Finding the kernel of a Jacobian matrix

Question

The function $f: \mathbb{R}^n \backslash\{0\} \rightarrow \mathbb{R}^n$ is defined by $f(x) = \frac{x}{|x|}$. After finding the Jacobian matrix $\partial f(x)$, I want to find the kernel of $\partial f(x)$. I have a feeling that nullity($\partial f(x)) = 1$ because I think the hyperplane $\{v\in \mathbb{R}^n: x \cdot v = 0\}$ is the image of $\partial f(x)$ and that has dimension $n-1$. I found the entries in the Jacobian matrix to be $\partial_j f_i(x) = \frac{|x|^2\delta_{ij}-x_ix_j}{|x|^3}$.

I've been playing around with $n = 2$ and $n = 3$ to try and understand the problem better - solving $\partial f(x)v=0$ by row-reduction, and I've always managed to find $v$ in these cases, mostly because they're simple to work with. However, I'm not sure how to proceed now - either I perform row-reduction on the $n\times n$ Jacobian matrix and get a vector $v$ acting as the basis of the kernel, or there's some small theorem or fact which reduces the problem to a simpler one.

For the $n=3$ case, I found that, if $\partial f(x)v = 0$, then $v = \begin{pmatrix} \frac{x_1}{x_3}, \frac{x_2}{x_3}, 1 \end{pmatrix}$, so I have a feeling that the $n$-dimensional case is going to be similar. Also, this seems suspiciously like $\begin{pmatrix} x_1, x_2, x_3\end{pmatrix}$, which would mean that, if $x \in \mathbb{R}^n$ is not an eigenvector of $\partial f(x)$, it is in the kernel.

I also managed to find that the hyperplane mentioned above is an eigenspace of $\partial f(x)$ with eigenvalue $\frac{1}{|x|}$. Does anyone know how to proceed, or am I going in the wrong direction?

Answer 1

You are absolutely on the right track! Note that there's no difference between $(x_1, x_2, x_3)$ being in the kernel and $\left( \frac{x_1}{x_3}, \frac{x_2}{x_3}, 1 \right)$ being in the kernel. So yes, your working hypothesis should be that the span of $x = (x_1,\dots,x_n)$ is the kernel of the Jacobian.

How would you show it? For starters, maybe let's factor out the $|x|^3$ from the matrix, it doesn't affect its kernel. Then, we have a matrix where $A_{ij} = -x_ix_j$ if $i \neq j$ and $x_1^2 + \dots + x_n^2 $ excluding $x_i^2$ for $i = j$. Now, recall the first component of $Ax$ is just the first row times the vector (i.e. for $Ax = b$ we have $b_i = \sum_j A_{ij}x_j$). So for example, lets use $i = 1$, then we get for $i \neq j$, the sum is $\sum_{i \neq 1} -x_j^2$. So, adding $A_{11}x_1= x_2^2 + \dots x_n^2$ to this, we get $0$. This is true for any row (see if you can write this to be more general!) so we get the $Ax = 0$ so $x$ is in the kernel.

You might have to think of a way to show that there is nothing else in the kernel however!

Answer 2

Fix $x \in \mathbb{R}^n \setminus 0$. We have $$Df(x) = \frac{|x|^2I - xx^T}{|x|^3}.$$ Pick $v_2, \dots, v_n \in \mathbb{R}^n$ so that $\frac{x}{|x|}, v_2, \dots, v_n$ is an orthonormal basis of $\mathbb{R}^n$. We have $$Df(x)\frac{x}{|x|} = \frac{x}{|x|^2} - \frac{x}{|x|^2} = 0,$$ and $$Df(x) = \frac{I}{|x|} \text{ on } \text{span}(v_2, \dots, v_n).$$ Thus $Df(x)$ has the matrix representation $$Df(x) = \begin{pmatrix} 0 & 0\\ 0 & \frac{I}{|x|} \\ \end{pmatrix}. $$ From this it is clear that $\ker Df(x) = \text{span}(x)$.