i have problems solving this Task. I know how to calculate the jacobian matrix but i dont understand how to create the function f.

Here is the task:

We define a function $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ with $f(x_{1},x_{2}) = \left( \begin{array}{cc} x_{1}-x_{2} \\ x_{1}x_{2}+x_{2} \end{array} \right)$.

Write functions $f_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}$ and $f_{2}: \mathbb{R}^{2} \rightarrow \mathbb{R}$, for $f=\left( \begin{array}{cc} f_{1} \\ f_{2} \end{array} \right)$. Calculate the jacobian matrix $Df(x)$ for $x \in \mathbb{R}^{2}$.

What does the seoncd sentence mean? As far as i understand, i have to do find the partial derivatives of $f_{1}$ and $f_{2}$, write that as a vector/matrix (function $f$) and calculate the jacobian matrix. Is this right?

I am not fully sure about this idea because if i calculate the partial derivatives for example $f_{1}$ i will get the gradient of $f_{1}$ which will be a 1x2 vector. How can i combine $f_{1}$ and $f_{2}$ in f and calculate the jacobian matrix? So $f$ should be a vector of both gradients. Right?!?

I tried to explain everything as best as i can but i think you can "see" my confusion in the test i wrote :-/

  • 1
    $\begingroup$ The Jacobian would be $$J_{(x_1,x_2)}(f)=\begin{pmatrix} \dfrac{\partial f_1}{\partial x_1} & \dfrac{\partial f_1}{\partial x_2} \\ \dfrac{\partial f_2}{\partial x_1} & \dfrac{\partial f_2}{\partial x_2} \end{pmatrix} = \begin{pmatrix} 1 & -1 \\ x_2 & x_1+1 \\ \end{pmatrix}$$ $\endgroup$ – Moo Nov 4 '17 at 12:40
  • $\begingroup$ so my idea was right? it is just the jacobian of the gradients? $\endgroup$ – hukachaka Nov 4 '17 at 12:41
  • $\begingroup$ Thank you very much. if this would be an answer instead of a comment i would have chosen this as best ;-) $\endgroup$ – hukachaka Nov 4 '17 at 12:46

$f_1(x_1,x_2)=x_1-x_2$ and $f_2(x_1,x_2)=x_1x_2+x_2$. The Jacobian matrix of $f$ is $$Jf(x_1,x_2)=\left(\begin{array}{cc} \frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\ \frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \\ \end{array}\right) = \begin{pmatrix} 1 & -1 \\ x_2 & x_1+1 \\ \end{pmatrix}\,.$$ The $i$th row of the Jacobian matrix is the gradient of the function $f_i$.

  • $\begingroup$ Thank you very much. Now i did understand. You answer was the fastest so i will chose this as best solution. $\endgroup$ – hukachaka Nov 4 '17 at 12:48

In general when you have a vector field $f \colon\mathbb{R}^2 \rightarrow \mathbb{R}^2$ this takes $(x,y) \in \mathbb{R}^2$ in input and returns a vector in $\mathbb{R}^2$, which can be written as $(x',y')$, where both $x'$ and $y'$ are functions of $x$ and $y$. The second sentence is just giving a name to $x'$ and $y'$, that is $f_1$ and $f_2$ respectively. So $$f_1(x,y) = x-y, \quad f_2(x) = xy+y.$$ The Jacobian is defined as the matrix $$ \left( \begin{matrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{matrix} \right) = \left( \begin{matrix} 1 & -1 \\ y & x+1 \end{matrix} \right) $$ so the raws are the gradients of $f_1$ and $f_2$.


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