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It is well known how to take the derivative of the determinant: let $A(s)$ be a family of square matrices smoothly parametrised by the variable $s$ (in other words, $A:\mathbb{R}\to \mathbb{R}^{N^2}$ is a smooth curve). Then the derivative of the determinant can be given by

$$ \frac{d}{ds} \det A(s) = \operatorname{Tr} \left(\tilde{A}(s) A'(s)\right)$$

where $A'(s) = \frac{d}{ds}A(s)$ and $\tilde{A}(s)$ is the adjugate matrix of $A(s)$. The above formula also generalises to the case that $A:\mathbb{R}^d\to \mathbb{R}^{N^2}$ is a smooth, $d$-parameter family of square matrices.

On the other hand, I have not come across a nice expression for the second derivative (Hessian) of the determinant of such a family. Just by using Leibniz rule, one term is obvious: $\operatorname{Tr}\left(\tilde{A}(s) A''(s)\right)$. However, I don't know of any nice expression of the derivative of the adjugate. Evaluating the adjugate component wise, we see that it should be linear in $A'(s)$.

So the question is: does anyone know of a nice form to present the derivative of the Adjugate matrix, or a nice form to present the Hessian of the determinant?

For my purpose, an answer to the following weaker form of the problem would also be satisfactory:

Let $A:\mathbb{R}^d\to \mathbb{R}^{N^2}$ be a smooth $d$-parameter family of symmetric matrices. Suppose $\det(A(0)) = 0$. Now, if we know that the kernel of $A(0)$ as dimension at least 2, then by a diagonalisation argument, the adjugate matrix $\tilde{A}(0)$ must vanish identically. And therefore the Hessian is determined by the term with the derivative of the adjugate. Then under the assumptions that $d \geq 2$ and $A(0)$ has nullity at least 2: (a) is it possible for the Hessian to be non-degenerate? (b) If so, what are some sufficient conditions to guarantee nondegeneracy?

(In the case $d = 1$ it is clear that the "Hessian" can be non-degenerate, just by taking $N = 2$ and $A(s) = s I$.)

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google provided which has a section on the derivative of the adjugate – yoyo Jul 8 '11 at 22:15
Ah, taking the limit from the non-singular case. That's an idea. Thanks. – Willie Wong Jul 9 '11 at 2:00

I'm not sure if this is as "nice" as a form as you had in mind, but it's pretty straightforward (conceptually, though maybe not symbolically). You just write the adjugate matrix as a matrix of determinants of $A$'s cofactors. i.e. $ \tilde{A} = [ \det M_{kl} ] $, and then apply the same rule you used the first time. Hence if

$$ \frac{\partial}{\partial x_i} \det A = \mathrm{Tr}\left( [ \det M_{kl} ] \frac{\partial A}{\partial x_i} \right)$$


$$ \frac{\partial^2}{\partial x_i \partial x_j} \det A = \mathrm{Tr}\left( \tilde{A} \frac{\partial^2 A}{\partial x_i \partial x_j} + \left[ \mathrm{Tr}\left( \tilde{M}_{kl} \frac{\partial M_{kl}}{\partial x_j} \right) \right]_{kl} \frac{\partial A}{\partial x_i} \right) $$

Note this expression includes the cofactor matrices' adjugate matrices. As indicated by the subscript, the indices $ k $ and $ l $ are held constant inside the square brackets.

Therefore the answer to your question (a) is "yes, it is possible for the Hessian to be nondegenerate" and as for (b), figuring and then articulating sufficient conditions would still be a pretty involved problem given the complexity of this formula.

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I should've probably clarified that I am only interested in the case where $d \geq 1$ and $A(0)$ has nullity 2. Let me edit that in the question. – Willie Wong Jul 9 '11 at 11:55
up vote 1 down vote accepted

To implement what anon wrote: take a basis in which $A(0)$ is diagonalized.

Then it is easy to check that $\partial \tilde{A}_{ij} = \partial \det A_{\hat{i}\hat{j}}$ where $A_{\hat{i}\hat{j}}$ is the submatrix with the $i$th row and $j$th column knocked out. Using that at least two of $A(0)$'s eigenvalues are 0, we see that for $i\neq j$, $A_{\hat{i}\hat{j}}$ has rank at most $N-2$, so has vanishing adjugate, and so $\partial\tilde{A}_{ij} = 0$.

In other words, $\partial\tilde{A}(0)$ is still a diagonal matrix. Now, if the nullity of $A(0)$ is strictly bigger than 2, then any submatrix $A_{\hat{i}\hat{j}}$ has nullity at least 2, and hence $\partial \tilde{A}(0) = 0$. And in this case $\partial^2 \det A (0) = 0$.

Restricting to the case $A$ has rank $N-2$, we can without loss of generality assume that $A_{11}(0) = A_{22}(0) = 0$. Then we have that for $i > 2$, $\partial\tilde{A}_{ii}(0) = 0$ again.

Computing $\partial\tilde{A}_{11}$, we see that it is equal to $$ \partial \det A_{\hat{1}\hat{1}}(0) = \partial A_{22}\prod_{i = 3}^N A_{ii}(0) $$

So in fact we have in this case $$ \partial^2_{\mu\nu} \det A(0) = (\partial_{\mu} A_{11} \partial_\nu A_22 + \partial_\nu A_{22} \partial_\mu A_{11}) \prod_{i = 3}^N A_{ii} $$ which actually answers completely the weak version of my question.

Observe that by the above formula, the Hessian of such a determinant matrix is rank at most 2. Therefore if the dimension $d\geq 3$, the Hessian must be degenerate (determinant 0). If $d = 1$, the necessary and sufficient condition for the Hessian to be non-degenerate is that $A_{11}'A_{22}' \neq 0$. In the case $d = 2$, the necessary and sufficient condition for non-degenerate Hessian is that $\nabla A_{11}$ and $\nabla A_{22}$ are both non-vanishing, and are linearly independent.

For the question in the title, in the case where $\det A \neq 0$, the comment given by yoyo gives a good description. Writing $\tilde{A} = (\det A) A^{-1}$, we can actually derive

$$ \partial^2_{\mu\nu} \det(A) = \operatorname{Tr} (\tilde{A}\partial^2_{\mu\nu}A) + \frac{1}{\det A} \left( \operatorname{Tr} X_\mu \operatorname{Tr} X_\nu - \operatorname{Tr}(X_\mu X_\nu)\right) $$

where $X_\mu = \tilde{A}\partial_\mu A$.

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