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I'm tripping over something elementary:

Suppose $f:\mathbb{R^2}\rightarrow X$ is linear, then $f(x+y)=f(x)+f(y)$ for all vectors $x$ and $y$. Now suppose that $f$ is also bilinear and in particular linear in the first argument, then

$$f\Big((a+b,c)\Big)= f\Big((a,c)\Big)+f\Big((b,c)\Big)$$

But now setting $x=(a,0)$ and $y=(b,c)$ contradicts linearity of f.

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migrated from mathematica.stackexchange.com Feb 11 '16 at 15:39

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a function cannot be linear and bilinear at the same time. Bilinearity is a kind of non linearity.

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Bilinear is nonlinear. It's linear in both main variables, but not in any superposition. Naively speaking, it's linear if you cut along $x$ or $y$ axis, but you're not allowed to rotate the frame (which is what a proper linear function allows, even requires, as linearity is independent of choice of coordinates). A typical example is bilinear interpolation (splines & such), which puts a smooth interpolation over a rectangular frame (a classic problem that a 3D quadrilateral is almost never planar).

Example slide

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A (non-trivial) linear function from $\mathbb{R}^n$ for $n > 1$ will never be a multilinear function (bilinear when $n=2$) if it is a linear function. Instead, a multilinear function is a linear function from a tensor product. So a bilinear function on $\mathbb{R}$ is a linear function $\mathbb{R}\otimes\mathbb{R} \to X$, but $\mathbb{R}$ is the unit for the tensor product, so $\mathbb{R}\otimes\mathbb{R} \cong \mathbb{R}$ which is very much not $\mathbb{R}^2$.

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