Second order derivative of the inverse matrix operator

Question

Let $f : Gl_{n}(\mathbb{R}) \to Gl_{n}(\mathbb{R})$ defined by $f(X)=X^{-1}$. Compute $f''(X)(H,K)$.

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I calculated $f'(X).H=-X^{-1}HX^{-1}$ so I tried to use some composition of linear functions but did not find the appropriate functions. Can anyone help me in this matter?

Answer 1

Let $f (\mathrm X) := \mathrm X^{-1}$. Hence,

$$\begin{array}{rl} f (\mathrm X + h \mathrm V) &= (\mathrm X + h \mathrm V)^{-1}\\ &= \left(\mathrm X \left(\mathrm I + h \mathrm X^{-1} \mathrm V \right) \right)^{-1}\\ &= \left(\mathrm I + h \mathrm X^{-1} \mathrm V \right)^{-1} \mathrm X^{-1}\\ &\approx \left(\mathrm I - h \mathrm X^{-1} \mathrm V + h^2 \mathrm X^{-1} \mathrm V \mathrm X^{-1} \mathrm V \right) \mathrm X^{-1}\\ &= f (\mathrm X) - h \mathrm X^{-1} \mathrm V \mathrm X^{-1} + h^2 \mathrm X^{-1} \mathrm V \mathrm X^{-1} \mathrm V \mathrm X^{-1}\end{array}$$

Thus, the 2nd directional derivative of $f$ in the direction of $\mathrm V$ at $\mathrm X$ is $\color{blue}{2 \mathrm X^{-1} \mathrm V \mathrm X^{-1} \mathrm V \mathrm X^{-1}}$.

Answer 2

Using the Taylor formula, (As Rodrigo de Azevedo and user1952009 did) you can calculate $f''(X)(H,H)$. If you want the gneral formula $f''(X)(H,K)$, then you can calculate the derivative (with respect to $X$, considering $H$ as a fixed vector) of $f'(X)(H)=-X^{-1}HX^{-1}$.

We obtain $f''(X)(H,K)=X^{-1}KX^{-1}HX^{-1}+X^{-1}HX^{-1}KX^{-1}$ (that is the "bilinearization" of the quadratic form $f''(X)(H,H)$).

Answer 3

Let $f (\mathrm X) := \mathrm X^{-1}$. The 1st directional derivative of $f$ in the direction of $\mathrm V_1$ at $\mathrm X$ is $\color{blue}{-\mathrm X^{-1} \mathrm V_1 \mathrm X^{-1}}$.

Let $\mathrm G (\mathrm X) := -\mathrm X^{-1} \mathrm V_1 \mathrm X^{-1}$. Hence,

$$\mathrm G (\mathrm X + h \mathrm V_2) \approx \mathrm G (\mathrm X) + h \left( \mathrm X^{-1} \mathrm V_1 \mathrm X^{-1} \mathrm V_2 \mathrm X^{-1} + \mathrm X^{-1} \mathrm V_2 \mathrm X^{-1} \mathrm V_1 \mathrm X^{-1} \right)$$

Thus, the 1st directional derivative of $\mathrm G$ in the direction of $\mathrm V_2$ at $\mathrm X$ is

$$\color{blue}{\mathrm X^{-1} \mathrm V_1 \mathrm X^{-1} \mathrm V_2 \mathrm X^{-1} + \mathrm X^{-1} \mathrm V_2 \mathrm X^{-1} \mathrm V_1 \mathrm X^{-1}}$$

which is the 2nd directional derivative of $f$ in the directions of $\mathrm V_1$ and $\mathrm V_2$ at $\mathrm X$. If $\mathrm V_1 = \mathrm V_2 = \mathrm V$, then the 2nd directional derivative of $f$ in the direction of $\mathrm V$ at $\mathrm X$ boils down to $\color{blue}{2 \mathrm X^{-1} \mathrm V \mathrm X^{-1} \mathrm V \mathrm X^{-1}}$.