Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

let $f\colon \mathbb R^{2}\times\mathbb R^{2} \to \mathbb R$ be a bilinear map,i.e,linear in each variable separately.Then for $(V,W)$ in $\mathbb R^{2}\times \mathbb R^{2}$,the derivative $Df(V,W)$ evaluated on $(H,K)$ in $\mathbb R^{2}\times \mathbb R^{2}$ is given by

share|cite|improve this question
Please: use LaTeX to write mathematics in this site! You can go the FAQ section and follow directions. – DonAntonio Dec 19 '12 at 9:01

1 Answer 1

By bilinearity we have $$f(V+H,W+K)=f(V,W)+f(V,K)+f(H,W)+f(H,K)\ .$$ Now $f(H,K)$ can be estimated as $$|f(H,K)|\leq |H|\>|K|\ \max_{X,Y\in S^1}f(X,Y)\leq c\bigl(|H|^2+|K|^2\bigr)\ .$$ Therefore we have $$\eqalign{f(V+H,W+K)-f(V,W)&=f(V,K)+f(H,W)+o\left(\sqrt{|H|^2+|K|^2}\right)\cr & \qquad\bigl((H,K)\to(0,0)\bigl)\ ,\cr}$$ and this is saying that $$Df(V,W).(H,K)=f(V,K)+f(H,W)\ .$$

share|cite|improve this answer
Christian Blatter..great answer as always!! – mathlover Nov 23 at 5:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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