Let $A\in M(\mathbb F)_{n \times n}$
Prove that the trace of A is minus the coefficient of $\lambda ^{n-1}$ in the characteristic polynomial of A.
I had several ideas to approach this problem - the first one is to develop the characteristic polynomial through the Leibniz or Laplace formula, and from there to show that the contribution to the coefficient of $\lambda ^{n-1}$ is in fact minus the trace of A, but every time i tried it's a dead end.
Another approach is to use induction on a similar matrix to ($\lambda I-A$) from an upper triangular form, which has the eigenvalues of A on its diagonal, and of course the same determinant and trace, to show that for every choice of n this statement holds.
I think my proof doesn't hold for all fields, so any thought on the matter will be much appreciated, or an explanation to why this statement is true.
Answer
The determinant is a sum of (signature-weighted) products of $n$ elements, where no two elements share the same row or column index. From this, it follows, that there is no term with $(n-1)$ terms on the diagonal (if $n-1$ terms of a product are on the diagonal, then the last one must be too, because all other rows and columns are taken). So... the only term that can possibly include a power of $\lambda^{n-1}$ is the product of the main diagonal. Therefore, the $\lambda^{n-1}$ coefficient of $\det A$ equals the $\lambda^{n-1}$ coefficient of $\prod_i (\lambda-A_{ii})$ for which it's easy to show, the coefficient equals $-\sum_i A_{ii}$.
This definition doesn't make assumptions about the field over which the matrix is defined, because the field operations + and * are used directly (with no assumptions about inverses and distribution laws).
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