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.
Best Answer
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.