I would appreciate it if someone could unpack/explain the following definition. In particular I don't really understand why the function fixes all the vectors but the $\mathbf{u}$ vector is removed? Also does $\mathbf{u} \in V$ or is $\mathbf{u} \in V^n$?

Let $V$ be linear space. A function $F:V^n \rightarrow R $, where the cartesian product $V^n=V \times V \times ...V$ contains $n$ copies of $V$, is said to be an $n$-multilinear form on $V$ provided that the function $\mathbf{u}\mapsto F(\mathbf{u}_1,...,\mathbf{u}_{i-1},\mathbf{u},\mathbf{u}_{i+1},...\mathbf{u}_n)$ is a linear transformation $V\rightarrow R$ for every $i$ and every (n-1)-tuple $(\mathbf{u}_1,...,\mathbf{u}_{i-1},\mathbf{u}_{i+1},...\mathbf{u}_n)$. We say that $F$ is alternating if $F(\mathbf{u}_1,...,\mathbf{u}_n)=0$ whenever there exist an index $i$, $1\le i \le n-1$, such that $\mathbf{u}_i=\mathbf{u}_{i+1}$.

Hence, $F$ is an $n$-multilinear form if $$F(...,\mathbf{u}_i,s\mathbf{u}+t\mathbf{v},\mathbf{u}_{i+1},...)=sF(...,\mathbf{u}_i,\mathbf{u},\mathbf{u}_{i+1},...)+tF(...,\mathbf{u}_i,\mathbf{v},\mathbf{u}_{i+1},...)$$ adn alternating if $F(\mathbf{u}_1,...,\mathbf{u}_n)=0$ whenever two adjacent vectors are equal.

  • $\begingroup$ $\mathbf u$ lies in $V$. The function of $n$ vectors $F(\mathbf u_1, \dots, \mathbf u_n)$ has to be linear w.r.t. each $\mathbf u_i\in V)$, and it must vanish if any two $\mathbf u_i,\;\mathbf u_j$ are equal (in the definition,if any two consecutive vectors are equal, but it's easy to show this implies the vanishing of the function for any two equal vectors). $\endgroup$ – Bernard Feb 14 '17 at 14:37

To answer your question at the end of the first paragraph: $u \in V$.

For the rest:

If you have a function of three variables, like $$ f(x, y, z) = 2x + 3y + z $$ you can hold two of them fixed and make a new function like this; $$ g(t) = f(3, t, 5). $$ In this case, $g(t)$ turns out to be $$ g(t) = 2\cdot 3 + 3t + 5 = 11 + 3t. $$

Is that clear? I could have set $x$ and $z$ to any values I liked; whatever values I chose, I'd get a function $g$. That function $g$ might have various properties: it might be constant. It might be linear. It might be quadrtic. It might be differentiable. And these properties might vary depending on the values you chose for $x$ and $z$ (although not in this example).

OK. You've got a function $F$ of $n$ vector variables. If you hold all but the first one fixed at values $u_2, \ldots, u_n$, you get a new function $$ g(t) = F(t, u_2, u_3, \ldots, u_n) $$ (where $t$ here is a vector rather than a real number as in the previous example).

More precisely, for any choice of $u_2 \ldots u_n$, you get a function $g$. And for any such choice, you can look at $g$ and ask "Is this a linear function of $t$?"

You could do the same thing with the second slot, and define $$ g(t) = F(u_1, t, u_3, \ldots, u_n) $$ and ask, for each possible choice of $u_1, u_3, \ldots, u_n$, whether the resulting function was linear.

And you could do the same thing for each of the other slots. If the answer is always "yes", then we say that $F$ is multilinear.

As a very simple example, the function

$$ F(x, y) = 2xy $$ is a multilinear function on $\Bbb R \times \Bbb R$. You might want to check this claim carefully.

On the other hand, $F$ is not alternating, for $F(1, 1) = 2 \ne 0$. In fact, there's no alternating multilinear function on $\Bbb R \times \Bbb R$ except the constant function 0.

Post-comment addition

OP asks about "alternating"; in particular, if you have an $n$-linear function $F$ with the property that any time an argument is repeated, you get zero, why is it true that if you swap two arguments, the value negates?

To see this, I'm going to fix everything except $u_1$ and $u_2$, so I'll write $$ g(a, b) = F(a, b, u_3, \ldots, u_n) $$ and then look at the function $g$. You could do the same thing for any two arguments (the 3rd and 14th, say, or $i$th and $j$th), but making it really concrete like this helps.

Because $F$ is multilinear, we know that $g$ is bilinear, i.e., $$ g(a_1 + c a_2, b) = g(a_1, b) + c g(a_2, b) \\ g(a, b_1 + c b_2) = g(a, b_1) + c g(a, b_2). $$

And because $F$ has the "alternating" property, we know that $g(v, v) = 0$ for any vector $v$.

Suppose that $a$ and $b$ are any vectors, and let $v = a + b$. Let's see what $g(v, v) = 0$ tells us: \begin{align} 0 &= g(v, v) \\ &= g(a+b, a+b) \\ &= g(a, a+b) + g(b, a + b) \\ &= g(a, a) + g(a, b) + g(b, a) + g(b, b). \end{align} Now because $g(u,u) = 0$ for any $u$, we know the first and last terms are zero, so we have \begin{align} 0 &= g(a, a) + g(a, b) + g(b, a) + g(b, b)\\ &= 0 + g(a, b) + g(b, a) + 0\\ 0 &= g(a, b) + g(b, a) \end{align} from which it follows that $g(a, b) = - g(b, a)$.

So why do we define "alternating" to be "always gives zero when there's a repeated argument" instead of $g(a, b) = - g(b, a)$? Because the little proof I just gave isn't very difficult, and it's often easier to PROVE the "repeated args lead to zero" claim than the negation claim. Or maybe just because it once seemed clever to someone...I don't really know.

  • $\begingroup$ Thank you for your through reply John. Your explanation on fixing variables has also helped me better understand my multi-variable calculus class. I am still struggling with the definition though. What is it that causes the sign to alternate if $F(\mathbf{u}_1,...,\mathbf{u}_n)=0$. Is there a reason for it or just something we accept by definition? $\endgroup$ – Eiraus Feb 15 '17 at 7:35
  • $\begingroup$ Nah...that's just a bit of sneakiness. It's not that $F(u_1,\ldots, u_n) = 0$. The claim is that $F(\ldots, a, \ldots, b, \ldots) = -F(\ldots, b, \ldots, a, \ldots)$, where the dots indicate stuff that doesn't change. Let me add something to my answer.... $\endgroup$ – John Hughes Feb 15 '17 at 12:04

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