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I heard somewhere that the biggest eigenvalue of $A^{-1}$ is reciprocal to the biggest eigenvalue of $A$. In which theorem is this stated? Also, what would be the proof of it?

Also, what happens if we restrict $A$'s entries to be all non-negative or all positive?

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    @DavidMitra Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138). If you do so, it is helpful to post it to [this chat room](http://chat.stackexchange.com/rooms/9141) to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see [here](http://meta.stackexchange.com/q/143113), [here](http://meta.math.stackexchange.com/q/1148) or [here](http://meta.math.stackexchange.com/a/9868).2013-09-17

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This post is made community wiki in order to remove this question from the "unanswered" list.


As noted by David Mitra in the comments we have for invertible $A$ and an eigenvector $x$ that $Ax=\lambda x\Rightarrow A^{-1}x=\lambda^{-1}x$. Thus, if $\lambda$ is an eigenvalue for $A$, then $\lambda^{-1}$ is an eigenvalue for $A^{-1}$. So, if all eigenvalues are positive then the biggest eigenvalue of $A^{-1}$ is the reciprocal of the smallest eigenvalue of $A$.