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. ConsiderOriginally Posted by paupsers
with basis
. The space spammed by
is the x-axis,y-axis, z-axis respectively. The space spammed by
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
is a subspace but it one one of those coordinate planes and so it cannot be spammed by a subset of the basis.
We considered a basisof
. Now there are
possible subsets of
. 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.
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}).
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.
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