Proving linearly independent vectors

Question

Let $u$, $v$, $w$ be three linearly independent vectors in $\mathbb{R}^3$, and let $A$ be a non-singular $3\times3$ matrix. Then vectors $Au$, $Av$, $Aw$ are also linearly independent.

This is a True/False question, that needs to be proved as well.

Thanks!

Answer 1

It's true. Suppose $c_1Au + c_2Av + c_3Aw = 0$ for some scalars $c_1,c_2,c_3$. Then $A(c_1u + c_2v + c_3w) = 0$. Since $A$ is non-singular, this implies that $c_1u + c_2v + c_3w = 0$. As $\{u,v,w\}$ is linearly independent, it follows that $c_1=c_2=c_3=0$, and hence $\{Au, Av, Aw\}$ is linearly independent.

Answer 2

Pre-multiplying by a non-singular(invertible) matrix induces an isomorphism between $\mathbb R^3$ and $\mathbb R^3$. isomorphisms send linearly independent sets to linearly independent sets, so yeah. The new vectors are independent.