Is Linear Independence Preserved Under Subsets?

[AlexChandler]

Homework Statement

Let V be a vector space and $$\{v_1,...,v_{n+1} \} \subset V$$ a set of linearly independent
vectors of V . Show directly: (Don't just quote a theorem!)

(a) The set $$\{v_1,...,v_{n} \}$$ is linearly independent.

(b) $$v_{n+1} \not \in span \{v_1,...,v_{n} \}$$

Homework Equations

$$r_1_v_1_ + ... + r_{n+1}v_{n+1} = 0 \Rightarrow r_1=...=r_{n+1} = 0$$

The Attempt at a Solution

I have a feeling that I am doing something horribly wrong by saying this. But...

(a) We are given that

$$r_1_v_1_ + ... + r_{n+1}v_{n+1} = 0 \Rightarrow r_1=...=r_{n+1} = 0$$

since we know that $$r_{n+1} = 0$$

we must have

$$r_1_v_1_ + ...+ r_n v_n + 0 v_{n+1} = 0 \Rightarrow r_1=...=r_n =r_{n+1} = 0$$

then

$$r_1_v_1_ + ...+ r_n v_n = 0 \Rightarrow r_1=...=r_n = 0$$

(b) suppose $$v_{n+1}$$ is an element of $$span\{v_1,...,v_{n} \}$$

then

$$v_{n+1} = r_1_v_1_ + ...+ r_n v_n$$

then we have

$$r_1_v_1_ + ...+ r_n v_n - v_{n+1} =0$$

since we know that $$\{v_1,...,v_{n+1} \}$$ is linearly independent, this last equation must be impossible. Thus our initial assumption must be incorrect, and we must have:

$$v_{n+1} \not \in span \{v_1,...,v_{n} \}$$

I feel a bit more confident on part b, but not completely. We have not really focused much on proofs this semester in Linear Algebra, but I have a feeling they will be emphasized on the final. Any comments would be much appreciated.

 

[lanedance]

both of them look good to me
a) definition of linear independence, shows a subset of a linearly independent set mut aslo be linearly independent
b) definition of linear independence shows no vector in a linearly independent set can be re-written as a linear combination of others vectors in the set