How many Hecke operators span the Hecke algebra?

Linear algebra, span, vector space and subspaces?

• why is the span S of any set of vectors in a vector space V a subspace of V?

• Answer:

  Let v1,...,vn be any set of n vectors in vector space V. S = span of {v1,...,vn} is defined as the set of all linear combinations of the vectors v1,...,vn. To prove that S is a subspace of V, we need to prove that i) 0 is in S, ii) S is closed under addition, and iii) S is closed under scalar multiplication. i) This part is easy. Note that 0 = 0v1+...+0vn, which is a linear combination of the vectors v1,...,vn and is therefore in S. ii) For any vectors x and y in S, we can write, for scalars c1,...,cn and d1,...,dn, x = c1v1+...+cnvn and y = d1v1+...+dnvn. Then x + y = (c1v1+...+cnvn) + (d1v1+...+dnvn) = (c1+d1)v1+...+(cn+dn)vn and so x + y is a linear combination of v1,...,vn and is therefore in S as well. So S is closed under addition. iii) For any vector x in S and any scalar r, we can write, for scalars c1,...,cn, x = c1v1+...+cnvn. Then rx = r(c1v1+...+cnvn) = (r*c1)v1+...+(r*cn)vn and so rx is a linear combination of v1,...,vn and is therefore in S as well. So S is closed under scalar multiplication. We conclude that the span S of any set of n vectors is a subspace of V. Lord bless you on this Easter Sunday!