Originally Posted by
alexandrabel90 let A be an nxn matric and the trace of A is defined by tr(A)= a_11+a_22...+a_nn . Prove that the coeff of x^(n-1) in the characteristic polynomial is -tr(A)
Another method (the first is similar to that of Dr. Revilla's) is that you can first recall that $\displaystyle p_A(x)$ is monic and it's roots are precisely the mulitiset spectrum $\displaystyle \sigma\left(A\right)$ of $\displaystyle A$. and so
$\displaystyle \displaystyle p_A\left(x\right)=\prod_{\lambda\in\sigma\left(A\r ight)}(x-\lambda)$
We can then recall Vieta's formula so that (remembering that $\displaystyle p_A(x)$ is monic) that $\displaystyle -\text{coeff}(x^{n-1})=\Pi_1(\lambda_1,\cdots,\lambda_n)$ where $\displaystyle \Pi_1(\lambda_1,\cdots,\lambda_n)$ is the first symmetric polynomial. Thus,
$\displaystyle \displaystyle -\text{coeff}(x^{n-1})=\Pi_1(\lambda_1,\cdots,\lambda_n)=\sum_{\lambd a\in\sigma(A)}\lambda=\text{tr}(A)$