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$f: M_n(\mathbb{R}) \times M_n(\mathbb{R}) \to M_n(\mathbb{R}) , f(X,Y) = XYX.$

Find $Df(A,B)(X,Y) \forall A,B,X,Y \in M_n(\mathbb{R}).$

My attempt :

$f(X+H, Y+K) - f(X,Y) = (X+H)(Y+K)(X+H)-XYX = XYX + XYH + XKX + XKH +HYX + HKX +HKH -XYX = XYH + XKX + XKH +HYX + HKX +HKH.$

Then $Df(A,B)(X,Y) = XYH + XKX + XKH +HYX + HKX +HKH$ ?

I am not entirly sure , any help is welcome.

Algo
  • 2,322
  • I think there is are some typos where you mixed up $H,K$ and $A,B$. The first line: The l.h.s doesnt depend on $A,B$ but the r.h.s does. Similarly in the bottom line. – Alex Aug 27 '23 at 14:39
  • @Alex fixed , by the way the sign Df is meaning the gradient of f ? – Algo Aug 27 '23 at 14:43
  • By the product rule, $d(XYX)=(dX)YX+X(dY)X+XY(dX)$. In more explicit notation, it means the Frechet derivative of $f$ at $(X,Y)$ along the vector $(\xi,\eta)$ is $Df_{(X,Y)}[(\xi,\eta)]=\xi YX+X\eta X+XY\xi$. If you want to parse more carefully why one can use the product rule as such, see here for the careful statement of the general product rule (although I phrased it there for $\omega$ a bilinear map, one can easily extend to multilinear maps). – peek-a-boo Aug 27 '23 at 16:25
  • Ok, here is a more detailed explanation I’ve written for derivatives of multilinear maps. What is the derivative of the scalar product function?. In your question, $f$ is the composition of the trilinear map $(X,Y,Z)\mapsto XYZ$ with the linear map $(X,Y)\mapsto (X,Y,X)$. So using the derivatives of (multi)linear maps and the chain rule gives you the full explanation. – peek-a-boo Aug 27 '23 at 16:32

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