3} . Now there are  2^3 = 8 possible subsets of  B . We considered all of them as possible bases except the empty set. Then we showed what they span. So then you can choose a subspace which is not spanned by any of those subsets. These are counterexamples.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by paupsers View Post
    "Therefore, any subspace that is distinct from these will not be spammed by a subset of this basis."

    Can you (or someone) please elaborate on this part? I'm trying to understand your logic here...
    The only subspaces spammed by i or j or k are the axes. The only subspaces spammed by i,j or i,k or j,k are the coordinate planes. If you pick any subspace that is different from these spaces, then you have found one that cannot be spammed by the subspace of {i,j,k}. For example, the plane x+y+z=0, is a subspace of R^3 but since it is not one of the axes or the coordinate planes it means it cannot be spammed by a subspace of this basis (that basis {i,j,k}).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Mar 2009
    Posts
    168
    I guess I understand. I was just thrown off by the idea that the xy-plane is the "span" of {i, j} for example. I've only ever known span by the formal definition of "the set of all linear combinations of the vectors of a subset S." But I guess that would define the xy-plane in this case.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. seemingly simple limit proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 2nd 2010, 12:09 PM
  2. A seemingly simple proof
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: July 8th 2010, 09:09 PM
  3. seemingly simple derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 17th 2009, 06:05 PM
  4. Trouble with a seemingly simple integral!
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 26th 2009, 02:08 AM
  5. Another seemingly simple integral
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 25th 2009, 01:44 PM

Search Tags


/mathhelpforum @mathhelpforum