complex symmetric singular matrix equivalence I'm having trouble proving this equivalence:

Given $A$ a $3\times 3$ complex symmetric matrix. Let $q \in \mathbb{C}^3$ be such that $Aq\neq 0$ and $\langle Aq, q\rangle = 0$. Then, the following are equivalent:
  
  
*
  
*$\exists\; v\in \mathbb{C}^3$ linearly independent of $q$ such that $$\langle Aq, v\rangle = \langle Av, v \rangle = 0;$$
  
*$\det A=0$.

Here $\langle (v_1,v_2,v_3)^T, (u_1,u_2,u_3)^T\rangle = v_1u_1+v_2u_2+v_3u_3$.
Comment: $ 2 \implies 1$ is direct. Proving $ 1 \implies 2$ in $\mathbb{R}^3$ is easy since we have that $\langle v, v\rangle = 0 \iff v=0$ but this isn't true for $\mathbb{C}^3$ (take $(1,i,0)$). Any ideas? I'm guessing the symmetry of $A$ is crucial in the proof.
 A: Firstly, let's prove that $A$ is a symmetric operator: for all $x,y\in\mathbb{C}^3$ we have
$$\langle Ax, y\rangle = (Ax)^Ty = x^TA^Ty = x^TAy = \langle x, Ay \rangle$$ 
as $A^T = A$ (here $^T$ means transpose). 
Now let's consider first condition. From $\langle Aq, v\rangle = \langle Av,v\rangle  = 0$ we get that for all $\alpha, \beta \in \mathbb{C}$
$$
\langle A(\alpha q + \beta v), v\rangle = 0. \tag1
$$
Then, using already proved symmetry of $A$ and symmetry of given "dot product" we get that 
$$\langle Av, q\rangle = \langle q, Av\rangle = \langle Aq,v\rangle  = 0.$$
Also we know that $\langle Aq,q\rangle = 0$, so by the same way for all $\alpha, \beta \in \mathbb{C}$
$$
\langle A(\alpha q + \beta v), q\rangle = 0. \tag2
$$
Let's denote linear span $\mathcal{L}\{q,v\}$ of vectors $q$ and $v$ as $L$. $(1)$ means that subspace $AL\subset \mathbb{C}^3$ is orthogonal to the vector $v$, $(2)$ means that it is orthogonal to the vector $q$, so $\dim(AL\cap L) = 0$. Now, if $v$ and $q$ are linearly independent, then $\dim L = 2$ and $\dim AL \leq 3 - \dim L = 1$, i.e. $\dim AL < \dim L$ which implies $\det A = 0$ as $L$ is a subspace of $\mathbb{C}^3$. 
If, vice versa, we know that $\det A = 0$ then $\dim\ker A > 0$ and there exist such vector $v\neq 0$ that $Av = 0$ (obviously $v$ linearly independent of $q$ as $Aq \neq 0$). Thus $\langle Aq, v\rangle = \langle q, Av\rangle = \langle q, 0 \rangle = 0$ and $\langle Av, v \rangle = \langle 0, v \rangle = 0$.
QED. 
A: Suppose the contrary that $A$ is nonsingular. By Takagi factorisation, $A=C^TC$ for some nonsingular matrix $C$. If we redefine $q\leftarrow Cq$ and $v\leftarrow Cv$, the premise of the question and condition 1 would imply that the two redefined vectors are linearly independent vectors in $\mathbb C^3$ but $q^Tq=q^Tv=v^Tv=0$. You may try to prove that this is impossible. This shouldn't be hard.
