Salvaging $A-B$ and $A+B$ invertible imply $A,B$ invertible, for $A,B\in \mathbb{C}^{n\times n}$

Question

Recently, a professor posed the following question on a midterm:

If $A,B\in \mathbb{C}^{n\times n}$ and $A-B$, $A+B$ are invertible, prove that $A,B$ are invertible.

The question was meant to be, 'prove or disprove', as taking $A=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ and $B=A^T$ provides a counterexample. But I'm left wondering if we could add some condition to make the statement true, maybe with information about the Gershgorin discs or the spectrum of the matrix? In short, my question is: is there a condition $\mathcal{C}$ such that if $A-B$ and $A+B$ are invertible and satisfy $\mathcal{C}$, then $A$ and $B$ are invertible as well?

Answer 1

The problem is equivalent to asking when $C\pm D$ are invertible given that $C,D$ are. Which again is equivalent to when $I\pm C^{-1}D$ are invertible.

A well-known sufficient condition for this is when the spectral radius (magnitude of largest eigenvalue) of $C^{-1}D$ is strictly less than $1$. This would be guaranteed if

• $C,D$ commute and $\rho(D)<\rho(C^{-1})^{-1}$, i.e., the largest eigenvalue of $D$ is smaller in magnitude than the smallest of $C$ (or vice versa);
• $\|D\|<\|C^{-1}\|^{-1}$ (or vice versa), e.g. when $(\sum_{ij}|d_{ij}|^2)(\sum_{ij}|\tilde{c}_{ij}|^2)<1$ where $\tilde{c}_{ij}$ are the coefficients of $C^{-1}$.