Wedge product and linearly independent functionals

I have the following problem; let $V$ be a linear space, $\dim V\ge 4$. For $i=1...4$, $\varphi_i$ are linearly independent functionals. Show that: $\varphi_{1}\wedge \varphi_{2}+\varphi_{3}\wedge \varphi_{4}$ cannot be presented as $\psi_{1}\wedge\psi_2$.

My solution is: suppose that there exist functionals $\psi_1, \psi_2$ such that $$\varphi_{1}\wedge \varphi_{2}+\varphi_{3}\wedge \varphi_{4}=\psi_1\wedge\psi_2.\qquad (1)$$ Next, I use the formula $$\varphi\wedge\psi=\frac{1}{2}(\varphi\otimes\psi-\psi\otimes\varphi),\qquad (2)$$ so I plug (2) into (1) and obtain $$\varphi_{1}\otimes\varphi_{2}-\varphi_{2}\otimes\varphi_{1}+\varphi_{3}\otimes\varphi_{4}-\varphi_{4}\otimes\varphi_{3}=\psi_{1}\otimes\psi_{2}-\psi_{2}\otimes\psi_{1}.$$ For simplicity I assume that $\dim V=4$ and pick basis $(e_1,...,e_4)$ and $(e^1,...,e^4)$ in $V$ and its dual $V^*$ respectively having $(e^j,e_i)_{V^*\times V}=\delta_{ij}$. Since $\varphi_{i}$ are linearly independent so let $e^j=\varphi_j$ and as a result $\varphi_{j}(e_{i})=\delta_{ij}$. Next, I pick $(e_{1},x)$ and plug it into (2) (I do not write $x$ after computations) and then $(e_2, x)$ and so on. I have $$\varphi_{2}=\psi_{1}(e_1)\psi_2-\psi_{2}(e_1)\psi_1,$$ $$-\varphi_{1}=\psi_{1}(e_2)\psi_2-\psi_{2}(e_2)\psi_1,$$ $$\varphi_{4}=\psi_{1}(e_3)\psi_2-\psi_{2}(e_3)\psi_1,$$ $$-\varphi_{3}=\psi_{1}(e_4)\psi_2-\psi_{2}(e_4)\psi_1.$$ I need to get a contradiction. As a result from the above considerations I have $\varphi_{i}\in \text{Lin}\{\psi_1, \psi_2\}$. Does it lead to the contradiction?

• You proved that four l.i. vectors belong to a dim two subspace... – Rafa Budría May 11 '17 at 12:16
• I get it. In fact $\text{dim}\,\text{Lin}\{\psi_1, \psi_2\}\le 2$. – zorro47 May 11 '17 at 19:10
• It seems to me pretty contradictory :) – Rafa Budría May 11 '17 at 19:19