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: For the first, 
$$A = P^{-1} M P$$
Where M is a (upper triangular) matrix with eigenvalues of A as diagonal elements. This is what it means to say that  A is always similar to its Jordan form.
Use $Tr(AB)=Tr(BA)$
$$Tr(A)= Tr( P^{-1} M P) = Tr(MPP^{-1})=Tr(M)=\sum_n\lambda_n$$
b) Let $B=A-kI$ with eigenvalues be $\chi_n$
Eigenvalues are determined by solutions of 
$$|B-\chi I|=0$$
or,
$$|A-(\chi+k)I|=0$$
but since you know $$|A-\lambda I|=0$$ you get $\chi_n = \lambda_n-k$
Let $Y$ be an eigenvector of $B$. So $BY=\chi Y$. Now plug stuff in for $B$ and $\chi$ and see what you'd get.
A: 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 $\lambda^{n-1}$.
