Inverse of sum of a identity plus a symmetric matrix

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.

• Are you missing a $-1$ in the exponent of the middle expression? Commented Jul 17, 2020 at 13:48
• @TSH yes thanks, it was missing also in the last equality
– lukk
Commented Jul 17, 2020 at 13:50

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 $$(QDQ^{-1}-\lambda I)^{-1} = (Q^{-1})^{-1}(D - \lambda I)^{-1}Q^{-1} = Q(D-\lambda I)^{-1}Q^{-1}.$$

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 $$Q^{-1}=Q^{T}$$.

• I accepted the other one, but thanks for your answer. @PM.
– lukk
Commented Jul 17, 2020 at 18:45