# 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.

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For (a), see en.wikipedia.org/wiki/Trace_%28linear_algebra%29 . For (b), use the equation that defines eigenvalues and eigenvectors. – joriki Mar 15 '11 at 8:05

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.

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What does the Jordan form has to do with the upper triangular form? Also, you might want to elaborate on why an upper triangular has its eigenvalues on the diagonal. – wildildildlife Mar 15 '11 at 10:09
@Christopher: given that you're asking something about the TRace of a matrix, you might have suspected that Tr stands for Trace :) – wildildildlife Mar 15 '11 at 10:54
@Christopher "stuff" is the expressions got for B and $\chi$ earlier. @wild...life This is a question on trace which usually appears in textbooks after introducing diagonalizability, etc. So i assumed some familiarity with that. – Please Delete Account Mar 15 '11 at 21:31

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 $\lambda^{n-1}$.

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@Christopher: If you find it hard to deal with the general $n \times n$ case right away, try it first for $n=2$ ($\det(\lambda-a,-b;-c,\lambda-d)=(\lambda-a)(\lambda-d)-(-b)(-c)=\dots$) and $n=3$ (similarly, using en.wikipedia.org/wiki/Rule_of_Sarrus). Hopefully, you will then see a pattern that you can generalize to higher $n$. – Hans Lundmark Mar 15 '11 at 12:23