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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!

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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

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  • $\begingroup$ First of all, thanks a lot for your quick answer! I cannot see why $\;\mathcal A\;$ can be written in the form you suggested. I don't understand why $\;ff^T\;$ follows from $\;\frac {\partial f}{\partial x_i}\cdot\frac {\partial f}{\partial x_j}\;$ $\endgroup$
    – kaithkolesidou
    Commented Aug 22, 2017 at 11:10
  • $\begingroup$ @kaithkolesidou Oh I am sorry, I thought $f_{x_i}$ just meant the $i$-th element and not the $i$-th partial derivative. The issue with the latter notation is : partial derivative of which component? That is, writing $f_{x_i}$ suggest you are indexing a vector (since you are using only one index). But the collection of the partial derivatives of the vector $f$ is a matrix (and you write it out correctly when computing the frobenius norm) , so this is a little confusing. Could you clarify it? :) $\endgroup$
    – Ant
    Commented Aug 22, 2017 at 11:20
  • $\begingroup$ @kaithkolesidou My point was that in general a matrix in the form $a_{i,j} = f_i f_j$ (where $f$ is a vector, being indexed by a single index) can be written as $$a = ff^T$$ $\endgroup$
    – Ant
    Commented Aug 22, 2017 at 11:22
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    $\begingroup$ well, you shouldn't be sorry! I should have been more specific...This exactly is my problem here. Since $\;f\;$ is a matrix, what does $\;f_{x_i}\;$ stand for? I think I'll ask my professor..However, you've been very helpful!:) $\endgroup$
    – kaithkolesidou
    Commented Aug 22, 2017 at 11:32
  • $\begingroup$ @kaithkolesidou you're welcome! :) $\endgroup$
    – Ant
    Commented Aug 22, 2017 at 12:08

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