Let $W$ be a finite dimensional $K$ vector space and $W^*$ its dual space. For $V := W \oplus W^*$ the mapping

$$ V \times V \to K,((a, \varphi),(b, \psi)) \mapsto \langle(a, \varphi),(b, \psi) \rangle := \varphi(b)-\psi(a) $$

is given. Show, that $\langle , \rangle$ is a skew-symmetric, non-degenerate bilinear form.

I have tried to show it's a bilinear form but I'm stuck in rearanging, I'm not even sure I'm on the right track:

$$ \begin{align} ((\lambda a_1 + \mu a_2, \varphi),(b, \psi) ) \mapsto \langle(\lambda a_1 + \mu a_2, \varphi),(b, \psi)\rangle \\ :=\varphi(b)-\psi(\lambda a_1 + \mu a_2) \\ = \varphi(b)-\lambda\mu\psi(a_1 + a_2) \\ = ..... \\ = \lambda\mu\varphi(b) - \lambda\mu\psi(a_1 + a_2) \\ = \lambda\mu(\varphi(b)-\psi(a_1 + a_2)) \\ = \lambda (\varphi(b)-\psi(a_1)) + \mu(\varphi(b)-\psi(a_2)) \\ = \lambda\langle(a_1, \varphi),(b,\psi)\rangle + \mu\langle(a_2, \varphi),(b,\psi)\rangle \end{align} $$

I need help showing it's bilinear and non-degenerate


1 Answer 1



For the bi-linearity prove

$$\langle(\lambda a_1+\mu a_2, \lambda\varphi_1+\mu \varphi_2),(b, \psi) \rangle=\lambda\langle( a_1, \varphi_1),(b, \psi) \rangle+\mu\langle( a_2,\varphi_2),(b, \psi) \rangle$$

and the same thing for the second component. It's easy to see that the map is skew-symmetric.

Finally let's prove that this bilinear form (we denote it $\Phi$) is non degenerate: Recall that $$\ker\Phi=\{(a,\varphi)\;|\; \Phi((a, \varphi),(b, \psi))=0\;\forall (b, \psi)\in W\times W^*\}$$ so let $(a,\varphi)\in\ker\Phi$ then for $\psi=0^*$ we have $$\varphi(b)=0,\quad\forall b\in W\Rightarrow \varphi=0^*$$ and for $b=0$ we have $$\psi(a)=0,\quad \forall \psi\in W^*\Rightarrow a=0$$ hence we proved that $$\ker\Phi=\{(0,0^*)\}$$ which allows us to conclude.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .