From eigenvalues alone no. If you have eigenvalues and a full set of eigenvectors, you know the matrix. If there are not enough eigenevctors, then the answer will again be no.

Even quicker: $M_m(R)$ is free of rank $m^2$ over its centre (and rank of free modules is an invariant for commutative rings).

Schemes have open coverings by affine schemes, which are locally ringed spaces, so schemes are locally ringed spaces. There are trivial examples of ringed spaces which are not locally ringed. Consider the one point space $X=\{x_0\}$. Given any ring $R$ there is a sheaf of rings on $X$ with global sections $R$. If $R$ is not local, these ringed space is not a locally ringed space. Of course, such examples are nothing like the ones that arise geometrically.

A duplicate of math.stackexchange.com/questions/4368/… where it has already had a reply.

Eliminating 13: if three planes meet in a line the create 6 regions; the last at most doubles that. If two planes are parallel they make 3 regions and the others at most quadruple it. If three planes meet pairwise in three parallel lines they create 7 regions. A plane not parallel to these lines doubles the number of regions, a plane parallel to them creates at four more regions. Thus all triples of planes meet in one point. If all planes have a point in common, the arrangement is symmetric about it and the number of regions is even. Otherwise we are in general position and get 15 regions.

MathSciNet doesn't list any papers by the author of this preprint.

And at a more banal level, one can get any finite cyclic group in the same manner :-)

One can get any group as $\pi_1(X)$.

I'm not sure what you are getting at, but $\eta^4$ is an honest modular form of weight $2$, but not for the whole modular group $\mathrm{SL}_2(\mathbb{Z})$ but for a congurence subgroup thereof.

Agreed, Andy