0
$\begingroup$

Let $A=\mathbf{J}\left(\frac{x^2-y^2}{x^2+y^2},\frac{xy}{x^2+y^2}\right)$, then the $\text{det}A=0$ for every point in $\mathbb{R}^2$ except $\vec 0$ where the function is undefined. Its rank can only be $0$ or $1$. At first I thought its rank was $0$ since it sent the basis vectors to $\vec 0$, but I suspect this may be wrong since $A$ is not linear (given that it changes with its arguments). The book says its rank is $1$ because its domain does not include the zero vector, but why does not including a zero vector imply the rank of the transformation is $1$?

  • 0
    the determinant of $A$ is not dependant on a point – 2017-02-24
  • 0
    @JorgeFernándezHidalgo but why does that imply its rank is $1$? – 2017-02-24
  • 0
    The rank is the dimension of the range, so the only matrix with rank $0$ is the zero matrix. – 2017-02-24
  • 0
    By the way, the rank in the case is in the sense of [differential topology](https://en.wikipedia.org/wiki/Rank_(differential_topology)). In general, the rank of a smooth map can vary from point to point, but there is a nice class of maps with constant rank. – 2017-02-24
  • 0
    @SpamIAm so the fact that the domain does not include the zero vector imply that $A \ne \mathbf 0$ and therefore it cannot have rank $0$? – 2017-02-24
  • 0
    I don't think the fact that the domain of the function is $\mathbb{R}^2 \setminus \{(0,0\}$ has anything to do with it. As I said, the only matrix of rank $0$ is the zero matrix. Is there a point $(x,y)$ such that $A(x,y) = 0$? If not, then the rank is $1$. – 2017-02-24

0 Answers 0