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Let $E = M_{3}(\mathbb{R})$ and $f: E \rightarrow E$ which $f(X) = X^3$ is $f^{'}(X)H = 3X^2 H$ the derivative for this function?

I tried to prove that $r(H) \rightarrow 0$ where $r(H) = -X^3 + (X+H)^3 - 3X^2 H$ but i dont get in anywhere.

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    $\begingroup$ @RodrigodeAzevedo The set of $3\times 3$ matrices with real entries (I think). $\endgroup$ – Winther May 18 '16 at 22:40
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The directional derivative of $f (\mathrm X) = \mathrm X^3$ in the direction of $\mathrm V$ is

$$\lim_{t \to 0} \frac{1}{t} \left( f (\mathrm X + t \mathrm V) - f (\mathrm X) \right) = \lim_{t \to 0} \frac{1}{t}\left( (\mathrm X + t \mathrm V)^3 - \mathrm X^3 \right)$$

Since

$$(\mathrm X+ t \mathrm V)^3 = \mathrm X^3 + t (\mathrm X^2 \mathrm V + \mathrm X \mathrm V \mathrm X + \mathrm V \mathrm X^2) + t^2 (\mathrm X \mathrm V^2 + \mathrm V \mathrm X \mathrm V + \mathrm V^2 \mathrm X) + t^3 \mathrm V^3$$

we obtain

$$\lim_{t \to 0} \frac{1}{t} \left( f (\mathrm X + t \mathrm V) - f (\mathrm X) \right) = \mathrm X^2 \mathrm V + \mathrm X \mathrm V \mathrm X + \mathrm V \mathrm X^2$$

If $\mathrm X$ and $\mathrm V$ commute, then $\mathrm X^2 \mathrm V + \mathrm X \mathrm V \mathrm X + \mathrm V \mathrm X^2 = 3 \mathrm X^2 \mathrm V$.

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    $\begingroup$ Or, in terms of commutators, $3X^2V+3X[V,X]$ $\endgroup$ – lisyarus Aug 1 '16 at 11:06
  • $\begingroup$ @lisyarus I get $$3 \mathrm X^2 \mathrm V + 3 \mathrm X [\mathrm V, \mathrm X] = 3 \mathrm X \mathrm V \mathrm X$$ $\endgroup$ – Rodrigo de Azevedo Aug 1 '16 at 15:26
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    $\begingroup$ I'm really sorry, that was a mistake. It should be $3X^2V+2X[V,X]+[V,X]X=3X^2V+X[V,X]+[V,X^2]$. Not that nice now :( $\endgroup$ – lisyarus Aug 1 '16 at 15:55
  • $\begingroup$ @lisyarus $$3 \mathrm X \mathrm V \mathrm X + [[\mathrm V, \mathrm X], \mathrm X]$$ would also work. $\endgroup$ – Rodrigo de Azevedo Aug 1 '16 at 16:32
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If you're looking for a Fréchet derivative approach, we're looking for a bounded, linear map $A_X : E \to E$ that satisfies the following limit:

$$ \lim_{ V \to 0\text{ in } M_{3}(\Bbb{R})} \dfrac{\|f(X+V)-f(X)-A_{X}V \|_E}{\|V\|_E} = 0 $$

The argument of the limit can be re-written as: $$ \dfrac{\| X^2\color{red}{V} + X\color{red}{V}X + XV^2 + \color{red}{V}X^2 + VXV + V^3 - A_{X}V\|_{E}}{\|V\|_E} $$

We're looking for a map $A_X$ that's $\color{red}{ \text{linear in $V$} }$ so we can hazard a guess that: $$ A_{X}V = X^2V + XVX + VX^2 $$ By successive applications of the triangle inequality (in the numerator) and using the operator norm for a matrix that represents the linear map $\| \cdot \|_E = \| \cdot \|_{\text{op}}$ the result should follow.

It'll help with cancellation where you want terms going to $0$ that for linear maps $A$ & $B$: $$ \| AB \|_{\text{op}} \leqslant \|A\|_{\text{op}} \|B\|_{\text{op}} $$

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