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I'm learning about Frechet derivatives of matrices from Bhatia's Matrix Analysis.

I want to compute the derivative of the inverse function using the formula for directional derivatives. I've seen on this website and elsewhere that the Frechet derivative of the inverse functon $g(A) = A ^{-1} $ is given by

$$Dg(A)(B) = - A ^{-1} B A ^{-1}. \tag{2}$$

I want to understand how to derive this from the directional derivative forumla for computing the Frechet derivative. Paraphrasing from Bhatia's Matrix Analysis (Eq. X.54): Let $f: \mathcal Y \to \mathcal Z$ be a continuous function. The derivative of $f$ at $A$ (if it exists) is a linear map denoted by $Df(A)$ from $\mathcal Y$ to $\mathcal Z$. It's action on an arbitrary $B \in \mathcal Y$ is then $$ Df(A)(B) = \frac {\operatorname d}{\operatorname d t} \bigg\vert_{t = 0} f(A + tB). \tag{2}$$

How does one derive Eq. (1) starting from Eq. (2)? I;ve seen answers (such as this one) where the $Dg(A)(B)$ is obtained by starting from the fact that $AA ^{-1} = I$ and then differentiating both sides, but I would like to understand it from the view of directional derivatives.

Afham
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    Assuming that $A$ is invertible, note that for small $H$, $A+H$ is invertible, and $(A+H)^{-1} = ((I+HA^{-1}) A))^{-1} = A^{-1} (I - HA^{-1} + (HA^{-1})^2-...$, from which is follows that $Dg(A)H = -A^{-1}HA^{-1}$. – copper.hat Sep 28 '23 at 18:30
  • To get intuition, you might think about differentiating (implicitly) the equation $AA^{-1}=I$ using the definition above. $\dfrac d{dt}\Big|_{t=0}(A+tB)(A+tB)^{-1}=0$. – Ted Shifrin Sep 29 '23 at 16:09
  • @TedShifrin, Yes, I knew how to do this with product rule on $AA ^{-1} = I$, but wanted to understand using the definition of directional derivative, which I now understand due the excellent answers! – Afham Sep 29 '23 at 20:13

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$$Dg(A)(B) = \frac{d}{dt}\Bigg|_{t=0}(A+tB)^{-1} = A^{-1}\frac{d}{dt}\Bigg|_{t=0}(I+tBA^{-1})^{-1}$$

Note that writing $A^{-1}$ is legal because we assumed $A$ is an element of the domain of $g$. Provided $BA^{-1}$ has small enough norm (which is fine because we can simply choose any direction $B$ that is small enough) then we have the Neumann series expansion, which generalises the familiar power series of $1 + x + x^2 \ldots$

$$ = A^{-1}\frac{d}{dt}\Bigg|_{t=0}\sum_{n=0}^\infty (-1)^n t^n (BA^{-1})^n = -A^{-1}BA^{-1}$$

Theo Diamantakis
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