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Let $A$ be an invertible matrix. Then is $A + A^{-1}$ invertible for any $A$?

I have a hunch that it's false, but can't really find a way to prove it. If you give a counterexample, could you please explain how you arrived at the counterexample? Thanks.

This isn't HW, and I don't really have any work to show.

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    $\begingroup$ $A+A^{-1}$ is invertible if and only if $I+A^2=A(A^{-1}+A)$ is invertible. So, the whole point is whether $i$ or $-i$ is an eigenvalue of $A$. $\endgroup$ – user1551 Nov 9 '13 at 22:21
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Let $A=[i]{{{{{{{{{}}}}}}}}}$.

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    $\begingroup$ This is a good one! $\endgroup$ – Carsten S Nov 9 '13 at 22:04
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    $\begingroup$ Who uses 1 x 1 matrices? $\endgroup$ – user85798 Nov 9 '13 at 23:23
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    $\begingroup$ I do. I don't really like saying a matrix is a regular number. But if you have a problem with this, wait until you see a $0\times 0$ matrix. $\endgroup$ – Git Gud Nov 9 '13 at 23:25
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    $\begingroup$ Or worse, a $0 \times 3$ matrix! $\endgroup$ – user14972 Nov 9 '13 at 23:44
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    $\begingroup$ @GitGud, you are a man of refined taste. $\endgroup$ – Will Jagy Nov 10 '13 at 20:05
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As the other answers show, the answer is no. However, if $A$ is symmetric it is true (Hermitian in the complex case): Let $A$ have eigenvalues $a_i$, and note that $A+A^{-1}$ has eigenvalues $a_i + a_i^{-1}$. These are zero if and only if $a_i^2 = -1$. But this is not possible since $a_i$ is real. Hence $A + A^{-1}$ is invertible.

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There is a complex number $i\ne0$ with the property that $i+i^{-1}=0$, or put otherwise $i^2+1=0$. It behaves exactly like $$\begin{pmatrix}0&-1\\1&0\end{pmatrix},$$ a rotation by $\pi/2$.

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$\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$

Diagonal matrix is a good thing to try out first, especially when their inverses are simple.

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