1
$\begingroup$

I find it impossible to remember that the Jacobian of $f: \mathbb R^n \to \mathbb R^m$ is

$$ \begin{pmatrix} {\partial f_1 \over \partial x_1} & {\partial f_1 \over \partial x_2} & \dots & {\partial f_1 \over \partial x_n} \\ \vdots & \dots & \vdots & \vdots\\ {\partial f_m \over \partial x_1} & \dots & \dots & {\partial f_m \over \partial x_n} \end{pmatrix}$$

and not

$$ \begin{pmatrix} {\partial f_1 \over \partial x_1} & {\partial f_2 \over \partial x_1} & \dots & {\partial f_m \over \partial x_1} \\ \vdots & \dots & \vdots & \vdots\\ {\partial f_1 \over \partial x_n} & \dots & \dots & {\partial f_m \over \partial x_n} \end{pmatrix}$$

How to memorise this? Is there any reason why it's defined the first way and not the other?

$\endgroup$
  • $\begingroup$ for $f:\mathbb{R}^n\rightarrow\mathbb{R}$ the gradient is horizontal $\endgroup$ – Thoth May 29 '15 at 1:39
  • $\begingroup$ The reason it's horizontal is because vectors are vertical by default and thus you have $\nabla f(x_0)x$ as the linear approximation to $f$ at $x_0$ without having to take a transpose. $\endgroup$ – Thoth May 29 '15 at 1:44
  • $\begingroup$ @Tyroshipleasurebarge Thank you for your comment. $\endgroup$ – a student May 29 '15 at 2:09
3
$\begingroup$

Since $f : \mathbb{R}^n \to \mathbb{R}^m$, then the Jacobian is also a function $Jf : \mathbb{R}^n \to \mathbb{R}^m$. This means that we will multiply the matrix $Jf$ by a $n \times 1$ column vector. Now a matrix product is only defined if the number of columns of the lefthand matrix is the same as the number of rows of the righthand matrix, so we must have $$ (m \times \underbrace{n) \times (n}_{=} \times 1) $$ which results in an $m \times 1$ column vector in $\mathbb{R}^m$. Thus $Jf$ must be $m \times n$, hence has $m$ rows and $n$ columns, as in your first formula.

$\endgroup$
4
$\begingroup$

Consider a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$. You of course have the gradient:

$$\nabla f= \left( \frac{\partial f}{\partial x_1},..., \frac{\partial f}{\partial x_n} \right)$$

If you have a function $f: \mathbb{R}^n \rightarrow \mathbb{R} ^m$, you have $m$ functions like above. Just stack the gradients in the matrix.

$\begin{pmatrix} \nabla f_1 \\ \nabla f_2 \\ ... \\ \nabla f_m\end{pmatrix}$

You could also remember that the jacobian must take a $n$-vector to a $m$-vector, hence it must be a $m \times n$ matrix.

$\endgroup$
  • 1
    $\begingroup$ I didn't know, I thought the gradient was vertical. $\endgroup$ – a student May 29 '15 at 2:10
  • 2
    $\begingroup$ Looking at Wikipedia it seems that the gradient is vertical: The partial derivatives are the coefficients of a vector and vectors are vertical. Right? $\endgroup$ – a student May 29 '15 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.