A question about multivariable functions and bilinear maps

Question

How can a bilinear map can be considered to be linear in each of its argument s when there is only one argument to the function $f:X\times X\rightarrow Y $ and that is an element of the set $X\times X$. I can't quite extend the notion of homomorphism $f:X\rightarrow Y$ where $f$ accepts arguments from the vector space only, not an element from the cartesian product of two vector spaces. I understand the definition of the latter, but I can't extend it when it comes to accepting an argument from $X\times X$.

Answer 1

Let $X$ and $Y$ be vector spaces over a field $F$. To say that a function $f:X \times X \to Y$ is "linear in each of its arguments" means that the functions $x \mapsto f(x,v)$ and $x \mapsto f(v,x)$ are linear for all $v \in X$.

If you don't like the phrase "linear in each of its arguments", you could use the term "bilinear" instead. Perhaps that would eliminate the confusion.

Answer 2

If you really don't like working with bilinear functions you could avoid them altogether (sort of).

Let $V, W, Z$ be vector spaces. A bilinear map $f : V \times W \to Z$ corresponds to a linear map $\tilde{f} : V \otimes W\to Z$, where $V \otimes W$ is the tensor product of vector spaces $V$ and $W$:

Let $\{e_i\}_{i\in I}$ be a basis for $V$, and $\{f_j\}_{j\in J}$ be a basis for $W$.

Then $V \otimes W := \mathrm{span}\{e_i\otimes f_j : i \in I, j \in J\}$.

$v\otimes w$ is pretty much just a fancy notation for $(v, w)$, except that the operations $+$ and $\cdot$ do not obey the classic properties you would have for $X \oplus Y$:

$$(v_1, w_1) + (v_2, w_2) = (v_1+v_2, w_1+w_2)$$ $$\lambda (v_1, w_1) = (\lambda v_1, \lambda v_2)$$

For these objects, however, we have:

$$(v_1, w) + (v_2, w) = (v_1+v_2, w)$$ $$(v, w_1) + (v, w_2) = (v, w_1+w_2)$$ $$\lambda (v, w) = (\lambda v, w) = (v, \lambda w)$$

In other words, the operation $\otimes : X \times Y \to X \otimes Y$ is bilinear.

Now for a bilinear map $f : V \times W \to Z$, define $\tilde{f} : V \otimes W \to Z$ on its basis:

$$\hat{f}(e_i \otimes f_j) := f(e_i, f_j)$$

Lo and behold, $\tilde{f}$ is a linear map.

To answer your question, your bilinear map $f : X \times X \to Y$ could be interpreted as a linear function on $X \otimes X$, not $X \times X$.