# biggest eigenvalue of inverse matrix dependent on original matrix?

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 $A$, $Ax=\lambda x\iff x=\lambda A^{-1}x$. – David Mitra Dec 14 '12 at 15:28
@DavidMitra So if the biggest eigenvalue of $A$ is 2, 1/2 may not be the biggest eigenvalue of $A^{-1}$? – czal Dec 14 '12 at 15:30
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 $A$". – David Mitra Dec 14 '12 at 15:38
czal: What happens if $A=\begin{bmatrix}1&0\\0&2\end{bmatrix}$? What about $\begin{bmatrix}-1&0\\0&2\end{bmatrix}$ or $\begin{bmatrix}-1&0\\0&-2\end{bmatrix}$? – Jonas Meyer Dec 14 '12 at 15:41
@DavidMitra Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Sep 17 '13 at 20:28

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 $A$.