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How can I prove that covectors $\omega^1, ..., \omega^k$ are linearly independent iff their wedge product $\omega^1\wedge ...\wedge \omega^k$ is not zero?

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"If part:" Suppose that $\omega^1\wedge ...\wedge \omega^k$ is not zero. Suppose that $a_i\in\mathbb{R}$ where $1\leq i\leq k$ such that $\tag{1}a_1\omega^1+\cdots+a_k\omega^k=0,$ the zero covector. Take the wedge product with $\omega^1\wedge \omega^2\wedge\cdots\wedge \omega^{i-1}\wedge\omega^{i+1}\wedge\cdots\wedge\omega^k$ with $(1)$, we get $a_i\omega_i\wedge\omega^1\wedge \omega^2\wedge\cdots\wedge \omega^{i-1}\wedge\omega^{i+1}\wedge\cdots\wedge\omega^k=(-1)^{i-1}a_i\omega^1\wedge ...\wedge \omega^k=0,$ where $1\leq i\leq k$. Therefore, if $\omega^1\wedge ...\wedge \omega^k\neq 0$, we must have $a_i=0$ for $1\leq i\leq k$. By definition, $\omega^1, ..., \omega^k$ are linearly independent.

After typing the "if part", I find that I forgot about the proof of "only if part". Hope that someone can give the full proof.

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    for the only if part: if they are independent, extend them to a basis, take $X_i$'s ($i=1\dots n$) to be the dual basis, and compute $\omega^1, ..., \omega^k(X_1,X_2,...,X_k)=1$ as you suggest2011-12-30