Let $\;f:\mathbb R^n \rightarrow \mathbb R^m\;$ and $\;G:\mathbb R^m \rightarrow \mathbb R_{+}\;$ and consider the $\;n\times n\;$ tensor $\;\mathcal A=(a_{ij})_{1\le i,j \le n}\;$ where $\;a_{ij}=f_{x_i} \cdot f_{x_j} -{\delta}_{ij}(\frac{1}{2} {\vert \nabla f \vert}^2+G(f))\;$
NOTE: $\; \cdot \;$stands for the Euclidean inner product and $\;\vert \cdot \vert\;$ is the Frobenius Norm of the matrix.
I want to prove that the entries of this tensor will be like:
$\; a_{11}=(f_{x_1})^2-1(\frac{1}{2} (f_{x_1})^2+\dots+\frac{1}{2}(f_{x_n})^2+G(f))\;$, $\;a_{12}=f_{x_1} \cdot f_{x_2}\;$, etc.
My attempt:
Since $\; \nabla f =(\frac{\partial f_i}{\partial x_j})_{1\le i \le m, 1\le j \le n}\;$, I computed the Frobenius norm of $\; \nabla f\;$ and I found $\;\frac{1}{2} {\vert \nabla f \vert}^2=\frac{1}{2} (f^1_{x_1})^2+\dots+\frac{1}{2}(f^1_{x_n})^2+\dots+\frac{1}{2}(f^m_{x_1})^2+\dots+\frac{1}{2}(f^m_{x_n})^2\;$ where $\;f^i_{x_j}=\frac{\partial f_i}{\partial x_j}\;$.
In addition, I know $\;{\delta}_{ij}=\begin{cases} 1\;if\;i = j\\ 0\;if\;i\neq j\\ \end{cases}\;$ Writing down all the above, I get:
- $\;a_{11}=(f_{x_1})^2-1(\frac{1}{2} (f^1_{x_1})^2+\dots+\frac{1}{2}(f^1_{x_n})^2+\dots+\frac{1}{2}(f^m_{x_1})^2+\dots+\frac{1}{2}(f^m_{x_n})^2+G(f))\;$
- $\;a_{12}=f_{x_1} \cdot f_{x_2}\;$
- etc.
My question:
I think I'm missing something but I don't know what! Are the above calculations right or wrong?
Any help would be valuable because I've been stuck here for days...
Thanks in advance!