Could anyone explain in simple words (and maybe with an example) what the difference between the gradient and the Jacobian is?

The gradient is a vector with the partial derivatives, right?

  • 7
    Actually, the gradient is subset of the Jacobian. – cheesyfluff Nov 8 '15 at 19:10

These are two particular forms of matrix representation of the derivative of a differentiable function $f,$ used in two cases:

  • when $f:\mathbb{R}^n\to\mathbb{R},$ then for $x$ in $\mathbb{R}^n$, $$\mathrm{grad}_x(f):=[\frac{\partial f}{\partial x_1}\frac{\partial f}{\partial x_2}\dots\frac{\partial f}{\partial x_n}]|_x$$ is the matrix $1\times n$ of the linear map $Df(x)$ exprimed from the canonical base of $\mathbb{R}^n$ to the canonical base of $\mathbb{R}$ (=(1)...). Because in this case this matrix would have only one row, you can think about it as the vector $$\nabla f(x):=(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2},\dots,\frac{\partial f}{\partial x_n})|_x\in\mathbb{R}^n.$$ This vector $\nabla f(x)$ is the unique vector of $\mathbb{R}^n$ such that $Df(x)(y)=\langle\nabla f(x),y\rangle$ for all $y\in\mathbb{R}^n$ (see Riesz representation theorem), where $\langle\cdot,\cdot\rangle$ is the usual scalar product $$\langle(x_1,\dots,x_n),(y_1,\dots,y_n)\rangle=x_1y_1+\dots+x_ny_n.$$
  • when $f:\mathbb{R}^n\to\mathbb{R}^m,$ then for $x$ in $\mathbb{R}^n$, $$\mathrm{Jac}_x(f)=\begin{bmatrix}\frac{\partial f_1}{\partial x_1}&\frac{\partial f_1}{\partial x_2}&\dots&\frac{\partial f_1}{\partial x_n}\\\frac{\partial f_2}{\partial x_1}&\frac{\partial f_2}{\partial x_2}&\dots&\frac{\partial f_2}{\partial x_n}\\ \vdots&\vdots&&\vdots\\\frac{\partial f_m}{\partial x_1}&\frac{\partial f_m}{\partial x_2}&\dots&\frac{\partial f_m}{\partial x_n}\\\end{bmatrix}|_x$$ is the matrix $m\times n$ of the linear map $Df(x)$ exprimed from the canonical base of $\mathbb{R}^n$ to the canonical base of $\mathbb{R}^m.$

For example, with $f:\mathbb{R}^2\to\mathbb{R}$ such as $f(x,y)=x^2+y$ you get $\mathrm{grad}_{(x,y)}(f)=[2x \,\,\,1]$ (or $\nabla f(x,y)=(2x,1)$) and for $f:\mathbb{R}^2\to\mathbb{R}^2$ such as $f(x,y)=(x^2+y,y^3)$ you get $\mathrm{Jac}_{(x,y)}(f)=\begin{bmatrix}2x&1\\0&3y^2\end{bmatrix}.$

  • how is your answer different from this one - math.stackexchange.com/questions/336640/… ? – gansub Jan 20 '17 at 4:30
  • @gansub : The two ones are really similar indeed! I think that it has not been tagged as duplicate since the two questions sound differently, but except an example my answer does not bring other informations than Branimir Ćaćić's one. – Balloon Jan 20 '17 at 8:06
  • not really similar IMHO. He says it is a column vector. You are saying gradient is a row vector. Clarification ? – gansub Jan 20 '17 at 8:07
  • @ gansub : I think he does the same clarification that me : you can think about the gradient as a matrix (more precisely a row vector), or as a vector of $\mathbb{R}^n$ (which is often written as $(x_1,\dots,x_n)\in\mathbb{R}^n$ to not overload the paper, but which means $$\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}\in\mathbb{R}^n,$$ and so corresponds indeed to a column vector). – Balloon Jan 20 '17 at 8:10
  • 1
    Yes, we have the equality $\mathbb{R}^n\simeq\mathcal{M}_{n\times 1}(\mathbb{R})$. I would prefer to write "we have the canonical isomorphism" between the two spaces, since the first notation really makes me think about $\mathbb{R}^n$ as a vector space whereas the second one makes me think about $\mathbb{R}^n$ as a coordinate space. – Balloon Jan 20 '17 at 8:20

The gradient vector of a scalar function $f(\mathbf{x})$ that maps $\mathbb{R}^n\to\mathbb{R}$ where $\mathbf{x}=<x_1,x_2,\ldots,x_n>$ is written as $$\nabla f(\mathbf{x})=\frac{\partial f(\mathbf{x})}{\partial x_1}\hat{x}_1+\frac{\partial f(\mathbf{x})}{\partial x_2}\hat{x}_2+\ldots+\frac{\partial f(\mathbf{x})}{\partial x_n}\hat{x}_n$$

Whereas the Jacobian is taken of a vector function $\mathbf{f}(\mathbf{x})$ that maps $\mathbb{R}^n\to\mathbb{R}^m$, where $\mathbf{f}=<f_1,f_2,\ldots,f_m>$ and $\mathbf{x}=<x_1,x_2,\ldots,x_n>$. The Jacobian is written as

$$J_\mathbf{f} = \frac{\partial (f_1,\ldots,f_m)}{\partial(x_1,\ldots,x_n)} = \left[ \begin{matrix} \frac{\partial f_1}{\partial x_1} & \cdots & \frac{\partial f_1}{\partial x_n} \\ \vdots & \ddots & \vdots \\ \frac{\partial f_m}{\partial x_1} & \cdots & \frac{\partial f_m}{\partial x_n} \end{matrix} \right]$$

Note that when $m=1$ the Jacobian is same as the gradient because it is a generalization of the gradient.

The Jacobian determinant can be used for changes of variables because it can be viewed as the ratio of an infinitesimal change in the variables of one coordinate system to another. This requires that the function $\mathbf{f}(\mathbf{x})$ maps $\mathbb{R}^n\to\mathbb{R}^n$, which produces an $n\times n$ square matrix for the Jacobian. For example

$$\iiint_R f(x,y,z) \,dx\,dy\,dz = \iiint_S f(x(u,v,w),y(u,v,w),z(u,v,w))\left|\frac{\partial (x,y,z)}{\partial(u,v,w)}\right|\,du\,dv\,dw$$

where the Jacobian $J_\mathbf{g}$ is taken of the function

$$\mathbf{g}(u,v,w)=x(u,v,w)\hat{\imath}+y(u,v,w)\hat{\jmath}+z(u,v,w)\hat{k}$$

and the areas $R$ and $S$ correspond to each other.

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