The inverse of the matrix $\widetilde{A} = A^2 + b^2 I$ comes up in an equation of my work, where $A \in \mathbb{R}^{n \times n}$, $b>0$ and $I$ is the $n$-dimensional identity matrix. Now I would like to know what conditions $A$ has to satisfy s.t. the inverse of $\widetilde{A}$ exists and maybe see a few different approaches to answer this.
My attempt:
The inverse does not exist iff $-b^2$ is equal to an eigenvalue of $A^2$. Because the eigenvalues of $A^2$ are the squares of the eigenvalues of $A$, the matrix is not invertible iff the condition $$\pm i b = \lambda$$ is satisfied, where $\lambda$ is an eigenvalue of $A$. Because $b>0$, the above given condition can only be met when at least one eigenvalue of $A$ is purely imaginary. Therefore if $A$ has no purely imaginary eigenvalues, matrix $\widetilde{A}$ is invertible for all $b>0$. For the general case, $\widetilde{A}$ is invertible for almost all $b>0$ (where for at most $n$ different values of $b$ the inverse does not exist).
Is this a valid explanation?