Linearity of Multilinear Maps

Question

If $f: \mathbb{R^n} \rightarrow \mathbb{R^m}$, with $n>1$, is a multilinear map, is $f$ linear? I think $f$ is only linear for the special case that the range of $f$ consists of a single element, $\vec 0$. Is this correct?

Answer 1

Multilinearity of $f$ means to be linear in each argument and makes only sense as extra property next to linearity for $n \ge 2$. Thus for arbitrary $a, b \in \mathbb{R}$ and $x, x' \in \mathbb{R}^n$. $$ f(\ldots, a x_i + b x_i', \ldots) = a f(\ldots, x_i, \ldots) + b f(\ldots, x_i', \ldots) $$ Now we apply it at once, for all arguments of $f$ $$ f(a x_1, \ldots, a x_n) = a f(x_1, a x_2, \dots a x_n) = \cdots = a^n f(x_1, \ldots, x_n) \quad (*) $$ (Note: This holds for GPerez's earlier counter example $\mbox{det}$, which we agree on is multilinear)

But for $f$ to be linear in the single argument vector $x = (x_1, \ldots, x_n)$ we want $$ f(a x) = a f(x) \quad (**) $$ If $f \ne 0$, this would already require $a^n = a$, which is not true for arbitrary $a$ in every field, like $\mathbb{R}$.

Proof:

$n > 1$ was given. Multilinearity, see equation $(*)$, gives $$ f(2x) = 2^n f(x) $$ Linearity requires $$ f(2x) = 2f(x) $$ thus $$ 0 = (2^n - 2) f(x) = 2(2^{n-1} - 1) f(x) = c f(x) $$ with $c \ne 0$. This implies $f(x) = 0$ for arbitrary $x \in \mathbb{R}^n$ thus $f = 0$.

Answer 2

No, consider the function $f : \mathbb{R},\mathbb{R} \rightarrow \mathbb{R}$ given by $f(x,y) = xy$. Then $f$ is clearly multilinear. Its also non-linear. To see this, observe that $f$ satisfies: $$f[a \cdot (x,y)] = f(ax,ay) = (ax)(ay) = a^2(xy) = a^2 f(x,y)$$ So it clearly doesn't satisfy the linearity condition that $f[a \cdot (x,y)] = a \cdot f(x,y)$.

In regards to the comment about your notation being ambiguous; I recommend the notation $f : \underbrace{\mathbb{R}, \ldots, \mathbb{R}}_n \rightarrow \mathbb{R}^m.$ This is standard notation from multicategory theory.