# 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

## 2 Answers

Yes. The inverse of A has the inverse eigenvalues.

Note that $$Av=\lambda v\implies v=\lambda A^{-1}v\implies A^{-1}v=\frac 1\lambda v$$.

Finding eigenvalues close to zero is known to be sensitive to rounding errors though. It's also problematic if eigenvectors are close to being dependent.

• Would you elaborate on the last paragraph? A matrix with a cluster of eigenvalues is close to a matrix with a multiple eigenvalue. The condition number of an eigenvalue depends on the angle between the left and right eigenvectors. – Carl Christian Mar 12 '20 at 22:26
• 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
• @samie I'm saying that if an eigenvalue is close to zero, that many algorithms will kind of break down, and give bad results, effectively multiplying rounding errors. We need special robust algorithms to keep that under control. – Klaas van Aarsen Mar 18 '20 at 14:37

Yes. If $$A.v=\lambda v$$, then $$A^{-1}.(\lambda v)=A^{-1}.(A.v)=v$$. But then $$A^{-1}.v=\frac1\lambda v$$.