Let $A$ be a matrix over a field of characteristic different than two. For convenienve, let's suppose $A \in \mathbb{C}^{n\times n}$.
Let $A$ be strictly upper triangular and $B = A - A^T$ the corresponding skew-symmetric matrix. Consider the following two bilinear forms: $$\alpha_A : \mathbb{C}^n\times\mathbb{C}^n \to \mathbb{C}: (v,w) \mapsto v^TAw$$ and $$ \alpha_B : \mathbb{C}^n\times\mathbb{C}^n \to \mathbb{C}: (v,w) \mapsto v^TBw.$$
Is the following statement correct?
$$\ker \alpha_B \subset \ker\alpha_A,$$ where $\ker\alpha = (\mathbb{C}^n)^\perp = \{ v\in \mathbb{C}^n \mid \forall w\in \mathbb{C}^n : \alpha(v,w) = 0 \}$.
The statement is not true. Let $$A=\begin{pmatrix}0 & 1 &1\\ 0 & 0 &1 \\ 0 &0 & 0\end{pmatrix}$$ Then $$B=\begin{pmatrix}0 & 1 &1\\ -1 & 0 &1 \\-1 &-1 & 0\end{pmatrix}$$ The condition $v^TCw=0$ for any $w$ is equivalent to $C^Tv=0.$ Let $v=(1,-1,1).$ Then $B^Tv=-Bv=0,$ but $A^Tv\neq 0.$