Derivative of matrix Bilinear map Let A be an invertible  $n\times n$matrix. Define a function $ F:\mathbb{R}^n \rightarrow \mathbb{R}^n$ by $F(x,y)=<Ax,y>$. $<x,y>$ denotes inner product.Let DF(x,y) be derivative of F at (x,y) . Then


*

*If $x\ne 0$ then $DF (x,0) \ne 0$

*If $y\ne 0$ then $DF(0,y) \ne 0$

*If $(x,y) \ne (0,0)$, then $DF(x,y) \ne 0$

*If $x=0$ or $y=0$, then $DF(x,y)=0$
This question is about derivative of a bilinear form. But I know that derivative at (x,y) evaluated at some point (h,k) will be F(x,k)+F(h,y).
But just derivative at (x,y) will be a linear tranformation. How to tackle this.

The Frechet derivative at $(h,k)$ is $DF(h,k)=\langle Ax,k\rangle+\langle Ah,y\rangle.$
Then, $DF(x,0)(h,k)=\langle Ax,k\rangle$ so clearly $DF(x,y)$ is not the $0\  \textit{transformation}$  if $x\neq 0.$
The others are proved similarly, except $4.$ should be $DF(0,0)=0.$
