Let $(V, \langle .,. \rangle)$ be an euclidean vector space. For a linear map $F: V \to V$ we define $|F|^2 = \sum_i |F(e_i)|^2$ for an orthonormal basis $\{e_i\}$. We write $\hat{F} : \Lambda^2 V \to \Lambda^2 V, \quad \hat{F}(X \wedge Y) = F(X) \wedge Y + X \wedge F(Y).$
Then we want t show that $|\hat{F}|^2 = 2|F|^2 - \operatorname{tr}(F)^2.$
We have \begin{align*} |\hat{F}|^2 &= \sum_{i<j} |\hat{F}(e_i \wedge e_j)|^2 = \frac12 \sum_{i,j} |\hat{F}(e_i \wedge e_j)|^2 \\ &= \frac12 \sum_{i,j} \langle F(e_i) \wedge e_j + e_i \wedge F(e_j), F(e_i) \wedge e_j + e_i \wedge F(e_j) \rangle \\ &= \frac12 \sum_{i,j} |F(e_i) \wedge e_j|^2 + |e_i \wedge F(e_j)|^2 + 2\langle F(e_i) \wedge e_j, e_i \wedge F(e_j) \rangle \end{align*} With the induced inner product on $\Lambda^2 V$ we get \begin{align*} &|F(e_i) \wedge e_j|^2 = |F(e_i)|^2 - \langle F(e_i), e_j \rangle^2\\ &|e_i \wedge F(e_j)|^2 = |F(e_j)|^2 - \langle F(e_j), e_i \rangle^2 \\ &2\langle F(e_i)\wedge e_j, e_i \wedge F(e_j) \rangle = 2\langle F(e_i), e_i \rangle \langle e_j, F(e_j) \rangle - 2\langle F(e_i), F(e_j) \rangle \delta_{ij} \end{align*} so \begin{align*} |\hat{F}|^2 =\frac12 \left( (\sum_j |F|^2) + (\sum_i |F|^2) + 2 (\operatorname{tr} F)^2 - 2|F|^2 - 2\sum_{i,j} \langle F(e_j), e_i \rangle^2 \right) \end{align*}
Is that correct? How do I proceed?