Re: determinant question
From: Dave Rusin (rusin_at_vesuvius.math.niu.edu)
Date: 01/25/05
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Date: 25 Jan 2005 22:40:48 GMT
In article <ct3tr6$28d$1@joe.rice.edu>,
Nathan <beethoven3322@hotmail.com> wrote:
>The subspace of the set of real nxn matrices with zeros all along the
>top row has dimension n^2-n and the determinant of every element is
>zero. Is it true that any subspace with dimensions n^2-n+1,n^2-n+2,...,
>or n^2-1 will have non-zero matrices of both zero and nonzero
>determinant in it?
When you say (sub)space, do you mean _vector_ space? Then any
subspace contains the zero matrix, so yes, there must be matrices of
zero determinant. A much more interesting question is, what is
the dimension of the largest linear subspace of M_n(R) all of whose
non-zero entries have non-zero determinant. The answer is known
(Adams/Lax/Phillips, Proc. Amer. Math. Soc. 16 (1965) 318-322):
if 2^m is the largest power of 2 dividing n then the dimension is
8 floor(m/4) + 2^{ m mod 4 }. You can check that this equals n
itself when n = 1, 2, 4, 8 and indeed the existence of division
algebras of those dimensions proves that vectors subspaces this big
do indeed exist in those dimensions.
In the other direction you ask conversely how large is the largest linear
subspace contained in the variety of matrices of determinant zero.
I'm pretty sure this is indeed of dimension n as you suggest but I
don't recall how to prove that there can be nothing larger.
dave
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