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**NonCommAlg** we know that the sum of zeros of a polynomial $\displaystyle f(x)=x^n + c_1x^{n-1} + \cdots + c_n$ is $\displaystyle -c_1.$ now the eigenvalues of a matrix $\displaystyle A$ are the zeros of the polynomial $\displaystyle p(\lambda)=\det(\lambda I-A).$ so we only need

to prove that the coefficient of $\displaystyle \lambda^{n-1}$ in $\displaystyle p(\lambda)$ is equal to $\displaystyle -\text{tr}(A).$ this can be easily proved: if $\displaystyle A=[a_{ij}]$ is an $\displaystyle n \times n$ matrix, then:

$\displaystyle p(\lambda) = \det(\lambda I - A)= \begin{vmatrix} \lambda - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \lambda - a_{22} & \cdots & -a_{2n} \\ . & . & \cdots & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ -a_{n1} & -a_{n2} & \cdots & \lambda - a_{nn} \end{vmatrix}.$ it's easy to see that in the expansion of this determinant, all terms are polynomials (in $\displaystyle \lambda$) of degree at most $\displaystyle n-2$ except for the term

$\displaystyle q(\lambda)=(\lambda -a_{11})(\lambda - a_{22}) \cdots (\lambda - a_{nn}).$ thus the coefficient of $\displaystyle \lambda^{n-1}$ in $\displaystyle p(\lambda)$ is equal to the coefficient of $\displaystyle \lambda^{n-1}$ in $\displaystyle q(\lambda),$ which clearly is: $\displaystyle -(a_{11}+a_{22} + \cdots + a_{nn})= - \text{tr}(A). \ \ \ \Box$