Stuck on this proof for a custom concatenation operation:
let there be: $\vec a$,$\vec b$ ∈ $F^n$, $\vec c$,$\vec d$ ∈ $F^m$, λ∈F
⊗ is a vector concatenation operator:
$\vec a$ ⊗ $\vec b$ ∈ $F^{n+m}$. for example:
$$(2,0,3) ⊗ (1,2,3,4) = (2,0,3,1,2,3,4) $$
I have found that $λ(\vec a⊗\vec c) = (λ\vec a)⊗(λ\vec c)$
and : $(\vec a+\vec b)⊗(\vec c+\vec d) = (\vec a⊗\vec c) + (\vec b⊗\vec d)$
I need to prove that if {${\vec {a1},...,\vec {ak} }$} is linearly independent in $F^n$
and {${\vec {b1},...,\vec {bk} }$} is some vector set in in $F^m$
then {${\vec {a1} ⊗ \vec {b1} ,...,\vec {ak} ⊗ \vec {bk} }$} is linearly independent in $F^{n+m}$
i understand that if {${\vec {a1},...,\vec {ak} }$} is linearly independent then we pick scalars such that their sum is zero. then (${λ_1\vec {a1} ⊗ λ_1\vec {b1} +,...,+λ_k\vec {ak} ⊗ λ_k\vec {bk} }) = \vec 0 ⊗ $ (${λ_1\vec {b1}+,...,+λ_k\vec {bk}}$)
but that doesn't seem to prove anything..
It's the other way around:
Suppose $\lambda_1(a_1\otimes b_1)+\dots \lambda_k(a_k\otimes b_k) =0=0\otimes 0$.
Then conclude that $\lambda_1a_1+\dots +\lambda_ka_k=0$, hence each $\lambda_i =0$.