How Do You Prove L(L(A)) ⊆ L(A) Using Subspace Properties?

[yifli]

L(A) is the linear span of all the vectors in A, and A is a subset of a vector space V

Since the linear span of A always include A, it follows that $$L(A)\subset L(L(A))$$

But how do you prove the other direction?

 

[Unit]

Try taking any vector v in L(L(A)) and show that it is also an element of L(A). This will show that L(L(A)) ⊂ L(A).

 

[HallsofIvy]

Saying the v is in L(L(A)) means that it is a (finite) linear combination of vectors in L(A): $v= a_1 u_1+ a_2u_2+ \cdot\cdot\cdot+ a_nu_n$. But every vector in L(A) is a linear combination of vectors in A: $u_i= b_{i1}w_1+ b_{i2}w_2+ \cdot\cdot\cdot+ b_{im}w_m$. That is, $v= a_1(b_{11}w_1+ b_{12}w_+ \cdot\cdot\cdot+ b_{1m}w_m)p$$+ a_2(b_{21}w_1+ b_{22}w_2+ \cdot\cdot\cdot+ b_{2m}w_m)$$+ \cdot\cdot\cdot+ a_n(b_{n1}w_1+ b_{n2}w_2+ \cdot\cdot\cdot+ b_{nm}w_m)$

That can be rewritten as a linear combination of members of A.

 

[yifli]

Saying the v is in L(L(A)) means that it is a (finite) linear combination of vectors in L(A): $v= a_1 u_1+ a_2u_2+ \cdot\cdot\cdot+ a_nu_n$. But every vector in L(A) is a linear combination of vectors in A: $u_i= b_{i1}w_1+ b_{i2}w_2+ \cdot\cdot\cdot+ b_{im}w_m$. That is, $v= a_1(b_{11}w_1+ b_{12}w_+ \cdot\cdot\cdot+ b_{1m}w_m)p$$+ a_2(b_{21}w_1+ b_{22}w_2+ \cdot\cdot\cdot+ b_{2m}w_m)$$+ \cdot\cdot\cdot+ a_n(b_{n1}w_1+ b_{n2}w_2+ \cdot\cdot\cdot+ b_{nm}w_m)$

That can be rewritten as a linear combination of members of A.

Thank you for the reply.

But what if I can only use the following two properties to prove $$L(L(A))\subset L(A)$$
1. L(A) is a subspace which includes A
2. If M is any subspace which includes A, then $$L(A) \subset M$$