Does there exist an invertible matrix $A \in \mathbb{R}^{2 \times 2}$ such that $A+A^{-1}= O_2$? Does there exist an invertible matrix $A \in \mathbb{R}^{2 \times 2}$ such that $A+A^{-1}= O_2$? Why, or why not?
 A: Yes, there exists such a matrix. Why? Because
$$\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}^{-1} = \begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$$
For another example,
$$\begin{pmatrix}7 & -5 \\ 10 & -7\end{pmatrix}^{-1} = \begin{pmatrix}-7 & 5 \\ -10 & 7\end{pmatrix}$$
A: It shouldn't be too hard to find all examples.
If we consider
$$
A=\begin{pmatrix}
a & b\\
c & d
\end{pmatrix}
$$
Then we want $-A=A^{-1}$, but we have formulas for both, so what we want is:
$$
\begin{pmatrix}
-a & -b\\
-c & -d
\end{pmatrix}
=\frac{1}{ad-bc}
\begin{pmatrix}
d & -b\\
-c & a
\end{pmatrix}
$$
Looking at the off-diagonal elements we can conclude that $ad-bc=1$. And then looking at the other two elements we can conclude that $a=-d$. That gives a restriction on the product $bc$, but we can choose $b$ freely from $\mathbb R\setminus\{0\}$. So we have a bijection from the set of matrices with that property to $\mathbb R\times(\mathbb R\setminus\{0\})$.
A: If $A + A^{-1} = O_2$, then $A^2 + I_2 = O_2$ and, thus, $A^2 = - I_2$. Assuming that $A$ is diagonalizable, let $A = Q \Lambda Q^{-1}$ be the eigendecomposition of $A$. From $A^2 = - I_2$, we obtain $Q (\Lambda^2 + I_2) Q^{-1} = O_2$ and, thus, $\Lambda^2 = - I_2$. Hence, the spectrum of $A$ is $\{\pm i\}$. Which diagonalizable matrices have purely imaginary eigenvalues?
