- #1
AlexChandler
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Homework Statement
Let V be a vector space and [tex] \{v_1,...,v_{n+1} \} \subset V [/tex] a set of linearly independent
vectors of V . Show directly: (Don't just quote a theorem!)
(a) The set [tex] \{v_1,...,v_{n} \} [/tex] is linearly independent.
(b) [tex] v_{n+1} \not \in span \{v_1,...,v_{n} \} [/tex]
Homework Equations
[tex] r_1_v_1_ + ... + r_{n+1}v_{n+1} = 0 \Rightarrow r_1=...=r_{n+1} = 0 [/tex]
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
[tex] r_1_v_1_ + ... + r_{n+1}v_{n+1} = 0 \Rightarrow r_1=...=r_{n+1} = 0 [/tex]
since we know that [tex] r_{n+1} = 0 [/tex]
we must have
[tex] r_1_v_1_ + ...+ r_n v_n + 0 v_{n+1} = 0 \Rightarrow r_1=...=r_n =r_{n+1} = 0 [/tex]
then
[tex] r_1_v_1_ + ...+ r_n v_n = 0 \Rightarrow r_1=...=r_n = 0 [/tex]
(b) suppose [tex] v_{n+1} [/tex] is an element of [tex] span\{v_1,...,v_{n} \} [/tex]
then
[tex] v_{n+1} = r_1_v_1_ + ...+ r_n v_n [/tex]
then we have
[tex] r_1_v_1_ + ...+ r_n v_n - v_{n+1} =0 [/tex]
since we know that [tex] \{v_1,...,v_{n+1} \} [/tex] is linearly independent, this last equation must be impossible. Thus our initial assumption must be incorrect, and we must have:
[tex] v_{n+1} \not \in span \{v_1,...,v_{n} \} [/tex]
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.