Dear MathTalk,
	Suppose M is a matrix.
        Can we claim that the spectral radius of |M| <= the spectral
radius of M? (|M| is the matrix obtained by taking the absolute values
of the corresponding entries in M)
        Does it make any difference if we suppose that M has real as
opposed to complex entries?
        Many thanks,

        Dooogle

Actually, the reverse inequality is true.  You can apply Gelfand's formula
which says the spectral radius of a (complex) valued matrix is defined by:

rho(M) = lim_{k -> inf} || M^k ||^(1/k)


where || M || is *any* matrix norm.  This, combined with the fact that
|M^k| <= |M|^k, can be used to show rho(M) <= rho(|M|).

A good reference for these kinds of questions is the book Matrix Analysis
by Horn & Johnson.

Just so.  Although the spectral radius of M and "|M|" must agree in
the (trivial) 1x1 case, a simple 2x2 "counterexample" to dooogle-ga's
conjecture (using only real entries) is:

     /  1    1  \
 M = |          |
     \  1   -1  /

The spectral radius of M is sqrt(2), but if we take the absolute value
of its entries, we get spectral radius 2.

regards, mathtalk-ga

Many thanks for your help. Much appreciated.

Dooogle