Define $f : M(n \times n,\mathbb R) \to \mathbb R$ by $$f(A) = e^{\mbox{trace}(A)}$$ Calculate $Df(A)H$, where $H \in M(n \times n, \mathbb R)$.
I'm unsure on what $Df(A)H$ actually means. Is it the Frechet derivative of $f$ at $H$? And, if so, how would I go about calculating that?
The notation $Df(A)H$ means the value of the Fréchet derivative of the function $f$ in the point $A$ in the direction $H$, that is, the directional derivative of $f$ in $A$ in the direction $H$: $$ Df(A)H = \lim_{t\to 0} \frac{1}{t}[f(A+tH)-f(A)]. $$ Define $g:\mathbb{R}\to\mathbb{R}$ by $$ g(t) = f(A+tH) = e^{\mbox{trace}(A)+t\mbox{trace}(H)}, $$ and note that $g'(0)=Df(A)H$. Indeed, \begin{align*} g'(0) &= \lim_{t\to 0} \frac{g(t)-g(0)}{t} \\ &= \lim_{t\to 0} \frac{1}{t}[f(A+tH) - f(A)]\\ &= Df(A)H. \end{align*} Now, $$ g'(t) = \mbox{trace}(H)e^{\mbox{trace}(A)+t\mbox{trace}(H)}, $$ thus $$ Df(A)H = g'(0) = \mbox{trace}(H)e^{\mbox{trace}(A)} $$