Can it be determined that the sum of the diagonal entries, of matrix A, equals the sum of eigenvalues of A I have a question to ask down below, that I have been having some trouble with and would like some help and clarification on.
Suppose A  is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that:   
(a) The sum of the main diagonal entries of A, called the trace of A, equals the sum of the eigenvalues of A. 
(b) A $- ~ k$ I has the eigenvalues $\lambda_{1}-k, \lambda_{2}-k, \ldots, \lambda_{n}-k$ and the same eigenvectors as A.
Thank You very much.

A more direct way of showing (a) (which doesn't involve the Jordan normal form) is to look at the second highest term in the characteristic polynomial
$$\det(\lambda I - A) = (\lambda - \lambda_1)(\lambda - \lambda_2) \dots (\lambda - \lambda_n).$$
When you expand the left-hand side using permutations and products of entries of $A$, you will get minus the sum of the diagonal entries of $A$ as the coefficient of $\lambda^{n-1}$, and when you multiply out the right-hand side, you will get minus the sum of the eigenvalues as the coefficient of $\lambda^{n-1}$.
