What is the derivative of $f (\mathrm X) = \mathrm X^3$?

Question

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.

Answer 1

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 $\mathrm X^2 \mathrm V + \mathrm X \mathrm V \mathrm X + \mathrm V \mathrm X^2 = 3 \mathrm X^2 \mathrm V$.