Derivative is an alternating 1-tensor?

Question

I am reading through spivak, and he states, if $f: \mathbb{R^n} \to \mathbb{R}$ is differentiable, then $Df(p): \mathbb{R^n} \to \mathbb{R}$, and since this is linear, we have that $Df(p) \in \Lambda^1(\mathbb{R^n})$. I don't follow this exactly, I know that since it is linear, $Df(p) \in \mathcal{J}^1(\mathbb{R^n})$, but I don't see how it is alternating.

Answer 1

If $T\in \mathcal J^k(V)$ is a $k$-tensor, then it is a function that takes as input $k$ vectors $v_1,\ldots,v_k\in V$ and gives back a number $r\in \Bbb R$. We say that $T$ is alternating if for any such vectors and any $i,j$ with $1\leq i<j\leq k$ we have $$ T(v_1,\ldots,v_i,\ldots,v_j,\ldots,v_k)=-T(v_1,\ldots,v_j,\ldots,v_i,\ldots,v_k) $$Now if $f$ is a $1$-tensor, then it fulfills the alternating property vacuously: in order for $f$ to not be alternating, there must be $i,j$ with corresponding $v_i$ and $v_j$ that makes a "witness" to $f$ not alternating. But we can't have $1\leq i<j\leq 1$. Therefore there is no witness, and $f$ is alternating.