Making sense of matrix derivative formula for determinant of symmetric matrix as a Fréchet derivative? I am familiar with the derivation of $D_A(\det(A)) := \dfrac{\partial \det(A)}{\partial A}$ as a Fréchet derivative by considering $$D_A[H] = \det(A+H) - \det(A) \; as \; \|H \| \rightarrow 0 = \det(A) \mathrm{tr} (A^{-1}H) = \det(A) A^{-T} \cdot H$$ for a general square matrix $A$. I have used the Frobenius inner product for writing the trace as an inner product. This leads to the same answer as that given in The Matrix Cookbook.
But if $A$ is symmetric , how do I compute the Fréchet derivative  to be $$ \det(A) (2A^{-1} - \operatorname{diag}(A^{-1}) )$$
I am curious to know if we can derive this formula  that is quoted from The Matrix Cookbook in various posts, see for example --
What is the derivative of the determinant of a symmetric positive definite matrix?
 A: This doesn't look right to me.
Let $J = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, $A=A^{-1} = J$, $H=J$.
Then $D \det (A)(H) = -2$, whereas the other formula would give $-4$ (since $A$ is zero on the diagonal).
Commentary:
Let me abuse notation and use $S^n$ to denote the set of symmetric
$n \times n$ matrices.
The Fréchet derivative is the best linear approximation to a function
at a given point, it is unique and does not depend on any basis. In
particular, the if the given point lies in some subspace (for
example, $S^n \subset \mathbb{R}^{n \times n}$),
the derivative is the same. There may be some simplifications in
the formula that depend on characteristics of the subspace, but the
resulting derivative will be the same.
Since the Cookbook formula gives a value that differs from the
derivative, it cannot be the Fréchet derivative of the map
$\det : S^n \to S^n$.
My understanding of what the Cookbook means by structured
matrices is that
there is some map $p$ from $\mathbb{R}^{n \times n}$ into the
collection of structured matrices, and the (Fréchet) derivative
of the composition $\det \circ p$, evaluated
at a point of $ S^n$, is what the Cookbook gives.
For symmetric matrices, a parameterisation that is consistent with
the Cookbook is $p:\mathbb{R}^{n \times n} \to S^n$,
$p(X) = X+X^T - \operatorname{diag} X$, which would give rise to
the formula (using the chain rule, linearity & symmetry of $p$ and properties of $\operatorname{tr}$):
\begin{eqnarray}
D (\det \circ p) (X)(H) &=& D \det(p(X)) ( D p(X)(H)) \\
&=& D \det(p(X))( p(H)) \\
&=& \det(p(X)) \operatorname{tr} ( p(X)^{-1} p(H)) \\
&=& \det(p(X)) \operatorname{tr} ( p(X)^{-1} (H+H^T - \operatorname{diag} H)) \\
&=& \det(p(X)) \operatorname{tr} ( 2p(X)^{-1} H - p(X)^{-1} \operatorname{diag} H)) \\
&=& \det(p(X)) \operatorname{tr} ( 2p(X)^{-1} H - \operatorname{diag} (p(X)^{-1})H) \\
&=& \det(p(X)) \operatorname{tr} ( (2p(X)^{-1} -\operatorname{diag} (p(X)^{-1}) H) \\
\end{eqnarray}
Note, however, this is the 
Fréchet derivative
of the composition $\det \circ p$, not the Fréchet derivative
evaluated at a point of $S^n$ in the direction $H$. The parameterisation is very relevant.
In terms of the example above, we have 
$D \det (J)(J) = -2$, but $D (\det \circ p) (J) (J) = -4$.
The reason for the extra factor of $2$ is that $p(J) = 2J$ since
the parameterisation $p$ doubles the non diagonal elements of
the perturbation $p$.
That is, $\det(J+H) \approx -1 -2 \langle J, H \rangle$ but
$(\det \circ p)(J+H) \approx -2 -4 \langle J, H \rangle$, so while both are Fréchet derivatives, they are
derivatives of different functions evaluated at different points.
