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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.

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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 $T.$

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  • $\begingroup$ Yes you are right about it being the discriminant, if it is diagonal then it is the vandermonde determinant. And if it is upper triangular I believe (though cannot show) you again get the vandermonde. However I am struggling to show this. $\endgroup$ Commented Apr 28, 2021 at 20:31
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    $\begingroup$ On further reflection, since the determinant is basis independent you can change to the basis of eigenvectors and what you have said constitutes on a proof. Thank you. $\endgroup$ Commented Apr 28, 2021 at 21:20

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