Re: Elementwise rank preserving map

toni.lassila@xxxxxxxxx writes:

Let f: R -> R be such that the elementwise matrix map f(A) preserves
rank, where A in R^{n x m} with n,m >= 2. In other words, for all A in
R^{n x m} we have rank(A) = rank(f(A)). Is is true that f is then
multiplication with a scalar?

In the n x 2 and 2 x m cases, what you need is f(x) = k g(x) where k is a
nonzero constant, g(0) = 0 and g is a one-to-one homomorphism of the
multiplicative group of nonzero reals into itself. For example, you could
have g(x) = signum(x) |x|^p for any nonzero real p (but there are also more
exotic, nonmeasurable examples).
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.