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Let $F : Gl_n(\mathbb{R}) \rightarrow Gl_n(\mathbb{R})$ is given by $a \mapsto a^{-1}$ Show that the Frechet derivative is given by $D_A F(h) = -a^{-1} h a^{-1}$.

Work done so far:

\begin{align*} ||F(u_0 + h) - f(u_0) + u_0^{-1}hu_0^{-1}|| &= || \Sigma_{k = 0}^{\infty}(-1)^k(u_0^{-1}h)^ku_0^{-1} - u_0^{-1} + u_0^{-1}hu_0^{-1}|| \\ &= ||\Sigma_{k \geq 2} (-1)^k(u_0^{-1}h)^ku_0^{-1}|| \end{align*}

I am not sure the last element approaches zero as h goes to zero. If someone could explain that part.

jlammy
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joe blacksmith
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  • Hint: (1) wlog let $u_0 = I$. (2) eventually $|h| < 1$. Therefore you can estimate the right side by $\sum_{k\ge 2} |h|^k$. – user251257 Jan 23 '21 at 01:36
  • why can we assume that $u_0 = I$. I see the rest is just geometric series. – joe blacksmith Jan 23 '21 at 01:38
  • you can factor out $u_0$ from $u_0 + h$ and use the fact that matrix norm is sub-multiplicative – user251257 Jan 23 '21 at 01:41
  • Can you show me the details as an answer and I will accept it. – joe blacksmith Jan 23 '21 at 01:44
  • Take a look at this answer for an outline. The point is that if you choose $h$ with small enough norm so that $\lVert u_0^{-1}\rVert\lVert h \rVert<1$, then you have an absolutely convergent (geometric) series, and using submultiplicativity of operator norm, you can bound that term as $C\cdot\lVert h\rVert^2$ for some finite constant $C$, namely $\lVert u_0^{-1}\rVert^3\sum_{k\geq 2}\lVert u_0^{-1}\rVert^{k-2}\cdot \lVert h\rVert^{k-2}$. Hence, the remainder goes to $0$ faster than $\lVert h\rVert$ as $h\to 0$. – peek-a-boo Jan 23 '21 at 03:19
  • ^ my bad, I should have written $C(||h||)$... but this is no issue since $C(||h||)$ is a decreasing function of $\lVert h\rVert$. – peek-a-boo Jan 23 '21 at 03:32

2 Answers2

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If you know the product rule -- derivative of $fg$ at $A$ is $H\mapsto f'(A)(H)\cdot g(A) + f(A)\cdot g'(A)(H)$ -- then you can apply it to the function $G=\mathrm{id}\cdot F$, i.e. $G(A)=AA^{-1}=I_n$ to get \begin{align*} 0=G'(A)&=\mathrm{id}'(A)(H)\cdot F(A)+\mathrm{id}(A)\cdot F'(A)(H)\\ &= H\cdot F(A)+A\cdot F'(A)(H), \end{align*} which rearranges to $F'(A)(H)=-A^{-1}HA^{-1}$.

jlammy
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  1. We may assume $u_0 = I$ wlog, as multiplication is continuous and $$ (u_0 + h)^{-1} - u_0^{-1} + u_0^{-1} h u_0^{-1} = u_0^{-1}\left( (I + hu_0^{-1})^{-1} - I + hu_0^{-1} \right). $$

  2. Using the Neumann series, we have $$ (I+h)^{-1} - I + h = \sum_{k\ge 2} (-h)^k = (-h)^2 \sum_{k\ge 0} (-h)^k = h^2 (I+h)^{-1} = o(\|h\|), $$ as $$ \frac{\| h^2 (I+h)^{-1} \|}{\|h\|} \le \frac{\|h\|^2 \|(I+h)^{-1}\|}{\|h\|} = \|h\|\|(I+h)^{-1}\| \to 0 \cdot 1. $$

Notes: If you already know that $F$ is differentiable, then you can simply apply the product rule as in @jlammy's answer. $F$ is in fact smooth, as it is a rational function by Cramer's rule.

user251257
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