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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.
Let $A=[i]{{{{{{{{{}}}}}}}}}$.
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.
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 $\pi/2$.
$\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$
Diagonal matrix is a good thing to try out first, especially when their inverses are simple.
