-1
$\begingroup$

Let $E$ and $F$ be vector spaces. The a mapping $\phi : E\times F \to \Gamma $ satisfying $$\phi(\lambda x_1 + \mu x_2, y) = \lambda \phi (x_1, y) + \mu \phi (x_2,y),$$ $$\phi(x,\lambda y_1 + \mu y_2) = \lambda \phi (x,y_1) + \mu \phi (x,y_2),$$ is called a bilinear function in $E \times F$.

From this definition of bilinear function, one can conclude that \begin{align} \phi(x,y) &= \phi(x+x-x, y)\\ & = \phi (x,y) + \phi (x,y) + \phi (-x,y)\\ & = \phi (x,y) + \phi (x,y) + (-1) \phi (x,y)\\ & = \phi (x,y) + \phi (x,y) + \phi (x,-y)\\ & = \phi (x+x+x,y). \end{align}

Since I can also write $x = x + x - x + x - x\dotsb $ infinitely many times, a bilinear function has to satisfy $\phi (x,y) = \phi (x+x+x+\dotsb, y)$ regardless of how I expressed $x$, and this really made me uncomfortable.

My question is, first of all, is this really the case with bilinear functions, i.e is my conclusion correct. Secondly, isn't this causing any problem in defining a map with the given condition. I mean, as I've said, this kind of freedom really bothers me, so I'm looking for an explanation to satisfy this conditions while defining a bilinear function.

$\endgroup$
  • $\begingroup$ How did you go from $\phi (x,y) + \phi (x,y) + \phi (-x,y) = \phi (x,y) + \phi (x,y) + \phi (x,-y)$? $\endgroup$ – user370967 Jun 3 '17 at 12:21
  • $\begingroup$ @Math_QED See my edit.It comes from the first property of the definition. $\endgroup$ – onurcanbektas Jun 3 '17 at 12:22
  • 1
    $\begingroup$ @Math_QED I think that would be via $\phi(-x,y) = -\phi(x,y) = \phi(x,-y)$. $\endgroup$ – SvanN Jun 3 '17 at 12:22
  • $\begingroup$ Okay I just realised $\endgroup$ – user370967 Jun 3 '17 at 12:22
  • $\begingroup$ But $\phi (x,y) + \phi (x,y) + \phi (x,-y)= \phi (x+x+x,y)$. How does that hold? $\endgroup$ – user370967 Jun 3 '17 at 12:22
4
$\begingroup$

The error is in the step $\phi(x, y) + \phi(x,y) + \phi(x,-y)$; $\phi$ is a linear map in the second argument when holding the first fixed, so this actually sums to $\phi(x, y+y-y) = \phi(x,y)$.

$\endgroup$
  • $\begingroup$ You are right.Thanks. $\endgroup$ – onurcanbektas Jun 3 '17 at 12:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.