Let $V$ be a vector space and $k$ a nonnegative integer.
Let $T$ be a $k$-linear function from $V^k$ to $\mathbb R$, where $k$-linearity means that $T$ is linear in its $i^{th}$ variables for $i=1,2, ... , k$. (Such a $T$ is just called a $k$-tensor.)
Now suppose $T$ is a decomposable $k$-tensor, by which we mean $\exists \ell_i\in V^*$ for $i=1,2, ..., k$, s.t. $T(v_1,..., v_k)=\ell_1(v_1)\cdots \ell_k(v_k)$, where $V^*$ means the space of linear functions from $V$ to $\mathbb R$.
We define an operator in this way: Let $v\in V$ and $T$ be a $k$-tensor. Define a $(k-1)$-tensor $$\iota _v T(v_1,...,v_{k-1}):=\sum_{r=1}^k (-1)^{r-1} T(v_1,...,v_{r-1},v,v_r,...,v_{k-1})\,.$$
Now there is a property saying that, if $k=1$, then for every $v\in V$ and every decomposable $k$-tensor $T$, $\iota_v(\iota_v T)=0$ is trivially true.
I cannot really see the triviality.
Since by definition every $0$-tensor is a real number and we know the operator $\iota_v$ sends a $k$-tensor to a $(k-1)$-tensor and now $k=1$, we can infer that $\iota_v T$ is a real number as well as a $0$-tensor. But then, $\iota_v$ sends $\iota_v T$ to a $(-1)$-tensor, which I don't know what is and is not even defined in my textbook.
My question is that, how to derive that if $k=1$, then for every $v\in V$ and every decomposable $k$-tensor $T$, $\iota_v(\iota_v T)=0$?
Edit: In the case that $k=1$, the decomposable condition can be ignored because every $1$-tensor is automatically decomposable. I introduced this condition was because of the general case in the context in my textbook.
As @Vries commented, if we use the convention that empty summation is $0$, then everything is clear. A reference can be found on this Wiki page.
I confused myself with another similar convention. There is another convention saying that, the infimum of the empty is considered as $+\infty$(c.f. the footnote on p. $350$, Real Analysis, $4^{th}$ edition, Royden).
And actually as @Vries commented, we can always alter our definition of empty summation such that the base case(the case that $k=1$ is a base case in an induction process) is true, though if so, we may encounter inconsistency when we go further and cross with other subfields in math. But luckily, the usual convention works here and we still have the beautiful consistency.