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?

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!