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Consider a two-form $gamma in Lambda^2(V)$ where $V$ is a real vector space. Now I would like to know the necessary and sufficient conditions for $gamma$ to be expressible as an exterior product of two one-forms, $gamma=alpha wedge beta,, alpha,beta in Lambda(V)$.

Obviously, a necessary condition for the decomposition to hold is that any exterior power of $gamma$ vanishes, $gammawedge...wedgegamma = 0$. But it is not clear to me whether this condition is sufficient.

While Plücker relations give the general theory, a direct answer to your question is the following: a 2-form $gamma$ is decomposable (is a product of two 1-forms) iff $(iota_v gamma) wedge gamma = 0$ for any vector $v$, where $iota_v gamma$ is the usual contraction of a vector with a 2-form, that is, $(iota_v gamma)(u) = gamma(v,u)$ for any other vector $u$.

The proof is straightforward. Pick some vector $v$ such that $alpha = iota_v gamma ne 0$. If it doesn't exist, then $gamma = 0$, which is the trivial case. Then $alpha wedge gamma = 0$ implies* that there exists $beta$ such that $gamma = alpha wedge beta$ and you are done.

* If the implication is not obvious, extend $e_1 = alpha$ to a basis $e_i$ in $Lambda(V)$ and expand $gamma$ in the induced basis $e_iwedge e_j$. Clearly, $e_1wedge gamma = 0$ iff the only non-zero coefficients in the expansion are in front of the terms $e_1 wedge e_j$.

Let $e_1$, ..., $e_n$ be a basis of $V$. Let $gamma = sum c_{ij} e_i otimes e_j$, so $C = (c_{ij})$ is a skew symmetric matrix. Then $gamma$ factors as $alpha wedge beta$ if and only if $C$ has rank $leq 2$.

Proof: Let $C$ be a skew symmetric matrix, we must show that $C$ can be written as $a b^T - b a^T$ for vectors $a$ and $b$ if and only if $mathrm{rank}(C) leq 2$. The condition is clearly necessary, since $mathrm{rank}(a b^T) = mathrm{rank}(b a^T) leq 1$.

Choose a $2$-dimensional subspace $mathrm{Span}(v_1, v_2)$ containing the image of $C$. Since $C^T = -C$, this space also contains the image of $C^T$. Replacing $C$ by $SCS^T$, we may assume that $v_1$ and $v_2$ are the first two basis vectors.

Then the conditions on the images of $C$ and $C^T$ say that $C$ is $0$ outside the intersection of the first two rows and the first two columns, and skew symmetry further says that the upper left $2 times 2$ block is of the form $left( begin{smallmatrix} 0&c \ -c&0 end{smallmatrix} right)$. For such matrices, the claim is obvious. $square$

The OP asks whether it is enough to ask that $gamma wedge gamma$ vanish (well, the OP asks about all higher wedge powers vanishing as well, but that clearly follows from $gamma wedge gamma=0$.) The answer is yes. Any $2$-form can be written as $u_1 wedge v_1 + u_2 wedge v_2 + cdots + u_r wedge v_{r}$ where $2r$ is the rank of the matrix $C$ and $u_1$, ..., $u_r$, $v_1$, ..., $v_{r}$ are linearly independent; this is more or less the classification of skew symmetric bilinear forms. Then
$$gamma^{wedge k} = k! sum_{1 leq i_1 < i_2 < cdots < i_k leq r} u_{i_1} wedge v_{i_1} wedge u_{i_2} wedge v_{i_2} wedge cdots wedge u_{i_k} wedge v_{i_k}.$$
This is clearly nonzero if and only if $k leq r$. In particular, $r leq 1$ if and only if $gamma wedge gamma=0$.

The condition that $gamma wedge gamma =0$, expanded in coordinates, states that the $4 times 4$ principal Pfaffians of $C$ vanish. It is a nice high school algebra challenge to show that this implies the $3 times 3$ minors of $C$ vanish, which is the more standard algebraic test for a matrix to have rank $leq 2$.