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Let us define a function $f$ from $M(n,\mathbb{R})$ to $M(n,\mathbb{R})$ by treating $M(n,\mathbb{R})\approx\mathbb{R}^{n^2}$, by $$f(X)=e^X+X$$ where $$e^X=1+X/{1!}+X^2/{2!}+\dots$$ I want to find the (Frechet) derivative of $f$.

We know, if derivative exists at $X$, then $$f(X+H)-f(X)=f'(X)H+r(H)$$ where $r(H)/\|H\|\to \bf{0}$ as $H\to \bf{0}$.

So I went on to find the difference, but couldn't figure out the linear part ($f'(X)H$) and the remainder part ($r(H)$).

Any help is appreciated.

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2 Answers 2

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The Frechet derivative is linear, so the only hard part is to find the derivative of the exponential (as $f$ is the exponential plus the identity; the identity is linear and so it is its own derivative). This seems to be well-known to the experts: after a quick search, the derivative is described here (Theorem 3.1) and here (see the second slide).

Translating those two into your notation, you have $$ f'(X)H=H+\sum_{k=1}^\infty\frac1{k!}\sum_{j=1}^kX^{j-1}HX^{k-j}, $$ or $$ f'(X)H=H+\int_0^1e^{X(1-s)}He^{Xs}\,ds. $$

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Thanks @Martin Argerami for the nice answer. I am very glad to know the literature behind this. –  Kumara Feb 27 '12 at 5:46

Well $$e^{X+H}-e^{X}=1+(X+H)/{1!}+(X+H)^2/{2!}+\dots-(1+X/{1!}+X^2/{2!}+\dots)$$

How theses series are absolutely convergent we have

$$e^{X+H}-e^{X}=(1-1)+\frac{(X+H)-X}{1!}+\frac{(X+H)^2-X^2}{2!}+\dots$$

But note that $$(X+H)^n=X^n+\sum_{i=1}^n\prod_{j=1}^{n}{X_{ij}}+o(H)$$

Where $X_{ij}=H$ if $i=j$ and $X_{ij}=X$ if $i\neq j$.

And them we may state that $$f'(X)H=\sum_{n=1}^{\infty}\frac{1}{n!}\sum_{i=1}^n\prod_{j=1}^{n}{X_{ij}}$$

And whenever $H$ commutes with $X$ we get $f'(X)H=e^X H$.

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Thanks @chessmath for the elegant answer. –  Kumara Feb 27 '12 at 5:49

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