This question concerns a matrix $A\in\mathbb{C}^{n\times n}$, whose inverse we know the action of on a particular vector $v\in\mathbb{C}^{n}$. If we know that $A^{-1}v = u$, is there any way to express $(A+\alpha I)^{-1}v$? We have that $\alpha \in \mathbb{C}$ but it is not bounded in magnitude and $A$ is any matrix.
2 Answers
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The only case in which you have enough information is if $u$ is a scalar multiple of $v$. If $u = t v$ with $t \ne 0$, i.e. $A v = v/t$, then $(A + \alpha I) v = (1/t + \alpha) v$ so either $(A + \alpha I)^{-1}$ doesn't exist or $(A + \alpha I)^{-1} v = (1/t + \alpha)^{-1} v$. If $u$ is not a scalar multiple of $v$, you don't have enough information to determine $(A + \alpha I)^{-1} v$.
answered Aug 5, 2011 at 5:12
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2$\begingroup$ Also known as "eigenvector"? $\endgroup$ Commented Aug 5, 2011 at 7:13
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By adding a constant identity to $A$, you increase all of its eigenvalues by $\alpha$. I rewrite your transformation as $$(A+\alpha I)^{-1}v = X(\Lambda+aI)^{-1}X^{-1}v,$$ where $X$ is a set of eigenvectors for $A$ and $\Lambda$ is the eigevalues diagonal. We have that $u = X\Lambda^{-1}X^{-1}v$, or $$X\Lambda X^{-1}u = v.$$ We can obtain $$(A+\alpha I)^{-1}v=X(\Lambda+aI)^{-1}\Lambda X^{-1}u=X(I-a(\Lambda+aI)^{-1}) X^{-1}u=u-aX(\Lambda+aI)^{-1}X^{-1}u$$ due to $\frac{\lambda_i}{\lambda_i+\alpha}=1-\frac{\alpha}{\lambda_i+\alpha}$.
