Fréchet derivatives for $f: \mathbb{R}^{d\times d} \rightarrow \mathbb{R}^{d\times d}_{sim}$

Question

I am starting my studies at Fréchet derivatives and I saw the exercise below:

Let $\mathbb{R}^{d \times d}$ with the usual operator norm. We know it is a Banach space. Let $\mathbb{R}^{d\times d}_{sim} \subset \mathbb{R}^{d\times d}$ be the subspace of the symmetric matrices (those who satifies $A^t=A$) It is also a Banach space with the usual operator norm. Compute the Fréchet derivative of the map $f:\mathbb{R}^{d\times d} \rightarrow \mathbb{R}^{d\times d}_{sim}$ that maps $M \in \mathbb{R}^{d\times d}$ to $M^t\cdot M$.

I attempt to prove using the idea that, since both spaces are finite dimensional, I could see $\mathbb{R}^{d\times d}$ as $\mathbb{R}^{d^2}$ and, so, $Df(x)\cdot h = (D(f[i])(x)\cdot h)_{i=1}^{d^2}$, where this ugly notation means that I could compute the Fréchet-deriviative for each component and, later, join then into the vector $Df(x)$, however, I could not go further. Could anyone give me a hand? I am going completely to the wrong side?

Answer 1

$f(M+H) = f(M)+M^TH+H^TM + H^TH$ so $Df(M)(H) = M^TH+H^TM$.