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