**Lemma**. Suppose $\tau_1,\cdots,\tau_n$ are linear functionals on a linear space $V$, then they are linear dependent if and only if

\begin{align*}

\det(\tau_i(v_j))=0 \forall v_1,\cdots, v_n\in V.

\end{align*}

*Proof*. We need only prove the ONLY IF part.

If $\tau_1,\cdots,\tau_n$ are linearly independent, by [this post][1], there is some $v\in V$ such that

$$v\in (\bigcap_{i=1}^{n-1}\ker\tau_i)\backslash \ker \tau_n.$$

$\newcommand{diag}{\operatorname{diag}}$

Fix $v_n=v$, then $$\det(\tau_i(v_j))_{1\leq i,j\leq n-1}=\tau_n(v_n)^{-1}\det(\tau_i(v_j))_{1\leq i,j\leq n}=0.$$

Apply induction.

[1]: https://math.liveadvances.com/c-ding/equivalent-condition-of-linear-dependence-of-linear-functionals/