Show that an $n\times{n}$ invertible matrix A has the same eigenvectors as its inverse.
I can recall that the definition of a matrix and its inverse, together with the equation for the eigenvector $x$. But this proof I am not getting a concept to deal with it.
$(A-\lambda{I})x=0$
$(A^{-1}-\lambda{I})x=0$
Thank you!
Best Answer
$$Ax = \lambda x$$
$$x =\lambda A^{-1}x$$
$$\frac{1}{\lambda} x = A^{-1}x$$
Remark:
Having the same eigenvector doesn't mean the eigenvalue is the same. The equation $$(A^{-1}-\lambda I)x = 0$$ does not hold in general.