Let $b$ be an eigenvector of $A$ with eigenvalue $\lambda_1$ and let $\delta b$ be an eigenvector of $A$ with eigenvalue $\lambda_2$. $A$ is invertible. Show that ${||\delta x||\over ||x||}={|\lambda_1|\over |\lambda_2|}\cdot {||\delta b||\over ||b||}$. I guess $\delta$ is supposed to be some matrix and that the norm is $l_2$. What I entirely didn't understand is the solution, as follows:
"If $\lambda$ is an eigenvalue of $A$ then ${1\over \lambda }$ is an eigenvalue of $A^{-1}$. Therefore, ${||\delta x||\over ||x||}=$ ${||A^{-1}\delta b||\over ||A^{-1}b||}$ $={|\lambda_1|\over |\lambda_2|}{||\delta b||\over ||b||}$. I understand all but the second first equality. I have no idea how one replaces $x$ by $b$. Could you help me figure it out?