To original poster: Ignore my example, it doesn't work...although I do remember something from linear algebra about not every square matrix having a square root. I think that result is much harder than the problem intended.
Here is how complicated it can get with matrices: http://en.wikipedia.org/wiki/Square_root_of_a_matrix
Justin: if you want a matrix (over
say) without a square root, just get one with a negative determinant like
Makes sense...so my original idea was not wrong then. Thanks!