Derivative of trace squared and trace of a matrix squared?

Question

Going through the matrix cookbook and other resources I can't find an algebraic proof of the following, if $f:\mathbb{R}^{n\times n} \to \mathbb{R}$ and $F:\mathbb{R}^{n\times n} \to \mathbb{R}$ and $H\in\mathbb{R}^{n\times n}$such that $$ f(A) :=\left( \text{tr}[A]^2 \right) \text{ and } F(A):= \left( \text{tr}[A^2]\right). $$ It then asks to calculate $\partial_H f(A)$ and $\partial_H F(A)$ but I can't figure the algebra out and I'm stuck on both.

Answer 1

Most of the calculation relies on linearity of trace.

For the first one, \begin{align*} (D_Hf)(A) &= \lim_{t \to 0} \frac{\operatorname{tr}[A + tH]^2 - \operatorname{tr}[A]^2}{t}\\ &= \lim_{t \to 0} \frac{\left(\operatorname{tr}[A] + \operatorname{tr}[tH]\right)^2 - \operatorname{tr}[A^2]}{t}\\ &= \lim_{t \to 0} \frac{\operatorname{tr}[A]^2 + 2 \operatorname{tr}[A]\operatorname{tr}[tH] + \operatorname{tr}[tH]^2 - \operatorname{tr}[A^2]}{t}\\ &= \lim_{t \to 0} \frac{2 t \operatorname{tr}[A]\operatorname{tr}[H] + t^2\operatorname{tr}[H]^2 }{t}\\ &= 2\operatorname{tr}[A]\operatorname{tr}[H]. \end{align*}

For the second one, \begin{align*} (D_H F)(A) &= \lim_{t \to 0} \frac{\operatorname{tr}[(A + tH)^2] - \operatorname{tr}[A^2]}{t}\\ &= \lim_{t \to 0} \frac{\operatorname{tr}[A^2 + t(AH + HA) + t^2H^2] - \operatorname{tr}[A^2]}{t}\\ &= \lim_{t \to 0} \frac{\operatorname{tr}[A^2] + t\operatorname{tr}[(AH + HA)] + t^2\operatorname{tr}[H^2] - \operatorname{tr}[A^2]}{t}\\ &= \lim_{t \to 0} \frac{t\operatorname{tr}[(AH + HA)] + t^2\operatorname{tr}[H^2]}{t}\\ &=\operatorname{tr}[AH + HA]\\ &=2\operatorname{tr}[AH]. \end{align*}