I'm having trouble understanding the following proof.
If are vectors in a vector space , then:
(a) The set of all linear combinations of is a subspace of .
(b) is the smallest subspace of that contains in the sense that every other subspace of that contains must contain .
I understand how to prove (a).
Now I have questions regarding proof (b):
The proof of (b) is given like this:
Each vector is a linear combination of since we can write
Therefore, the subspace contains each of the vectors .
Let be any other subspace that contains
Since is closed under addition and scalar multiplication, it must contain all linear combinations of . Thus contains each vector of
( My Question ) How does the author deduce that contains all linear combinations of in the last sentence of above proof?
How does the statement that " is closed under addition and scalar multiplication" make the statement that " contains all linear combinations of '' true?