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I have a function that that takes a matrix and returns a scalar, $f : \mathbb{R}^{m\times n} \rightarrow \mathbb{R}$. I know how to calculate the derivative of this function with respect to the matrix argument. For example, if $f(X) = -\log \det X$, then $\nabla_X f(X) = -X^{-1}$. If $f(X) = tr(X^{-1} Y)$, $\nabla_X f(X) = -X^{-1} Y X^{-1}$ (assuming $X$ and $Y$ are symmetric for now).

What I don't know how to do is compute the second derivative (Hessian?) of $f$ with respect to $X$. I'm assuming it'll be messy, as the result might require four indices, if I'm not mistaken?

Here's the reason I'm curious: I'm trying to prove a function $f : \mathbb{R}^{m\times n} \rightarrow \mathbb{R}$ (specifically $f(X) = \log \det X + tr(X^{-1} Y)$) is convex; one way to do so is to prove that $\nabla^2_X f(X) \succeq 0$ everywhere. At least, this is true for functions $g : \mathbb{R}^n \rightarrow \mathbb{R}$, I haven't been able to find any generalizations of this to matrices, but I think it must be true for the correct notion of "Hessian."

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Your function $f$ is perfectly well defined on the open subset $GL(n,\mathbb{R})$ of $M_n(\mathbb{R})$, consisting of all invertible matrices, but not on $M_n(\mathbb{R})$. Moreover, $GL(n,\mathbb{R})$ is not a convex subset of $M_n(\mathbb{R})$, since the straight line between $I_n$ and $-I_n$ contains $0$, so that you can't use the (generalised) second derivative test to conclude convexity of $f$ on all of $GL(n,\mathbb{R})$... –  Branimir Ćaćić Apr 5 '13 at 8:48
    
Your second $f$ is not convex. Consider the scalar case with $Y=-1$. –  user1551 Apr 5 '13 at 13:07

1 Answer 1

Many times, convexity for these types of functions can be checked by proving convexity of $f(Z+tV)$ in $t$, for arbitrary $Z,V$ (see Boyd & Vandenberghe "Convex Optimization" p.74).

The general Hessian for $\log\det(X)$ would require 4 indices. See p.644 for an approximation.

(http://www.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf).

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