Question

factor of x^2 + 16

Answer 1

If the subsets can be equal to the original set, my proof would work. In formal terms, this works if \(L \subseteq U\), but not if \(L \subsetneq U\).

You're given a subset S of U that spans U, and a subset L of S that is linearly independent. Recall that to be a basis, you must be linearly independent, and span the vector space. Since S spans U, and there exists a subset L of S that is linearly independent, choose the subset B of S such that B is the largest linearly independent subset of S containing L. This may or may not be L.

Since it is the largest one you could choose, it must have n vectors that are all linearly independent. Thus, it must span U. Since it spans U, and is linearly independent, it must be a basis for U.