MIT Linear Algebra, Lecture 5: Vector Spaces and Subspaces

You're viewing a comment by bobovski and its responses.

bobovski Permalink
January 22, 2010, 21:03

Nate, there is an algebraic structure on R^2 that behaves a lot like a vector space structure for which Z^2 *is* a sub-thingy. Instead of multiplying vectors in R^2 by real scalars, you allow only multiplication by integers. This gives R^2 the structure of a Z-module, and Z^2 is a submodule of R^2 with this structure.

Reply to this comment

Reply To This Comment

(why do I need your e-mail?)

(Your twitter name, if you have one. (I'm @pkrumins, btw.))

Comment Help

Type first 3 letters of your name: (just to make sure you're a human)

Please preview the comment before submitting to make sure it's OK.