# Is $A + A^{-1}$ always invertible?

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.

• $A+A^{-1}$ 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 $A$. – user1551 Nov 9 '13 at 22:21

Let $A=[i]{{{{{{{{{}}}}}}}}}$.

• This is a good one! – Carsten S Nov 9 '13 at 22:04
• Who uses 1 x 1 matrices? – user85798 Nov 9 '13 at 23:23
• 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. – Git Gud Nov 9 '13 at 23:25
• Or worse, a $0 \times 3$ matrix! – user14972 Nov 9 '13 at 23:44
• @GitGud, you are a man of refined taste. – Will Jagy Nov 10 '13 at 20:05

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.