Suppose $V$ is a vector space , $V\cong \mathbb{R}^4$. Let $\omega_1, \omega_2, \omega_3, \omega_4\in Alt^1(V)$ be basis. The matrix $A=(a_{ij})_{4\times4}$ satisfies $A^T=-A$ and $\eta=\sum_{i<j}a_{ij}\omega_i\wedge\omega_j$. Prove that $\eta \wedge\eta=0$ if and only if $\det A=0$.
Direct computation shows $\eta \wedge\eta=0$ iff $$ \sum_{\sigma\in S(2,2)}a_{\sigma(1)\sigma(2)}a_{\sigma(3)\sigma(4)}sgn(\sigma)w_1\wedge w_2\wedge\omega_3\wedge \omega_4=0, $$ where $S(2,2)$ is the set of (2,2)-shuffle. I don't know how to go on. Also, we can write $\eta$ as $$ \eta=\sum_{i<j}a_{ji}\omega_j\wedge\omega_i, $$ thus the wedge product is of another form. But I still don't know how to connect it with the determinant of A.
Appreciate any help or hint!
Your computation is false. Note that if $\sigma\in S(2,2)$, then $\sigma$ must be an even permutation, since $S(2,2)=\{(12)(34), (13)(24), (14)(23)\}$, no matter what $\sigma$ is, we must have $a_{\sigma(1)\sigma(2)}a_{\sigma(3)(4)}=a_{12}a_{34}$.
Indeed $$\eta\wedge \eta = \sum_{i<j, l<k} a_{ij}a_{lk}\cdot \text{sgn}(\begin{pmatrix} i & j & l & k \\ 1 & 2 & 3 & 4 \end{pmatrix})\omega_1\wedge\cdots\wedge\omega_4$$ $$=2(a_{12}a_{34}-a_{13}a_{24}+a_{14}a_{23})\omega_1\wedge\cdots\wedge\omega_4$$
Meanwhile one can compute $\det A = (a_{12}a_{34}-a_{13}a_{24}+a_{14}a_{23})^2 $ such as WolframAlpha.
I had another ugly proof without establishing the relation between Pfaffian and det.
WLOG, we may assume $\{\omega_1, \cdots, \omega_4\}$ is the dual basis of the standard basis $\{e_1, \cdots, e_4\}$ of $\mathbb R^4$. Then note that $\eta(e_i, e_j)=a_{ij}$ for $i<j$, the quadratic form $\eta$ is represented by the matrix $A$, that is $\eta(x,y)=x^tAy=(x, Ay)$ for any $x,y\in\mathbb R^4$.
Now the if part is clear. If $A$ is singular, pick nonzero $v_1$ with $Av_1=0$ and extend it to be a basis $\{v_1, \cdots, v_4\}$. Hence $\eta(v_i, v_j)=(v_i, Av_j)=(-Av_i, v_j)=0$ as long as $i=1$ or $j=1$. And since $1$ must be among $\sigma(1), \cdots, \sigma(4)$, $$\eta\wedge\eta(v_1, \cdots, v_4) = (\text{Some normalization factor}) \sum_{\sigma\in S_4}\text{sgn}(\sigma)\eta(v_{\sigma(1)},v_{\sigma(2)})\eta(v_{\sigma(3)}, v_{\sigma(4)})=0$$
For the other direction, note that $iA$ is self-adjoint, so it can be unitarily diagonalized whose eigenvalues are real. Morever as $A$ is real for each eigenvalue $\lambda i$ of $A$, $\overline{i\lambda}=-\lambda i$ is also an eigenalue. Combing these, we may find an orthonormal basis $v_+, v_-=\overline{v_+}, w_+, w_-=\overline{w_+}$ whose eigenvalues are $ai, -ai, bi, -bi$. Note that $a=b$ is possible but doesn't matter.
Now we want to compute $(\eta\wedge\eta) (v_+, v_-, w_+, w_-)$. Note that complex inner product is slight different from the real one, and $\eta(x,y) = x^t A y = (x, A\bar y)$. For each summand $(x, A\bar y)(z, A\bar w)$ where $x,y,z,w$ is a permutation of $(v_+, v_-, w_+, w_-)$. As they are orthonormal, the only nonzero summand is when both $x=\bar y$ and $z=\bar w$ hold, and so on. Putting everything together, it can be show $(\eta\wedge\eta) (v_+, v_-, w_+, w_-)>0$. If we are more careful, we can compute the relationship between det and Pfaffian in the process as well.