$A$ is a rectangular matrix, and $(v_i)_1^k$ forms a basis of the rowspace. Is $(Av_i)_1^k$ linearly independent?

$$A$$ is a rectangular matrix of size $$m\times n$$ with real entries, and $$\{v_1,v_2,\ldots,v_k\}$$ forms a basis of the row space of $$A$$. Is the set $$\{Av_1,Av_2,\ldots,Av_k\}$$ linearly independent?

So we have that $$\{v_1,\ldots,v_k\}$$ is a basis of row space of $$A$$. We've to check if $$\{Av_1,\ldots,Av_k\}$$ is a linearly independent set. I started with $$\sum_{i=1}^k\alpha_i Av_i =0 \stackrel{??}{\implies} \alpha_i = 0$$ but couldn't conclude anything about $$\alpha_i's$$. I also haven't found a counterexample to the result yet, so it may be true.

When $$A$$ is square and invertible, I could prove the claim.

Any hints to point me in the right direction?

you're working over reals so you have a nice symmetric bilinear form given by the dot product, which is positive definite.

since $$\{\mathbf v_1, \mathbf v_2,\ldots,\mathbf v_k\}$$ forms a basis, it is a linearly independent set.

$$\{A\mathbf v_1, A\mathbf v_2,\ldots,A\mathbf v_k\}$$ is linearly dependent
$$\implies \mathbf 0 = \sum_{r=1}^k \alpha_r A\mathbf v_r = A\big(\sum_{r=1}^k \alpha_r \mathbf v_r\big)$$
where at least some $$\alpha_j\neq 0 \implies \mathbf w :=\big(\sum_{r=1}^k \alpha_r \mathbf v_r\big) \neq \mathbf 0$$ since $$\mathbf v_r$$ are linearly independent.,
$$\implies \mathbf w \in \ker A$$ but there must exist some x such that $$\mathbf x^T A=\mathbf w^T$$ giving
$$0 = \mathbf x^T\mathbf 0= \mathbf x^T \big(A\mathbf w\big) = \big(\mathbf x^T A\big)\mathbf w = \mathbf w^T \mathbf w \gt 0$$
(i.e. to further expand the logic $$\mathbf w^T\in \text{row space }A \iff \mathbf w\in \text{image }A^T$$ which it is since it is written as a linear combination of basis vectors for such space, which means there is at least one vector in pre-image, i.e. some $$\mathbf x$$ satisfying $$A^T\mathbf x = \mathbf w$$.)