3
$\begingroup$

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.

  • 3
    For (a), see http://en.wikipedia.org/wiki/Trace_%28linear_algebra%29 . For (b), use the equation that defines eigenvalues and eigenvectors.2011-03-15

2 Answers 2

5

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.

  • 0
    @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.2011-03-15
4

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 $\lambda^{n-1}$.

  • 0
    @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 http://en.wikipedia.org/wiki/Rule_of_Sarrus). Hopefully, you will then see a pattern that you can generalize to higher $n$.2011-03-15