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Let $V$ be a finite-dimensional vector space and $\beta$ be a bilinear form. According to the definition of the bilinear forms is possible to do the following: $$\begin{align}\beta(u,v)&=\beta(u+w-w,v)=\beta(u+w,v)-\beta(w,v)\\ &=\beta(u+w,v)+\beta(w,-v)=\beta(u+2w,v-v)\\ &= \beta(u+2w,0)=0\end{align}$$

I find it curious, I think there's some mistake but I can not identify it, I ask for your help in this matter.

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    $\begingroup$ $\beta(a+b,c+d)=\beta(a,c)+\beta(b,d)$ doesn't hold. $\beta(a+b,c+d)=\beta(a,c)+\beta(a,d)+\beta(b,c)+\beta(b,d)$ does. $\endgroup$
    – user65203
    Sep 2, 2016 at 16:27
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    $\begingroup$ Note: this is a valid proof that all bilinear, linear functions are 0. $\endgroup$ Sep 2, 2016 at 21:28

2 Answers 2

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The step $$\beta(u+w,v) + \beta(w,-v) = \beta(u+2w,v-v)$$is wrong. You must have one of the arguments fixed in both terms in order to apply linearity in the other one. That is, $\beta$ being bilinear in $u$ and $v$ is not the same as being linear in the pair $(u,v)$. For example, the multiplication in $\Bbb R$ is bilinear. You'd be saying that $$(u+w)v + w(-v) = (u+2w)(v-v)$$for all $u,v,w \in \Bbb R$, which is false (take $u=v=w=1$, say).

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If the function $\gamma\colon V\times V\to V$ satisfies \begin{align} &\gamma(u,v)+\gamma(x,y)=\gamma(u+x,v+y),\tag{1} \\ &\gamma(au,v)=a\gamma(u,v),\tag{2} \\ &\gamma(u,av)=a\gamma(u,v),\tag{3} \end{align} for all $u,v,x,y\in V$ and all scalars $a$, then $\gamma(u,v)=0$ for all $u,v$.

Indeed, $\gamma(u,0)=\gamma(u,0\cdot0)=0\gamma(u,0)=0$ and similarly $\gamma(0,v)=0$; therefore $$ \gamma(u,v)=\gamma(u,0)+\gamma(0,v)=0+0=0 $$

Note that this is basically the statement you proved. However, for a bilinear form property $(1)$ is not required. Rather, it is required that $$ \beta(u,x+y)=\beta(u,x)+\beta(u,y),\qquad \beta(u+v,x)=\beta(u,x)+\beta(v,x) $$ which is very different from the other property.

Are there maps $f\colon V\times V\to V$ that satisfy property $(1)$? Yes, for instance all linear maps $V\oplus V\to V$ do. However, they don't satisfy $(2)$ and $(3)$, but only $$ f(au,av)=af(u,v) $$

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