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Given the linear transformation $f$, we can construct the matrix $A$ as follows: on the $i$-th column we put the vector $f(\mathbf e_i)$ where $E = (\mathbf e_1, \ldots, \mathbf e_n)$ is a basis of $\mathbb R^n$.

Now, I read that for linear transformations, if we use the canonical basis of $\mathbb R^n$, the matrix $A$ is equal to $J f$ (the Jacobian matrix associated to $f$). It's indeed so for the few examples that I tried, but I cannot find a proof of this fact.

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2 Answers

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If $f:\mathbb{R}^n\to\mathbb{R}^p$ is linear, then you know that for all $a\in\mathbb{R}^n,$ you have that $Df(a)=f.$ As $Jf(a)$ is the matrix which represents $f$ in the canonical bases $\mathcal{C}_n$ and $\mathcal{C}_p$ of $\mathbb{R}^n$ and $\mathbb{R}^p,$ you will have that $$Jf(a)=[f]_{\mathcal{C}_p\mathcal{C}_n}=A,$$ with your definition of $A.$

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If I understand you correctly then what you are trying to prove is not true. The Jacobian of f is the best linear approximation to f at a given point. It is equal only if the function f happens to be linear. That is exactly the same as saying the derivative of f(x)= ax is the constant, a.

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