# Is a linear operator equal to it's jacobian?

In the course of proving something else, I was stuck and someone showed me that if the matrix which describes a linear operator $f: \Bbb R^n \to \Bbb R^n$ (and maybe $\Bbb C^n \to \Bbb C^n$?) is the jacobian matrix $J(f)(\mathbf x)$ wrt some basis, then the result I was looking for followed trivially. At first that seemed like it couldn't be the case, but then I thought that the Jacobian is basically the derivative, and the derivative is the best linear approximation -- and $f$ is already linear. So maybe $[f] = J(f)$?

I just don't see a how $$\begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} & \frac{\partial f_1}{\partial z} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} & \frac{\partial f_2}{\partial z} \\ \frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y} & \frac{\partial f_3}{\partial z}\end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} \frac{\partial f_1}{\partial x}x + \frac{\partial f_1}{\partial y}y + \frac{\partial f_1}{\partial z}z \\ \frac{\partial f_2}{\partial x}x + \frac{\partial f_2}{\partial y}y + \frac{\partial f_2}{\partial z}z \\ \frac{\partial f_3}{\partial x}x + \frac{\partial f_3}{\partial y}y + \frac{\partial f_3}{\partial z}z\end{pmatrix}$$

should equal $$\begin{pmatrix} f_1(x,y,z) \\ f_2(x,y,z) \\ f_3(x,y,z) \end{pmatrix}$$

Is this even true? If so, how would I prove it?

• This will not be true in general, but will be valid for functions which are homogeneous of degree 1 i.e. $f(\lambda x)=\lambda f(x)$. See Euler's homogeneous function theorem. – Semiclassical Dec 10 '14 at 3:14
• @Semiclassical $f$ is a linear operator, so it is both homogeneous of degree $1$ and additive. – user199053 Dec 10 '14 at 3:15
• Point. Then yes, it's true, and can be proven as an exercise in partial derivatives. – Semiclassical Dec 10 '14 at 3:17
• @Semiclassical I couldn't follow the proof in that link, but this one is much better. Thanks for showing me that theorem. I had never seen it before. :) – user199053 Dec 10 '14 at 3:20

Your idea about the derivative's being the best linear estimate is good (and leads to a proof, though you might need a bit more machinery than a standard multivariable calculus class would provide). For an explicit proof, write $f(x) = Ax$ for some matrix $A$, so that $y = f(x)$ has $y_i = A_{ij} x_j$. Then clearly $\frac{\partial y_i}{\partial x_j} = A_{ij}$. Unwinding the notation, we get $J(f) = (A_{ij}) = A = f$ itself.