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Note: I've seen that essentially the same question title has been posted here, but it does not answer my doubts.

Let $g: \mathbb{R}^n \to \mathbb R^m $ be a differentiable function. Then the total derivative of $g$ at $p \in \mathbb{R}^n $ is the linear function $Dg_p \in L(\mathbb{R}^n , \mathbb{R}^m )$. Therefore it can have a matrix representation once we specify bases for $ \mathbb{R}^n $ and $ \mathbb{R}^m $. Let $\{e_i\}_{i=1}^n, \{f_i\}_{i=1}^m$ be the standard bases of $\mathbb R^n, \mathbb R^m$ respectively.

In wikipedia I've seen that the Jacobian matrix $J$ is the matrix representation of $Dg_p$. There the Jacobian is defined as

$$ [J]_{i,j} = \partial_j g_i, $$

where $\partial _i := \partial/\partial e_i $ and $g_i := e_i^* \circ g $ are the component functions ($e^*_i$ is the dual vector of $e_i$). I'm interested in making the connection between $J$ and $Dg_p$.

By definition of partial derivatives, we know that $\partial _if (p) = Dg_p (e_i) $. Hence it is that the $i$th column of $J$ is simply $\partial _i f (p)$. This is also shown in the Wikipedia page.

However, the wikipedia also shows that

$$ J = \begin{pmatrix} \nabla ^Tg_1 \\ \vdots \\ \nabla ^T g_m \end{pmatrix} ,$$

where, according to Wikipedia, '$ \nabla ^T g_i$ is the transpose of the gradient of the $i$th component'.

I do not understand how this equality came about, neither do I understand how can we reach the statement $J_{ij} = \partial _j g_i $ from simply the fact that $g$ is differentiable at $p$ and its total derivative is $Dg_p$

Afham
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1 Answers1

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Per Definition of the Derivative we have $$ g(p+u) = g(p)+ Dg_pu + o(u). $$ Where i am using the little o notation for the Derivative.

Now take $u = e_j t$ for $t> 0$ and some $j \in \{1, \dots, n\}$. Then (using the above and the linearity of $Dg_p$) $$\lim_{t\to 0} \frac{g(p+e_jt)-g(p)}{t} = Dg_p e_j.$$ Let $i \in\{1, \dots, m\}$. Since $f^*_i$ is continuous and linear we receive: $$ \partial_j g_i (p)= f^*_i Dg_p e_j $$ and so the matrix representation of $Dg_p$ wrt standard basis is the matrix $(\partial_j g_i (p))_{i,j}$, which is the Jacobian.

jd27
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