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I know that $f(X) = \log \det X$ is concave on domain $S^n_{++}$, but what is the Hessian of f(X)? Is there any book I can refer to?

YF Yan
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  • Book? What for? Just calculate the partial derivatives of $\log{\sqrt{(x_1)^2 + \ldots + (x_n)^2}}$, then calculate the partial derivatives of those partials, and write down the Hessian matrix. – user539887 Apr 02 '18 at 08:41
  • I mean that X is a positive definite matrix not a vector. I am wondering the Hessian is a matrix or a 4-dimensional tensor. I know little about tensor or matrix calculus. – YF Yan Apr 02 '18 at 08:49
  • But you are looking for the Hessian of a real function defined on a subset of $\mathbb{R}^{n^2}$, namely the logarithm of the norm of a matrix. The gradient of that function is an $n^2$-dimensional vector, and the Hessian is some $(n^2 \times n^2)$-matrix. And if the norm you mean is the Euclidean norm, that is, $$ \lvert X \rvert = \Bigl(\sum\limits_{i = 1}^{n} \sum\limits_{j= 1}^{n} (x_{ij})^2\Bigr)^{1/2},$$ then just find the partials, etc. – user539887 Apr 02 '18 at 09:00

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I think people are confused because in your notation $$det(X)= |X|$$

Hint: To solve this we apply the chain rule. You have composed det with log.

The derivative of $\log$ is $d(\log(u))_y(h)=h/y$

The derivative $d(det(U))_{X}(H)= Tr(adj(X)XH)$

Applying chain rule gives:

$$ d(\log(det(U))_X(H)= \frac{1}{det(X)}Tr(adj(X)XH)$$

Now to complete you want to differentiate this expression with respect to X in the same way which I let you try, you will get a linear map in two matrices(it’s exactly the same idea), H and say K.

If you want a Hessian then you want to think about this as a map from $\mathbb{R}^{n^2}$ to $\mathbb{R}$ so plug in matrices H and K which form a natural basis (I.e. matrix with zero everywhere but position m,n) and read off the second order partials.

Edit:

To clarify, this second derivative is a set of $n^4$ numbers, I have suggested you arrange them as a hessian matrix by thinking of this as a map from $\mathbb{R}^{n^2}$ to $\mathbb{R}$. Note it is not a tensor as it won’t transform like one.

Mathmop
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