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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.

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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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  • $\begingroup$ it seems that I lack the algebra, not the theory. I have reached the same conclusion, but failed to show that the $\lambda^{n-1}$ coefficient of $\prod_i (A_{ii}-\lambda)$ equals minus the trace of A. $\endgroup$ – Yonatan Izutskiver Jan 18 '16 at 15:17
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    $\begingroup$ It's similar, again you have all posible products of one coefficient from each term of the product. All those of form $\lambda^{n-1}$ take n-1 terms of $\lambda$ and one other term (there are n combinations, with different diagonal term each time). $\endgroup$ – orion Jan 18 '16 at 15:40
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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 $n$.

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  • $\begingroup$ Sorry, I don't get how it is related to the trace of A... $\endgroup$ – Yonatan Izutskiver Jan 18 '16 at 15:15
  • $\begingroup$ Well start with $n=2$. Then we have $(x-\lambda_1)(x-\lambda_2)=x^2-(\lambda_1+\lambda_2)x+\lambda_1\lambda_2$. So we see that the coefficient of the $x$ term is $-(\lambda_1+\lambda_2)=-\operatorname{tr}A$. $\endgroup$ – Tim Raczkowski Jan 18 '16 at 15:20
  • $\begingroup$ but $\lambda_1$ and $\lambda_2$ aren't necessarily equal to $a_{1,1}$ and $a_{2,2}$ $\endgroup$ – Yonatan Izutskiver Jan 18 '16 at 15:23
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    $\begingroup$ This proof assumes that the characteristic polynomial can be factored into a product of linear terms. This can be done over an appropriate splitting field but if one doesn't want (or have the tools) to justify the passage to such a splitting field, orion's approach is preferable. $\endgroup$ – levap Jan 18 '16 at 15:27
  • $\begingroup$ You can use the fact that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ and that for any matrix there is a transition matrix $B$ such that $BAB^{-1}$ is an upper triangular matrix whose diagonal entries are the eigenvalues of the matrix. $\endgroup$ – Tim Raczkowski Jan 18 '16 at 15:33

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