Determinant of matrix on wedge product of vector spaces

Question

Let $T : V \to V$ be a linear map (matrix) on a vector space $V$, and let $V \wedge V$ be the second exterior power of the vector space. Let $e_1, \dots, e_n$ be an orthonormal basis for $V$. Consider the map $e_k \wedge e_j \mapsto e_k \wedge T e_j + e_j \wedge T e_k$. I want to find the determinant of this map $V \wedge V \to V \wedge V$. Is there an easy way to do this? It may simplify things to assume $T$ is upper triangular.

Answer 1

Assume $T$ is diagonal. Notice that if $T$ has two equal eigenvalues, then $e_k\wedge e_j \to 0,$ so the determinant is zero. This suggests that the determinant of your map is the discriminant of the characteristic polynomial of $T.$