Trouble understanding ij-element of a matrix 
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!

Yes it is correct. This is what the author meant. Note that the matrix $\mathcal A$ can also be written in more compact form as 
$$\mathcal A = ff^T - (\frac 12 \left |\nabla f\right| + G(f))I$$
where $I$ is the identity matrix
