Linear Algebra: Eigenvalues

[Julius]

I am new to this service. Maybe someone can help me.

A little background. I am not a student. I graduated with a degree in math a very long time ago. I began a math review as a hoppy a few years ago and have been working the problems in back of math textbooks I can usually get most problems but occasionally need help.

Problem details:
I only have the student solution manual for odd problems and this is an even problem. It is on page 298 problem 40 of Pooles' second edition "introduction to linear algebra" . ( I don't know is this is a good textbook for self teaching but I am better than half way through. )

The problem is this: Prove the trace of a matrix is equal to the sum of all it's eigenvalues.

I can see that the sum of the zeros of the characteristic polynomial is the sum of the coefficients of the lambda^(n-1) term for the expanded characteristic polynomial but how do I show this sum is always the same as the sum of the diagonals of a general nXn matrix?

The hints in the book are to expand the characteristic polynomial from det( A - lambda* ( Identity matrix) ) and isolate the lambda^(n-1) term and then compare the constant coefficients of both sides. )

I would be very grateful to anyone who can help me understand a solution.
Thank you
Julius

 

[galactus]

I will use L instead of the usual lambda.

The sum of the zeros of a polynomial \(\displaystyle f(x)=x^n + b_1x^{n-1} + b_{2}x^{n-1}+ \cdots + b_n\) is equal to\(\displaystyle -b_1.\). The eigenvalues of a matrix A are the zeros

of the polynomial \(\displaystyle p(L)=\det(L I-A).\).

So we need to prove the coefficient of \(\displaystyle L^{n-1}\) in \(\displaystyle p(L)\) is equal to \(\displaystyle -\text{tr}(A).\). Assume \(\displaystyle A=[a_{ij}]\) is an \(\displaystyle n \times n\) matrix, then:


\(\displaystyle p(L) = \det(L I - A)= \begin{vmatrix} \L - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \L - a_{22} & \cdots & -a_{2n} \\ . & . & L-a_{33} & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ -a_{n1} & -a_{n2} & \cdots & \L - a_{nn} \end{vmatrix}.\)

In the expansion of this determinant, all terms are polynomials (in \(\displaystyle L\)) of degree at most n-2 except for the term

\(\displaystyle q(L)=(L-a_{11})(L - a_{22})(L-a_{33}) \cdots (L - a_{nn}).\).

Thus, therefore, and whence, the coefficient of \(\displaystyle L^{n-1}\) in \(\displaystyle p(L)\)

is equal to the coefficient of \(\displaystyle L^{n-1}\) in \(\displaystyle q(L),\)

which is: \(\displaystyle -(a_{11}+a_{22} + a_{33}+ \cdots + a_{nn})= - \text{tr}(A).\)

So, we can shed the negative signs and get:

\(\displaystyle a_{11}+a_{22} + a_{33}+ \cdots + a_{nn}=\text{tr}(A)\)

 

[Julius]

Thank you Galactica, I appreciate your reply!

I am reading the proof. It is interesting. I am stuck on a point that I think I can work out on my own and will let you know once I understand all items in the proof. Let you know if I can't get through it.

Julius

 

[Julius]

To:
Galactica

I got it! It's easy once you see it. I want to thank you very much for helping me with this proof.

Thanx again and have a great day!
Jullius