2nd and 3rd derivative of $A^{-1}$

Question

Consider $$ f\colon R^{n\times n} \to R^{n\times n}, A \mapsto A^{-1}. $$ How can I find the 2nd and 3rd derivative of $f$ applied to $H$, $f'(A)[H]$ and $f''(A)[H,H]$ expanding $f(A+H)$ and collecting the terms in order 1 or 2 in $H$?

Warning: I need an hint, not a full solution, thank you.

score 1 · Answer 1 · answered Apr 14 '13 at 16:36

1

Hint: assuming $A$, $A+H$ are invertible, $$(A+H)(A^{-1} - A^{-1}HA^{-1}) = I + (HA^{-1})^2,$$ so $$(A+H)^{-1} = A^{-1} - A^{-1}HA^{-1} +O(H^2).$$

answered Apr 14 '13 at 16:36

Cocopuffs

10,307

Is invertibility of $H$ necessary? – Raskolnikov Apr 14 '13 at 16:37
@Raskolnikov, No, sorry $A+H$ needs to be – Cocopuffs Apr 14 '13 at 16:38

score 0 · Accepted Answer · answered Apr 14 '13 at 16:40

(Sorry, I forgot about the hint part.)

Use the fact that if $\|\cdot\|$ satisfies $\|AB\| \le \|A\| \|B \|$, then if $\|X\|<1$, $(I+X)^{-1} = \sum_{n=0}^\infty (-1)^n X^n$.

Then if $A$ is invertible, then for sufficiently small $H$ (in particular, if $\|H\| < \frac{1}{\|A^{-1}\|} $ you have $A+H$ is invertible.

Then $(A+H)^{-1} = ((I+H A^{-1})A)^{-1} = A^{-1}(I+H A^{-1})^{-1} = A^{-1} \sum_{n=0}^\infty (-1)^n (H A^{-1})^n$.

2nd and 3rd derivative of $A^{-1}$

2 Answers2