Proof that the trace of a matrix is the sum of its eigenvalues I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.
These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by 
$$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$
On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients,  we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.
By the Schur decomposition, any matrix $A$ is unitarily similar to an upper triangular matrix $T$. Being similar, $A$ and $T$ have the same trace and the same eigenvalues. Moreover, the diagonal entries of $T$ are equal to its eigenvalues (since $T$ is triangular). The stated result follows by calculating the trace of $T$. See https://www.statlect.com/matrix-algebra/properties-of-eigenvalues-and-eigenvectors.