In my math textbook, I found the following problem.
Given an $n$-dimensional vector space $V$ over a field $\mathbb{K},$ consider a nonzero map $f : V \times V \times \cdots \times V \rightarrow \mathbb{K}$ that is multilinear map for all integers $k \geq 2,$ where $k$ is the number of copies of $V$ in the product.
a.) What values can $k$ take?
b.) Prove that $f$ is not $\mathbb K$-linear.
What I'm generally confused about is what the problems ask.
If I pick a basis $B = \{v_1, \dots, v_n\}$ for each copy of $V,$ then I get $f(v_{1_{N_1}}, \dots, v_{k_{N_k}}) = C_{N_1 N_2 \cdots N_k}$ is a constant for one of the $n^k$ possible $k$-tuples of basis vectors, where $N_1, N_2, \dots, N_k \in \{1, 2, \dots ,n\}.$
So, as for a.), can't $k$ just take any value in $\mathbb{N}$?
As for b.), this is a multilinear map, which means it's linear in each argument separately, so is it not linear by virtue of having more than one argument?
a) Yes, $k$ can be any positive integer.
On the other hand, if $f$ is also assumed to be alternating (gives $0$ whenever two of its arguments are equal), then $k$ must be at most $n=\dim V$.
b) We have $f(2v_1,2v_2,\dots)=2^kf(v_1,v_2,\dots)$ whereas if $f$ were linear, we should get $f(2v_1,2v_2,\dots)=2f(v_1,v_2,\dots)$.
At least for the case when characteristic of $\Bbb K$ is $0$, this proves that $k$ must be $1$ (or $f$ must be constant $0$).