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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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    For invertible $A$, $Ax=\lambda x\iff x=\lambda A^{-1}x$.2012-12-14
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    @DavidMitra So if the biggest eigenvalue of $A$ is 2, 1/2 may not be the biggest eigenvalue of $A^{-1}$?2012-12-14
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    For invertible $A$, the above says $\lambda$ is an eigenvalue of $A$ if and only if $1/\lambda$ is an eigenvalue of $A^{-1}$. Your statement is false. You have to worry about signs. If all eigenvalues were positive, you could say "the biggest eigenvalue of $A^{-1}$ is the reciprocal of the smallest eigenvalue of $A$".2012-12-14
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    czal: What happens if $A=\begin{bmatrix}1&0\\0&2\end{bmatrix}$? What about $\begin{bmatrix}-1&0\\0&2\end{bmatrix}$ or $\begin{bmatrix}-1&0\\0&-2\end{bmatrix}$?2012-12-14
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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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