Inverse of sum of a identity plus a symmetric matrix

Question

Caveat: I know that this question has been already asked, and I already checked the Sherman Morrison formula.

---

Reading a paper, the authors are dealing with the expression $$(A-\lambda I)^{-1}v$$ where $A$ is a symmetric matrix with eigendecomposition $A=QDQ^{-1}$,$\lambda$ a real coefficient, and $v$ a vector.

They state $$(A-\lambda I)^{-1}v = (QDQ^{-1} - \lambda I)^{-1}v = Q (D - \lambda I)^{-1}Q^{-1}v$$

How can the last equality be proved? They just give it and write no justification, so it should be a known fact, but I'm puzzled honestly.

Answer 1

Factoring $Q$ on the left and $Q^{-1}$ on the right, we have $$ (QDQ^{-1}-\lambda I) = Q(D - \lambda Q^{-1}Q)Q^{-1} = Q(D - \lambda I)Q^{-1}. $$ Taking the inverse on both sides of the equality above gives the result you are looking for. In fact, remembering $(AB)^{-1} = B^{-1}A^{-1}$, you have $$ (QDQ^{-1}-\lambda I)^{-1} = (Q^{-1})^{-1}(D - \lambda I)^{-1}Q^{-1} = Q(D-\lambda I)^{-1}Q^{-1}. $$

Answer 2

Start at the end $$ Q (D - \lambda I)^{-1}Q^{-1} $$ and work backwards.

Apply $ B^{-1}A^{-1}=(AB)^{-1}$ and use $Q={(Q^{-1})}^{-1}$ $$ Q (D - \lambda I)^{-1}Q^{-1} \\= Q\, \left( Q(D - \lambda I) \right)^{-1} \\={(Q^{-1})}^{-1}\, \left( Q(D - \lambda I) \right)^{-1} $$

Again apply $ B^{-1}A^{-1}=(AB)^{-1}$

$$ \\ = \left( Q(D - \lambda I) Q^{-1}\right)^{-1}\\=(QDQ^{-1} - \lambda I)^{-1} $$ where we also utilise $$ Q\lambda I Q^{-1}=\lambda I $$ Note, when $A$ is real and symmetric, $Q$ will be orthogonal and $Q^{-1}=Q^{T}$.