Assume that $X$ and $A$ are $n\times n$ real matrices, with $A$ invertible (in fact, the identity matrix up to a non-zero factor).
Can we derive conditions on $X$ such that both $X+A$ and $X-A$ are invertible?
Conversely, knowing that both $X+A$ and $X-A$ are invertible, does this provide some information on $X$?

I tried to decompose the inverses with known formulas (Henderson-Searle, Woodbury) but it seems that my attempts only lead to circular arguments.


Write $\,A= \lambda\cdot\mathbb{1}\,$ for the fixed matrix, with $\mathbb 1$ denoting the $n\times n$ identity matrix and $\lambda$ a scalar factor. The condition "$X+A$ and $X-A$ are invertible"
is then equivalent to both $\,-\lambda\,$ and $\,\lambda\,$ being not eigenvalues of $X$.
(And this does not depend on $\lambda\ne 0$ or not.)

Another equivalent formulation is, by definition:
$\{\lambda, -\lambda\}$ lies in the resolvent set of $X$. The latter is the complement within $\mathbb C$ of the spectrum $\sigma(X)$.

This is all that can be said due to the following facts:

Every non-empty set of at most $\,n\,$ complex numbers appears as spectrum of some $n\times n$ complex matrix.

You may consider a diagonal matrix containing all the numbers from the given set.

When dealing with real matrices then in addition we have that non-real numbers in the spectrum show up in complex-conjugate pairs.

Given such a set you may obtain a corresponding block-diagonal matrix by setting in the diagonal all the real elements and for each conjugate pair $\,a\pm ib\,$ a submatrix $\left(\begin{smallmatrix}a&b\\ -b&a\end{smallmatrix}\right)$; note
that $\,\begin{pmatrix} a&b\\ -b&a\end{pmatrix} \begin{pmatrix} 1\\ \pm i\end{pmatrix} = (a\pm ib) \begin{pmatrix} 1\\ \pm i\end{pmatrix}\,$.

  • $\begingroup$ Thanks for your contribution. I hoped there were something missing in my approach, but this is probably the best one can aim. On the contrary, it is possible to do something better for what concern sufficiency conditions on $X$ in order to have $X\pmA$ invertible or the previous argument is in fact a "no-go" one? $\endgroup$ – Ivan Sep 9 '18 at 13:29
  • $\begingroup$ Just edited my answer to emphasize the equivalence of the given criteria. If I correctly understand your Q in the comment, then it should be covered by the answer too. Otherwise you could, and maybe should, consider to ask a new question. $\endgroup$ – Hanno Sep 9 '18 at 18:08

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