Find radical of hermitian form $\langle , \rangle _A$ 
Determine the radical of the hermitian form $\langle , \rangle _A$ over the field $\mathbb{C}^3$, where $$A = \begin{pmatrix} 1 & -i & -i \\ i & 2 & 1 \\ i & 1 & 1 \end{pmatrix}$$

Would it be sufficient to calulate the kernel of the matrix $A$ or is there another way to do this?

I suppose you define $\langle , \rangle _A$ as $\langle \vec x , \vec y \rangle _A = \vec x^TA\overline{\vec y}  $. I emphasize this as it is sometimes (particularly in physics) defined as $\langle \vec x , \vec y \rangle _A = \vec x^HA\vec y$. 
The (left) radical is defined as $L_A=\{\vec x \in \mathbb C^3 | (\forall \vec y \in \mathbb C^3)\langle \vec x, \vec y \rangle = \overline {\langle \vec y, \vec x \rangle} = \overline {\vec y^TA} \vec x=0  \}$. This means that $\overline{A}{\vec x}=0$. Therefore, we are looking for $\ker{\overline{A}} $.
The right radical $R_A=\{\vec y \in \mathbb C^3 | (\forall \vec x \in \mathbb C^3)\langle \vec x, \vec y \rangle = \vec x^T A \overline {\vec y} =0  \}=L_A$ in the case of hermitian form.
