Independence of the choice of base for the differential.

Let $f:\mathbb{R}^m\longrightarrow \mathbb{R}^n$ be a mapping, defined by differentiable functions which, generally speaking, are non-linear and map zero into zero: $$f(x_1,\cdots,x_m)=(\cdots,f_i(x_1,\cdots,x_m),\cdots), i=1,\cdots,n.$$ $$f_i (0,\cdots,0)=0$$ Associate with it the linear mapping $df_0:\mathbb{R}^m\longrightarrow \mathbb{R}^n$, called the differential of $f$ at the point $0$, according to the formula $$(df_0)(e_j)=\sum_{i=1}^n\frac{\partial f_i} {\partial x_j}(0)e'_i,$$ Where $\{e_j\}$, $\{e'_j\}$ are standard bases of $\mathbb{R}^m$ and $\mathbb{R}^n$. Show that if the bases of the spaces $\mathbb{R}^m$ and $\mathbb{R}^n$ are changed and $df_0$ is calculated using the same formulas in the new bases, then the new linear mapping $df_0$ coincides with the old one.

I want to show this without the matrix notation, but inspired by this I know that: $$[\bar{df_0}]=[id]_{B'}^{C'}[df_0][id]_C^B,$$ Where $C$ and $C'$ are the standard bases of $\mathbb{R}^m$ and $\mathbb{R}^n$, and $B$ and $B'$ are the new one. And $\bar{df_0}$ is the new linear mapping. But I am struggling to show it even to a particular case where the function is $f(x,y,z)=(x+y,xz-y)$. Could anyone give me any hints?

Obs.: $[id]_{X}^{Y}$ chance the base $Y$ to $X$.

$$\frac{\partial f_1} {\partial x}(x,y,z)=1\Rightarrow\frac{\partial f_1} {\partial x}(0)=1;\frac{\partial f_1} {\partial y}(x,y,z)=1\Rightarrow\frac{\partial f_1} {\partial y}(0)=1;$$ $$\frac{\partial f_1} {\partial z}(x,y,z)=0\Rightarrow\frac{\partial f_1} {\partial z}(0)=0.$$ $$\frac{\partial f_2} {\partial x}(x,y,z)=z\Rightarrow\frac{\partial f_2} {\partial x}(0)=0;\frac{\partial f_2} {\partial y}(x,y,z)=-1\Rightarrow\frac{\partial f_2} {\partial y}(0)=-1;$$ $$\frac{\partial f_2} {\partial z}(x,y,z)=x\Rightarrow\frac{\partial f_2} {\partial z}(0)=0.$$ The base changes are, considering $B=\{b_1,b_2,b_3\}=\{(1,1,1),(1,1,0),(1,0,0)\}$ e $B'=\{b'_1,b'_2\}=\{(1,1),(1,0)\}$: $$[id]^{C'}_{B'}=\begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}, [id]^{B}_{C}=\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ As we know it $[df_0]^C_{C'}=\begin{bmatrix} 1 & 1 & 0 \\ 0 & -1 & 0 \end{bmatrix}$ we get $[\bar{df_0}]^B_{B'}=[id]^{C'}_{B'}[df_0]^C_{C'}[id]^{B}_{C}=\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & -1 \end{bmatrix}$. As we compare $\bar{df_0}$ with $df_0$ we see that $\bar{df_0}(x,y,z)$ coincide with $[\bar{df_0}]^B_{B'}[(x,y,z)]$, but you just have to make the elements of the base. Thus: $$[\bar{df_0}]^B_{B'}[b_1]=\begin{bmatrix} 1 \\ -1 \end{bmatrix},$$ but: $$\bar{df_0}(b_1)=\frac{\partial f_1} {\partial x}(0)b'_1+\frac{\partial f_2} {\partial x}(0)b'_2=b'_1=(1,1).$$ Ie do not match and I do not know what I'm missing!

• I am not sure about your question. Are asking how $d\bar f_0$ is related to $df_0$ if $\bar f(y) = (B')^{-1} f(By)$ holds for every $y\in\mathbb R^m$? – user251257 Aug 26 '15 at 0:35
• $\bar{df_0}$ is just a notation for $df_0$ computed according to the formula on bases $B$ and $B'$. I want to prove that $\bar{df_0}$ coincide with $df_0$ (the differential computed according to the formula on standar bases). – donikvep Aug 26 '15 at 12:55
• Could you explain "$d\bar f_0$ is just a notation for $df_0$ computed according to the formula on bases $B$ and $B′$"? – user251257 Aug 26 '15 at 13:06
• $(\bar{df_0})(b_j)=\sum_{i=1}^n\frac{\partial f_i} {\partial x_j}(0)b'_i,$ – donikvep Aug 26 '15 at 16:10
• then you have $d\bar f_0 = B' d f_0 B^{-1}$ – user251257 Aug 26 '15 at 17:19