1. ## eigenspace help

A is the matrix $\displaystyle \begin{bmatrix} 2 & 2 & -2 \\ 2 & -1 & 4\\ -2 & 4 & -1 \end{bmatrix}$. You are given that A has eigenvalues -6 and 3, with algebaric multiplicities 1 and 2 respectively, and that the eigenspace $\displaystyle V_{3} = span{(2,-1,-2),(2,2,1)}$.

a) use the given eigenvalues to find the characteristic equation of A. (do not expand det(A-ILAMBDA), but the construct the equation from its roots.
Hence use the hamilton theorem to find A^{-1} .

I dont know where to being. Any help appreciated,

2. ## Re: eigenspace help

Hey Tweety.

Recall that the eigenvalues are the solutions to the characteristic equation which means they are the roots. Recall that if a polynomial has solutions a,b,c then we can factorize that polynomial as (x-a)(x-b)(x-c) = 0.

3. ## Re: eigenspace help

so the characteristic equation is $\displaystyle (\lambda+6)(\lambda-3)$ ?

4. ## Re: eigenspace help

You need to factor in the multiplicities of the roots. (Hint: You will have a squared term).

5. ## Re: eigenspace help

Thank you, but I am not sure how to factor in the multiplicties of the roots?

is it $\displaystyle 1(\lambda+6)2(\lambda-3)$ ?

is the equation $\displaystyle 2\lambda^{2}+6\lambda-36$

6. ## Re: eigenspace help

Basically its (lambda + 6)(lambda - 3)(lambda - 3) = characteristic polynomial = f(lambda)