# Elementary symmetric polynomials and matrices of 1-forms

Let $A$ be a $n \times n$ matrix of 1-forms (for example, a connection form). Note that $A \wedge A$ is not $0$, but by using the anti-symmetry of the wedge product applied to the entries of $A$ we can check that $\text{Tr}(A \wedge A)=0$.

Let $\sigma_i(A)$ denote the $i$-th symmetric polynomial of the eigenvalues of $A$, so $\text{det}(I+tA)=1+t\sigma_1(A)+t^2\sigma_2(A)+...+t^n\sigma_n(A)$. My claim above is equivalent to saying that $\sigma_1(A \wedge A)=0$ for any matrix $A$ of 1-forms. My question is, is it true that $\sigma_i(A \wedge A)=0$ for all $i$?

Thanks!

I figure out an answer to my own question: Note that $\sigma_i$ is an invariant polynomial, in the sense that for any invertible matrix $B$ we have $\sigma_i(B^{-1}AB)$; this follows since the determinant is invariant under similarity.
Let $A$ be a diagonal matrix of 1-forms, then $A \wedge A$ is a diagonal as well. Hence, $\sigma_i(A \wedge A)$ is just the $i-th$ elementary symmetric polynomial on the diagonal entries of $A \wedge A$. Now using the definition of the elementary symmetric polynomials and the anti-symmetry of the wedge product for 1-forms we get that $\sigma_i(A \wedge A)=0$.
Let $B$ is a invertible matrix of 1-forms and consider $B^{-1} \wedge A \wedge B$. By the invariance of $\sigma_i$ under similarity we get
$\sigma_i(A \wedge A)= \sigma_i(B^{-1} \wedge (A \wedge A) \wedge B)= \sigma_i((B^{-1} \wedge A \wedge B) \wedge (B^{-1} \wedge A \wedge B))$.