Let $A$ be a $n \times n$ invertible matrix.Define a function $F:\mathbb{R}^{n} \times \mathbb{R}^{n} \rightarrow \mathbb{R}$ by $F(x,y)=\langle Ax,y\rangle$. Let $DF(x,y)$ denotes the derivative of $F$ at $(x,y)$ which is a linear transformation from $\mathbb{R}^{n} \times \mathbb{R}^{n} \rightarrow \mathbb{R} $. Then
1) if $x\neq 0$,then $DF(x,0)\neq 0$
2) if $y\neq 0$,then $DF(0,y)\neq 0$
3) if $(x,y)\neq (0,0)$,then $DF(x,y)\neq 0$
4) if $x=0 \; or \;y=0$,then $DF(x,y)= 0$
I don't even know how to start the solution and also how to take derivative in higher dimensions.
Please help!
Thank you.
Total derivatives are linear maps, and a chain rule holds. If $F(x,y) =\langle Ax,y\rangle$, then $F$ is bilinear and $$DF(x,y)(h,k) = F(x,k)+F(h,y) = \langle Ax,k\rangle + \langle Ah,y\rangle.$$So:
1) $DF(x,0)(h,k) = \langle Ax,k\rangle$. Since $A$ is invertible and $x\neq 0$, take $k=Ax\neq 0$. So $DF(x,0)(0,Ax)>0$ and this item holds true.
2) Similarly, $DF(0,y)(A^{-1}y,0)>0$ and the conclusion is true.
3) Again, $DF(x,y)(A^{-1}y,Ax)>0$, so the conclusion is true.
4) False by (1) and (2).
To see why $$DF(x,y)(h,k) = F(x,k)+F(h,y),$$check that $$\lim_{(h,k)\to (0,0)} \frac{F(x+h,y+k)-F(x,y)-F(x,k)-F(h,y)}{\|(h,k)\|}=0,$$using that $F$ is bounded (i.e., there is $C>0$ with $\|F(h,k)\|\leq C\|h\|\|k\|$).