Why is the Jacobian matrix equal to the matrix associated to a linear transformation? Given the linear transformation f, we can construct the matrix A as follows: on the i-th column we put the vector f(ei) where E=(e1,⋯,en) is a basis of Rn.

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Why is the Jacobian matrix equal to the matrix associated to a linear transformation?  Given the linear transformation f, we can construct the matrix A as follows: on the i-th column we put the vector f(ei) where E=(e1,⋯,en) is a basis of Rn.

search633504 · Accepted Answer

Step 1  If f:Rn→Rp is linear, then you know that for all a∈Rn,  you have that Df(a)=f. As Jf(a) is the matrix which represents f in the canonical bases Cn and Cp of Rn and Rp, you will have that  Jf(a)=[f]CpCn=A,  with your definition of A.

Darrell Boone · Accepted Answer

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.

Why is the Jacobian matrix equal to the matrix associated

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