# $F: M_n \to M_n$, $F(m) := m^3$, what is the derivative?

We recall that $F: U \to Y$, with $U \subset X$ open and $(X, \|\cdot \|_X)$ and $(Y, \|\cdot|_Y)$ normed vector spaces, is differentiable at $x \in U$, if there is a continuous linear map $S_x: X \to Y$, which is the derivative of $F$ at $x$, such that$$\lim_{\|h\|_X \to 0} {{\|F(x + h) - F(x) - S_x(h)\|_Y}\over{\|h\|_X}} = 0.$$Let $M_n$ be the space of $n \times n$ real-valued matrices and define $F: M_n \to M_n$ by $F(m) := m^3$. How do I see that $F$ is differentiable at every $m \in M_n$? What is its derivative?

• Please add your thoughts/partial work when you ask a question, in order for us to help you more specifically. – Empiricist Dec 10 '15 at 6:27

## 1 Answer

Hint :-

$||(M+h)^3-M^3||\ =\ ||(M^2+Mh+hM+h^2)(M+h)-M^3||\ =\ \\ ||M^2h+MhM+Mh^2+hM^2+hMh+h^2M+h^3||$.

Since the derivative is a Linear Map so find the terms which are "Linear" in $h$ and bound the rest of the terms using the condition that $||h||\rightarrow 0$.