$A$ 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.

Suppose for contradiction
$\{A\mathbf v_1, A\mathbf v_2,\ldots,A\mathbf v_k\}$ 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$
which is a contradiction.

(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 $A^T\mathbf x = \mathbf w$.)


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