4
$\begingroup$

Hi there,

Consider linear endomorphisms ("endos") of a finite dimensional vector space.

How can those endos be characterized, for which said vector space has a basis with respect to which the endo has a matrix with only nonnegative entries?

Not all endos have this property: e.g., $x\mapsto -x$. More generally, necessary conditions can be derived from Perron-Frobenius theory.

Does anybody know this problem? Or its solution?

-DOT

$\endgroup$

1 Answer 1

6
$\begingroup$

This is related to the mathoverflow question Perron-Frobenius "inverse eigenvalue problem". Doug Lind's theory characterizes when an endomorphism $E$ can be extended to an endomorphism $\bar E$ of some larger vector space having a basis such that the matrix for the extended endomorphism is nonnegative: this can be done if and only if it satisfies the Perron-Frobenius condition, that the spectral radius of $E$ is an eigenvalue of $E$. Even for an automorphism of $R^3$, the dimension necessary for an extension with this property can be arbitrarily large (as shown by examples of Lind).

For dimensions 1 and 2, it's easy to see the Perron-Frobenius condition is sufficient.

You can think of the question geometrically in terms of projective space. For simplicity, take $E$ to be an automorphism. There's a sub projective space $M$ (which might be a single point) that mapped isometrically to itself, consisting of linear combinations of eigenvectors (real and complex) whose eigenvalue has maximal absolute value. Almost everything else is contracted toward $M$. The question is whether there's a simplex $T$ in projective space that contains its image.

For low dimensions it should be possible to describe the specific inequalities the set of eigenvalues need to satisfy by making use of the relatively simple geometry of the orbits of linear maps. For larger dimensions, I believe this question is too hopelessly complicated to expect a complete concrete description or feasible general algorithm.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .