Multilinear Forms as a Vector Space

Question

if we want to prove that the collection of all multilinear forms is a vector space over $F$, I am having some trouble wrapping my head around some fundamental concepts.

By definition, for some $f:V^k\to F$ which is multilinear, we have $c\cdot f(v_1,...,v_i,...,v_k)=f(v_1,...,c\cdot v_i,...,v_k)$ and $f(v_1,...,v_i,...,v_k)+f(v_1,...,v_i',...,v_k)=f(v_1,...,v_i+v_i',...,v_k)$ - from the link above.

1. From the first property doesn't this imply that $c\cdot f(v_1,...,v_i,...,v_k) = f(c\cdot v_1,...,v_i,...,v_k)=...=f(v_1,...,c\cdot v_i,...,v_k)=...=f(v_1,...,v_i,...,c\cdot v_k)$. There is no uniqueness to which specific $v_i$ the scalar is applied to?
2. To prove the collection of all such $f$, $A=\{f\}$ is a vector space, we need to prove that it is closed for addition and scalar multiplication, so for $f,f'\in A$, $(f+f')(v_1,...,v_k)$ is defined as $f(v_1,...,v_k)+f'(v_1,...,v_k)$. How exactly am I supposed to show that the sum of these two is a multilinear form?

My attempt to show that this is a vector space is for some $c\in F$, $c\cdot(f+f')(v_1,...,v_k)=(f+f')(v_1,...,c\cdot v_i,...,v_k)=f(v_1,...,cv_i,...,v_k)+f'(v_1,...,cv_i,...,v_k)=cf(v_1,...,v_k)+cf'(v_1,...,v_k)$.

This would be what's needed to show that it is a vector space but isn't this wrong because it assumes the definition of the sum of two multilinear functions is multilinear?

Any help is appreciated, thank you.

Best, Adam

Answer 1

The answer to your first question is: yes, the $c$ can be pulled into any of the arguments, since multilinear means that the map is linear in every argument.

For the seconds question, you indeed have to check that for multilinear $f,g\colon V^n\to F$ and some $c\in F$ both $f+g$ and $c\cdot f$ are multilinear again. We can just show that in general $c\cdot f+g$ is multilinear which yields the other two by choosing either $c=1$ or $g=0$. Note that for checking linearity we can use the same trick, when $f(\lambda\cdot v+w)=\lambda\cdot f(v)+f(w)$, we have additivity by choosing $\lambda=0$ and homogeneity by choosing $w=0$.

By definition of $c\cdot f+g$ we have \begin{align*} &(c\cdot f+g)(v_1,\dots,\lambda\cdot v_i+v_i',\dots,v_n) \\&\quad= c\cdot f(v_1,\dots,\lambda\cdot v_i+v_i',\dots,v_n) + g(v_1,\dots,\lambda\cdot v_i+v_i',\dots,v_n) \\ \end{align*} Now both $f$ and $g$ are multilinear, hence this is equal to \begin{align*} &= c\cdot\left(\lambda\cdot f(v_1,\dots,v_i,\dots,v_n) + f(v_1,\dots,v_i',\dots,v_n) \right)\\&\quad{}+ \lambda\cdot g(v_1,\dots,v_i,\dots,v_n) + g(v_1,\dots,v_i',\dots,v_n) \\ &= \lambda\cdot \left(c\cdot f(v_1,\dots,v_i,\dots,v_n) + g(v_1,\dots,v_i,\dots,v_n)\right) \\&\quad{}+ c\cdot f(v_1,\dots,v_i',\dots,v_n) + g(v_1,\dots,v_i',\dots,v_n) \end{align*} Using the definition of $c\cdot f+g$ again, we obtain \begin{align*} &= \lambda\cdot (c\cdot f+g)(v_1,\dots,v_i,\dots,v_n) + (c\cdot f+g)(v_1,\dots,v_i',\dots,v_n). \end{align*} Hence, we have shown that $c\cdot f+g$ is linear in the $i$th argument. Since $i$ was arbitrary this shows that $c\cdot f+g$ is indeed multilinear.

Answer 2

For 1, you are right that $c$ can multiply any of the $v_i$, and there is no contradiction. This just shows the action of $c\in F$ could be represented in different ways.

For 2, if $f,f'\in A$, by definition \begin{align} (f+f')&(v_1, \dots,v_i+v_i',\dots,v_k) = f(v_1,\dots,v_i+v_i',\dots,v_k) + f'(v_1,\dots,v_i+v_i',\dots,v_k) \\ & = f(v_1,\dots,v_i,\dots,v_k) + f(v_1,\dots,v_i',\dots,v_k) \\ & \hspace{0.5cm} + f'(v_1,\dots,v_i,\dots,v_k) + f'(v_1,\dots,v_i',\dots,v_k) \quad\text{by multilinearity of $f,f'$} \\ & = (f+f')(v_1,\dots,v_i,\dots,v_k) + (f+f')(v_1,\dots,v_i',\dots,v_k). \end{align}

\begin{align} c\cdot (f+f')(v_1, \dots,v_i,\dots,v_k) & = c\cdot [f(v_1, \dots,v_i,\dots,v_k)+f'(v_1, \dots,v_i,\dots,v_k)] \\ & = c\cdot f(v_1, \dots,v_i,\dots,v_k) + c\cdot f'(v_1, \dots,v_i,\dots,v_k)\\ & = f(v_1,\dots,cv_i,\dots,v_k) + f'(v_1,\dots,cv_i,\dots,v_k) \\ & = (f+f')(v_1,\dots,cv_i,\dots,v_k) \end{align}

These two together prove $f+f'$ is a multilinear form and $A$ is closed under addtion. It is very similar to show $A$ is closed under scalar multiplication. It follows the same process of using the definition, expand by multilinearity, and conclude.

Your attempt after 2 shows the action of $F$ is distributive: $c(f+f')=cf+cf'$. This is part of the definition of $F$-action, and I don't think you have to show it.