Trace and the coefficients of the characteristic polynomial of a matrix 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.
 A: Suppose the eigenvalues of $A$ are $\lambda_1,\ldots,\lambda_n$.  Then the factored form of the characteristic polynomial is
$$(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n).$$
Try using this with induction on $n$.
A: 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).
