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I am interested in the matrix fractional function (from $GL_n(\mathbb{R})$ to $\mathbb{R}$) : $S \mapsto x^\top S^{-1} x$, where $x \in \mathbb{R}^n$ is fixed.

By computing $f(S + E) - f(S)$ for a small $E$ I am able to show that its gradient at $S$ is $-S^{-1} x x^\top S^{-1}$, but I am stuck when it comes to computing its Hessian (which is a fourth order tensor). Any hint ?

EDIT: Pushing the Taylor expansion leads me to: \begin{align}x^\top(S + E)^{-1} x - x^\top S ^{-1} x &= x^\top(Id + S^{-1} E)^{-1} S^{-1} x -x^\top S^{-1} x \\&= x^\top(Id - S^{-1} E + S^{-1} E S^{-1}E) S^{-1} x -x^\top S^{-1} x +o(||E||^2) \\&= -x^\top S^{-1} E S^{-1} x + x^\top S^{-1} ES^{-1} E S^{-1} x \end{align}

and $-x^\top S^{-1} E S^{-1} x = - tr _,x^\top S^{-1} E S^{-1} x = - tr \, S^{-1} x x^\top S^{-1} E$ which gives me the gradient, and the expression of the second order term, but not the Hessian.

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  • $\begingroup$ Nice exercice see the answer below $\endgroup$ – Guy Fsone Oct 15 '17 at 0:39
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Use the Chain rule $$f(S) = g\circ \mathcal{I}(S)$$ with $\mathcal{I}(S) =S^{-1}$$ and $g(S) =x^\top Sx$ and see that, $$g: S\mapsto x^\top Sx $$ is linear so $Dg(S)(E) =x^\top Ex$.

We know that $$\mathcal{DI}(S)(V) = -S^{-1}VS^{-1}$$

Hence, for $H $ fixed,

$$\mathcal{D}f(S)(H) =\mathcal{D}g \circ \mathcal{DI}(S)(H) =\color{red}{x^\top S^{-1}HS^{-1} E x :=h\circ \mathcal{I}(S)}$$

Where $$h: A\mapsto -x^\top AHA x$$ Observe that $h$ is quadratic map hence

$$\mathcal{D}h(A)(K) = -x^\top KHA x-x^\top AHK x$$

Therefore, $$\mathcal{D}^2f(S)(H)(K) =\mathcal{D}h \circ\mathcal{DI}(S)(K) =-x^\top [\mathcal{DI}(S)(K)]HS x-x^\top SH [\mathcal{DI}(S)(K)] x \\ = \mathcal{D}^2f(S)(H)(K) = x^\top SH S^{-1}KS^{-1} x+ x^\top S^{-1}KS^{-1}HS x . $$

i.e

$$\color{red}{\mathcal{D}^2f(S)(H)(K) = x^\top SH S^{-1}KS^{-1} x+ x^\top S^{-1}KS^{-1}HS x }$$

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  • $\begingroup$ I was able to obtain this hormula by using a 2nd order development of f(S + Epsilon) and (Id + E)^-1 = Id - E + E^2 +o(E^2). However, this does not give me the expression of the Hessian directly. In fact, my true goal is to show that the Hessian is pos def. $\endgroup$ – P. Camilleri Oct 15 '17 at 8:51
  • $\begingroup$ Pleas by Hessian you mean the matrix? If yes I then I am curious to see how you did . since here we are in infinite dimensional space. I think the best way to give the hessian is ti give his quadratic form $\endgroup$ – Guy Fsone Oct 15 '17 at 9:09
  • $\begingroup$ I mean the 4th order tensor, since it's the Hessian of a function whose argument is a matrix. What do you mean, infinite dimension? See my edit for the computation. $\endgroup$ – P. Camilleri Oct 15 '17 at 11:43
  • $\begingroup$ Ok I see but it will very difficult to write the explicit expression of the matrix. But rather you have the quadratic associate with. It is enough to know everything about the Hessian matrix $\endgroup$ – Guy Fsone Oct 15 '17 at 11:53
  • $\begingroup$ Is it enough to show that the Hessian is positive definite everywhere ? $\endgroup$ – P. Camilleri Oct 15 '17 at 12:22
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Note that $x\rightarrow 1/x$ is convex only when $x>0$. Thus, the correct function to study is $f:A\in S^+\rightarrow x^TA^{-1}x$ where $S^+$ is the set of the $n\times n$ symmetric $>0$ matrices. Recall that $S^+$ is a convex open subset of $S$, the set of symmetric matrices..

Proposition. $f$ is convex.

Proof. There is a mistake in the calculation of Guy Fsone. Indeed $D^2f_A(H,H)=2x^TA^{-1}HA^{-1}HA^{-1}x$ for every $H\in S$ (recall that $S$ is the tangent space of $S^+$).

Note that $A^{-1}HA^{-1}HA^{-1}=(A^{-1}HA^{-1/2})(A^{-1}HA^{-1/2})^T$ is a symmetric $\geq 0$ matrix. Then, for every $(A,H)$, $D^2f_A(H,H)\geq 0$ and $f$ is convex. $\square$

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