# Imaginary part of a Hermitian form is non-degenerate

I'm stuck on one part of a proof, and any help would be greatly appreciated!

Let $\omega:V \times V \to \Bbb{C}$ be a positive definite Hermitian form. Define $\omega_r: V \times V \to \Bbb{C}$ and $\omega_i:V \times V \to \Bbb{C}$ by $\omega(v,w) = \omega_r(v,w) + i\omega_i(v,w)$. Prove that $\omega_i$ is non-degenerate.

The parts I already have done are showing $\omega_r$ is positive definite and symmetric, and $\omega_i$ is skew-symmetric, but I can't figure out how to prove the non-degeneracy of $\omega_i$

• Consider $\omega(v, iv)$. – Petr Naryshkin Oct 12 '17 at 19:46
• So $\omega(v,iv)= \omega_r(v,iv) + i\omega_i(v,iv)$, Looking at $\omega_i$, we have $\omega_i(v,iv) = i\omega_i(v,v) = 0$, since the skew-symmetry here guarantees that $\omega_i$ is alternating as well ($char(\Bbb{F})$ is not 2)? Is that right, or did I make a mistake? – Joe Oct 12 '17 at 20:12
• no, $\omega_i(v, iv) \ne i\omega_i(v, v)$. $\omega_i$ is not a Hermitian form with respect to complex numbers, only bilinear with respect to real numbers. But $\omega$ is and writing this identity gives us $\omega_r(v, iv) + i\omega_i(v, iv) = -i\omega(v, v) = -i(\omega_r(v, v) + i\omega_i(v, v))$. – Petr Naryshkin Oct 12 '17 at 20:19
• From the last equation (the first part) it follows that $\omega_i(v, iv) \ne 0$. – Petr Naryshkin Oct 12 '17 at 20:27
• Oh I see. So my mistake was treating a bilinear form as if it were sesquilinear. Thank you very much! – Joe Oct 12 '17 at 20:34