Re: One problem on linear algebra
From: William C Waterhouse (wcw_at_math.psu.edu)
Date: 10/08/04
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Date: Fri, 8 Oct 2004 20:19:52 +0000 (UTC)
In article <pan.2004.10.06.16.50.16.958099@lehigh.edu>,
"David L. Johnson" <david.johnson@lehigh.edu> writes:
> On Mon, 04 Oct 2004 07:29:29 -0700, moonlit wrote:
>
> > Let V be an n-dimensional vector space over R and let w be a NON ZERO
> > exterior k-form, 1<= k <=n-1 on V. Let us consider the following subspace
> > of V,
> > S = {x is in V | x#w = 0 }, where # denotes the exterior
> > product.
> >
> > Then, what is the maximum possible dimension of S?
>
> Let me first clarify what I think you mean. w is an element of the k-th
> exterior product of V (a k-form would be in the dual), right? And # is
> just the usual wedge product. Under those assumptions, I believe the
> maximum dimension of ker( x --> x#w ) occurs when w is decomposable,
> w=(v_1)#...#(v_k). In that case, then the maximum dimension would of
> course be k. The proof would be a messy case analysis, writing w in
> general as a sum of decomposable elements and dividing up the
> possibilities for x#(each term), with possible cancellations.
It's not really too messy. Take a basis x_1, x_2, ..., x_n of V such
that x_1,..., x_s are a basis of S. Write w as a sum of multiples of
the basic exterior products x_{i_1} # x_{i_2} # ... # x_{i_k}, where the
indices are increasing. If there are terms in which some x_j does not
occur, then those all yield independent terms in x_j # w. Thus
every nonzero term in our w (with this basis) must start with
x_1 # x_2 # ... # x_s. Hence of course s is indeed at most k, with
equality only when w is (up to scaling) the product of the k basis
vectors in S.
William C. Waterhouse
Penn State
PS. This is a "standard" fact, found (for instance) in Bourbaki's
Multilinear Algebra.
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