If so, what is vector addition on this space? Is it the usual pointwise addition of functions?

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    $\begingroup$ How is it possible to speak of a vector space without knowing how the operations are defined? $\endgroup$ – Azif00 Jun 2 at 22:58
  • $\begingroup$ I don't understand your point. I know that the vector space $V$ over the field $F$ has a vector addition and a scalar multiplication that satisfy the axioms of a vector space. Considering the set of all multilinear forms on $V$, the question is whether this set is a vector space, and what type of operations would be defined on the set. I'm not claiming that the set of multilinear forms on $V$ is a vector space, so I don't have to know how the operations are defined. That's the whole point of my question. $\endgroup$ – Silent Bat Jun 2 at 23:10

Yes, the "usual pointwise" operations make it into a vector space. More precisely (and slightly more generally), if $V_1, \dots, V_k$ are vector spaces over a common base field $F$, and $W$ is the set of all multilinear maps $V_1 \times \dots V_k \to F$, then define addition on $W$ by:

For every $\omega_1, \omega_2 \in W$ and all $(v_1, \dots, v_k) \in V_1 \times \dots \times V_k$, \begin{align} (\omega_1 + \omega_2)(v_1, \dots, v_k) := \omega_1(v_1, \dots, v_k) + \omega_2(v_1, \dots, v_k) \end{align}

Also, define scalar multiplication by

For every $c \in F$ and every $\omega \in W$ and every $(v_1, \dots, v_k) \in V_1 \times \dots V_k$, \begin{align} (c \cdot \omega)(v_1, \dots, v_k) &:= c \cdot \left[ \omega(v_1, \dots, v_k)\right] \end{align}

Next, you should verify that $\omega_1 + \omega_2$ and $c \cdot \omega$ are indeed still multilinear, so that the above operations are well defined. Then, $(W, +, \cdot, F)$ with the operations as defined above forms a vector space over $F$ (I leave it to you to verify the relevant axioms).

As a special case, you can choose $V_1 = \dots = V_k = V$ to all be the same vector space, so this is the situation you actually asked about.

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  • $\begingroup$ Also, I intentionally didn't clutter up the notation with things like $+_W, +_F$ etc to explicitly mention that the various $+$ and $\cdot$ appearing on the two sides of the equation mean different things, but if this point is unclear to you, let me know. $\endgroup$ – peek-a-boo Jun 2 at 23:22
  • $\begingroup$ Thank you very much for you elaborate answer! This is exactly what I was looking for. Very understandable. $\endgroup$ – Silent Bat Jun 3 at 0:41

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