Prove/disprove question regarding matrix-vector multiplication and linear independence

Question

If {Bv₁,...,Bv_k} is a linearly independent set in R^k where B is a k x n matrix, then {v1,...,v_k} is a linearly independent set in Rⁿ

Answer 1

B*v1, ..., B*vk linearly independent means that for a set of scalars c1, c2, ..., ck,

 

c1*(B*v1) + ... + ck*(B*vk) = 0 (the zero vector) if and only if c1 = c2 = ... = ck = 0 (the scalar zero)

 

For a set of scalars d1, d2, ..., dk, let d1*v1 + ... + dk*vk = 0.  We need to show d1 = ... = dk = 0 in order to show the vi vectors are linearly independent.

 

Multiply both sides by matrix B and distribute:

 

B*( d1*v1 + ... + dk*vk) = B*0 = 0

 

Then

 

d1*(B*v1) + ... + dk*(B*vk) = 0

 

But this expression is a linear combination of the linearly independent vectors B*v1, B*v2, ..., B*vk,

so to give 0, all the coefficients must be 0, i.e. d1 = d2 = ... = dk = 0.

 

Therefore, the vi are linearly independent.  They must also be in Rⁿ because of the rules of matrix multiplication.