Prove you can delete a coordinate from a linearly independent set of vectors and still be linearly independent

I'm trying to prove the following claim:

Let $m, n \in \mathbb{N}, m < n$, and let $v_1, \ldots, v_m \in \mathbb{R}^n$ be linearly independent. For $k \in \{1, \ldots, n\}$, let $P_k : \mathbb{R}^q \twoheadrightarrow \mathbb{R} \mathbf{e}_k \subseteq \mathbb{R}^q$ be the orthogonal projection onto the $k$th coordinate. Let $Q_k = I - P_k$, i.e. the projection onto the orthogonal completement of $\mathbf{e}_k$, which deletes the $k$th coordinate. Then there exists $k \in \{1, \ldots, n \}$ such that the set $\{Q_k v_1, \ldots, Q_k v_m\}$ is linearly independent.

If I think about it geometrically, it seems like this should be true. I suspect the proof of this claim would involve some clever rank-nullity and transpose and codimension setups, but I'm just not seeing what those are. I'd appreciate help with this claim. Or, if it's false, I'd appreciate a counterexample.

Thanks!

Solution 1:

If $\{Q_k v_j\}_j$ are linearly dependent, there exist $\alpha_j\in\mathbb{R}$ such that $Q_k(\sum_j\alpha_j v_j) = 0$ and $\sum_j\alpha_j v_j\not=0$. It implies that $e_k$ can be written as a linear combination of the $v_j$s. This cannot be true for all $k$, otherwise $\{v_j\}_j$ would generate $\mathbb{R}^n$, contrary to the hypothesis $m < n$.