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Assume that a matrix $B$ (real or complex, it's not important) has a null eigenvector x, namely $B x = 0$. Therefore, $B^{-1}$ does not exist (the kernel of $B$ has dimension greater or equal to 1). Define $A = \alpha I +B$, where $I$ is the identity and $\alpha$ is a number to be chosen such that $A^{-1}$ exists.

Is it true, in general, that $A^{-1}x = \alpha^{-1}x$? I guess that this statement is true, because $$x = A A^{-1}x= A^{-1} (Ax) = A^{-1} (\alpha x ) = \alpha ( A^{-1}x ) \, .$$

Now, just for curiosity: is it possible to find a more "explicit/direct" proof based on some known structural properties of $A^{-1}$?

Note: it is always possible to find a whole interval of values of $\alpha$ such that $A^{-1}$ exists, see this question about a slightly different case (here I add $\alpha$ only on the diagonal of $B$, not to every component).

Quillo
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    The unique solution of the linear system $Ay = x$ is $y = \alpha^{-1}x$, which can be checked directly. That's all that is required for a proof. It also has the advantage of generalizing to infinite dimensions. – Hans Engler Sep 22 '21 at 20:21
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    This is the most direct argument you can possibly give. Trying to find a formula for $A^{-1}$ is difficult unless a lot more is known about $B$. – Ted Shifrin Sep 22 '21 at 22:37
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    Inverting $A$ is related to the geometric series telescoping sum computation: $(aI + B)^{-1}x = \frac{1}{a}(I - -\frac{1}{a}B)^{-1}x = \frac{1}{a}\sum_{j = 0}^{\infty}(-1)^j\frac{1}{a^j}B^jx$, provided the sum converges, e.g. if $B^Jx = 0$ for some $J$. – Mason Sep 23 '21 at 05:09
  • @HansEngler thank you, your comment about infinite dimensions is very interesting and useful: I didn't realize that uniqueness could be so powerful in this context. – Quillo Sep 23 '21 at 06:52
  • @Mason this is exactly the sort of "direct" approach I was asking for. You can turn your comment into an answer, so I can accept it. Do you have a reference/link for the proof of the "matrix" version of the geometric series? – Quillo Sep 23 '21 at 06:54
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    @Quillo It follows from continuity of linear maps on finite-dimensional spaces that if $T \colon \mathbb{C}^n \to \mathbb{C}^n$ is linear and $\sum_{j = 0}^{\infty}T^jx$ converges, then $(I - T)\sum_{j = 0}^{\infty}T^jx = \sum_{j = 0}^{\infty}T^jx - \sum_{j = 1}^{\infty}T^jx = x$. There are many sufficient conditions for the series $\sum_{j = 0}^{\infty}T^jx$ to converge. For example, by completeness of $\mathbb{C}^n$, if $\sum_{j = 0}^{\infty}\lVert T^jx \rVert < \infty$, then it converges. In particular, when $\lVert T \rVert < 1$ it converges for every $x \in \mathbb{C}^n$. – Mason Sep 23 '21 at 21:08
  • Dear @Mason, please (if you have time) may you turn your comments into a brief answer? So that I can accept it and the question will be flagged as "answered" (I consider your comments a fully satisfying answer). Thank you! – Quillo Nov 16 '21 at 09:57

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