# Prove that if $V \cap \text{Span}\{u_1, u_2,…,u_k\} = \{ 0\}$, then $Au_1, Au_2,…,Au_k$ are linearly independent

Let $u_1, u_2,... , u_k$ be linearly independent vectors in $\Bbb R^n$.

Let $A$ be an $m × n$ matrix and $V$ the solution space of the homogeneous linear system $Ax = 0$

Prove that if$V \cap \text{Span}\{u_1, u_2,...,u_k\} = \{ 0\}$, then $Au_1, Au_2,...,Au_k$ are linearly independent.

I have an idea that to prove linearly independence of $Au_1, Au_2, ..., Au_k$ I have to show that the equation $c_1Au_1, c_2Au_2, ..., c_kAu_k = 0, c_1, c_2,...,c_k \in \Bbb R$ , only has the trivial solution $c_1=c_2=...=c_k=0$

any help is appreciated!

Hint: Write $c_1Au_1 + \dots + c_kAu_k = A(c_1u_1 + \dots + c_ku_k) = 0$ and conclude using that $V \cap \text{span}\{u_1,\dots,u_k\} = \{0\}$.
• I think a "$=0$" is missing in your first equation. – Marc van Leeuwen Oct 8 '16 at 9:03