In my lecture notes for my linear algebra course we got this question - Let $ \phi : \mathbb{R}^3 \rightarrow \mathbb{R}^3 $ be a linear mapping with eigenvalues 2,1 and -1. What is the trace of the mapping $\phi^{\wedge 2} : \Lambda^{2}\mathbb{R^3} \rightarrow \Lambda^{2}\mathbb{R^3}$ ?
The thing is, in the notes there isn't a single mention regarding the trace of an exterior power so I have no idea how to proceed. The only thing close to it would be this theorem :
Let $\phi : U \rightarrow U$ be a linear map, which has in the basis $\alpha = (e_1, \dots, e_n)$ of the space $U$ the matrix $A = (a^{i}_j)$. Then the following is true - $\phi^{\wedge n}(e_1, \dots, e_n) = det(A)*e_1 \wedge \dots \wedge e_n$.
Could anyone provide me some tips or solutions to it? Thank you in advance
The exterior product is the vector space where if $\boldsymbol{v}_1$, $\boldsymbol{v}_2$, $\boldsymbol{v}_3$ is a basis for $U$ then a basis for the exterior product of $U$ with itself is $\boldsymbol{v}_1\wedge \boldsymbol{v}_2$, $\boldsymbol{v}_2\wedge \boldsymbol{v}_3$, and $\boldsymbol{v}_1\wedge \boldsymbol{v}_3$, where $\boldsymbol{v}_i \wedge \boldsymbol{v}_j=-\boldsymbol{v}_j \wedge \boldsymbol{v}_i$. Then if $\phi(\boldsymbol{x}) = \sum_{i=1}^{3} a_i \boldsymbol{v}_i$ and $\phi(\boldsymbol{y}) = \sum_{i=1}^{3} b_i \boldsymbol{v}_i$ you would have $$ \phi^{\wedge 2}(\boldsymbol{x} \wedge \boldsymbol{y}) = \sum_{i,j=1}^{3} a_i b_j \boldsymbol{v}_i \wedge \boldsymbol{v}_j\, , $$ where you should simplify, eliminate and/or combine some of the terms $\boldsymbol{v}_i\wedge \boldsymbol{v}_j$ in the sum.