# EigenValues & EigenVectors proofs

Defennder
Homework Helper
It can't be closed, only locked. And there's no reason to lock it. Only certain users can affix a [SOLVED] tag to the thread title.

morphism
Homework Helper
Dick gave a perfectly fine proof of the fact that det(A) is the constant term of the characteristic polynomial.

I know that this is a ridiculous amount of time after the questions was asked, but I thought I would help with the first question.
as we all know with eigenvalues Ax=lambdax
the easiest way I have found to show that A^(-1)x=1/lambda x
Is to use matrix multiplcation to get A^(-1) on the right side and then get to your 1/lamda from there.

Dick
Homework Helper
Sure that works. Even though most of the people that have contributed to this messed up thread are dead, I'm sure they would have thanked you. :). Just kidding. Thanks.

Well I thought since I just did that problem in my matrix theory homework that anyone else who googled it like I did would appreciate the hint versus just having the answer given to them. Well maybe most kids won't but I would have! :)

HallsofIvy
As said the first time you posted this you don't have to do any evaluation or computation. If $\lambda_1$, $\lambda_2$, ..., $\lambda_n$ are are the eigenvalues of A, then its characteristic polynomial is $(x-\lambda_1)(x- \lambda_2)\cdot\cdot\cdot (x- \lambda_n)$. Multiplying that out, the constant term is clearly $\lambda_1\lambda_2\cdot\cdot\cdot\lambda_n|$. If you have set up the characteristic equation as $det(A- \lambda I)$= 0, as I learned, it is obvious that taking [/itex]\lambda= 0[/itex] gives just the determinant. If, as some people learn, you set it up as $det(\lambda I- A)$ it gives the negative of the determinant.