# Is every eigenvector of $A$ with eigenvalue $\lambda$ also an eigenvector of $A^{−1}$ with eigenvalue $\frac{1}{ \lambda}$?

$$A$$ is a matrix. I have implemented a function code in Python which produces a matrix $$A$$ and then I am working on the smallest eigenvalue of $$A$$, $$\frac{1}{ \lambda_{min}(A)}$$ or maximum eigenvalue of it's inverse $$A^{-1}$$, $$\lambda_{max}(A^{-1})$$. But sometimes the result is not same. Sometimes $$\frac{1}{\lambda_{min}(A)}$$ is low but $$\lambda_{max}(A^{-1})$$ becomes high. I don't know is it possible this condition or my implementation is incorrect !?

• Is $A$ assumed to be invertible? When you say "maximum" eigenvalue, do you mean maximum of $|\lambda|$ ?? – GEdgar Mar 12 '20 at 22:05
• @GEdgar I am not sure that $A$ is invertible or not, but if $A$ is not invertible so $\frac{1}{ \lambda_{min}(A)}$ should be large too, am I correct ? – samie Mar 18 '20 at 7:34
• @GEdgar you are correct – samie May 11 '20 at 12:40

Note that $$Av=\lambda v\implies v=\lambda A^{-1}v\implies A^{-1}v=\frac 1\lambda v$$.
• So you are saying that, $A$ is invertible and because of some eigenvalues close to zero, the $\lambda_{max}(A^{-1})$ is large ? but how about $\frac{1}{ \lambda_{min}(A)}$ ? I thought that eigenvalues close to zero in $\lambda_{max}(A^{-1})$ should be visible in $\frac{1}{ \lambda_{min}(A)}$ – samie Mar 18 '20 at 7:44
Yes. If $$A.v=\lambda v$$, then $$A^{-1}.(\lambda v)=A^{-1}.(A.v)=v$$. But then $$A^{-1}.v=\frac1\lambda v$$.