I am working on the following question and have managed to do most of it but unfortunately I am getting stuck on the last bit. I think part of my confusion lies with calculating second determinants (which are linear maps to a space of linear maps) of matrices (which are themselves linear maps). This is the question:
Let $V = M_{n×n}(\mathbb{R})$. By considering $\det(I + A)$ as a polynomial in the entries of
$A$, show that the function $\det : V → \mathbb{R}$ is differentiable at the identity matrix $I$ and that
its derivative there is the function $A → \text{tr} A$. Hence show that $\det$ is differentiable at any
invertible matrix $X$, with derivative $A → \det X \text{tr}(X^{−1}A)$. Compute the second derivative
of det at $I$ as a bilinear map $V × V → \mathbb{R}$, and verify it is symmetric.
I have done everything except for the last bit. However whatever I try for the last bit is not fruitful. I know that the second derivative at $I$ is a bilinear form whose matrix has as its entries the values of the second partial derivatives $D_{ij}(I)$ but I don't see how to compute the second partial derivatives. I know what the first partial derivatives are though as I worked them out for the first part of the question (at $I$ they are the trace of the matrix that represents the $i^{th}$ basis vector).
I also see how we can view the partial derivative as a function of $X$ but I don't see how to extend that to work out the limit $D_{ij}f(I) = \lim\limits_{t \to 0} \frac{D_{\mathbf{e}_j}f(I+t\mathbf{e}_i)-D_{\mathbf{e}_j}f(I)}{t}$
I don't think the solution to the first bit is important for the last part so I am not posting it up, however if you wish to see it please let me know and I will add it.
Thanks for all your help.
Best Answer
You have the first derivative, $D\det(X)(A) = (\det{X})\operatorname{tr}{(X^{-1}A)} $, so you want to find $D^2\det(X)(A,B) = D(D\det(X)(A))(B)$. You can do this using the multivariable product and chain rules, provided you know the first derivative of $X \mapsto X^{-1}$ at $X=I$. We have $$ (X^{-1}+\varepsilon K)(X+\varepsilon B) = I+\varepsilon (KX+X^{-1}B)+O(\varepsilon^2), $$ which suggests we put $ K=-X^{-1}BX^{-1} $. Then we have $$ (X+\varepsilon B)\left( (X+\varepsilon B)^{-1}-(X^{-1} - \varepsilon X^{-1}BX^{-1}) \right) = I -(I+\varepsilon BX^{-1})-\varepsilon BX^{-1}+O(\varepsilon^2) = O(\varepsilon^2), $$ and it follows that $DX^{-1}(B)=-X^{-1}BX^{-1}$.
Then the product rule gives $$ D((\det{X})\operatorname{tr}{(X^{-1}A)})(B) = D(\det{X})(B) \operatorname{tr}{(X^{-1}A)} + (\det{X})D(\operatorname{tr}{(X^{-1}A)})(B), $$ and trace is linear so the derivative passes inside and we find $$ D((\det{X})\operatorname{tr}{(X^{-1}A)})(B) = (\det{X}) \big( \operatorname{tr}{(X^{-1}B)} \operatorname{tr}{(X^{-1}A)} - \operatorname{tr}{(D(X^{-1})(B)A)}) \big) \\ = (\det{X}) \big( \operatorname{tr}{(X^{-1}B)} \operatorname{tr}{(X^{-1}A)} - \operatorname{tr}{(X^{-1}BX^{-1}A)}) \big). $$ This is symmetric because the trace is cyclic.