Recent content by Doesy

Ohhhh! Thanks heaps man, is there anyway I can +rep you or something?

Alright Cool, Here Goes! A\chi = \lambda\chi A^{-1}\chi = \lambda^{-1}\chi Let \lambda^{-1} = \mu A^{-1}(Ax)=A^{-1}\lambda x We can re-write this as \frac{A^{-1}(A\chi)}{\lambda} = \mu\chi A^{-1}A = I Here...

\mu = \frac{1}{\lambda} Correct?

Oops, my bad, I meant Identity. Ix gives us x again. So: \frac{I\chi}{\lambda\chi} = \mu Is that right? and Ix = x?

β1 = Π/2 Π/2 = 90° If that helps at all.

Inverse multiplied by the eigenvector will give us the original matrix?

Is A^-1*A the Inverse Matrix I?

So I can re-arrange as \frac{A^{-1}(A\chi)}{\lambda} = \mu\chi Is that right?

Do you mean I should have A^{2}x = A\lambdax and A^{-2}x = A^{-1}\mux ? How can I use this to show my Answer? Or do I substitute this second equation into the first?

Homework Statement A is an invertible matrix, x is an eigenvector for A with an eiganvalue \lambda \neq0 Show that x is an eigenvector for A^-1 with eigenvalue \lambda^-1 Homework Equations Ax=\lambdax (A - I)x The Attempt at a Solution I know that I need to find x and then apply...