Proof of $(\mathrm{Df}(X))(H)=XH+HX$

Question

Let $f:\mathrm{Mat}(n,n)\to \mathrm{Mat}(n,n)$ be the squaring map $f(X)=X \cdot X$ then $(\mathrm{Df}(X))(H)=XH+HX$ for $X,H \in \mathrm{Mat}(n,n).$ I tried the general rule, but I'm not suppose to use it here. How am I supposed to prove this?

Answer 1

Well, $$\begin{align} \frac{\| f(X+H) - f(X) - Df(X)(H) \|}{\| H \|} &= \frac{\| (X+H)^2 - X^2 - XH - HX \|}{\| H \|} \\ &= \frac{\| H^2 \|}{\| H\|} \\ &\leq \| H \| \end{align}$$ and, as $\| H \| \to 0$, the latter goes to zero also. So, indeed, $Df(X)$ satisfies the definition of the derivative, and since it is unique, you are done.

Answer 2

Here is a calculation of $(Df(X))(H)$ which is specific to

$f(X) = X^2; \tag 1$

we have

$f(X + H) - f(X) = (X + H)(X + H) - X^2$ $= X^2 + XH + HX + H^2 - X^2 = XH + HX + H^2; \tag 2$

$f(X + H) - f(X) - (XH + HX) = H^2; \tag 3$

$\Vert f(X + H) - f(X) - (XH + HX) \Vert = \Vert H^2 \Vert \le \Vert H \Vert^2; \tag 4$

since

$\dfrac{\Vert H \Vert^2}{\Vert H \Vert} = \Vert H \Vert \to 0 \; \text{as} \; \Vert H \Vert \to 0, \tag 5$

we conclude in accord with the usual definition of $Df(X)(H)$ that

$Df(X)(H) = HX + XH, \tag 6$

as per request.