Why don't these vectors form a basis at $r = 0$?

Question

Consider a point $P(x,y)$ in $\mathbb{R}^2$ with standard basis. The point is at $\mathbf{r} = (x,y) = (r\cos(\theta), r\sin(\theta)$ in polar coordinates. At $P(x,y)$, the vectors $\mathbf{e}_r = \dfrac{\partial{\mathbf{r}}}{\partial{r}} = \left( \cos(\theta), \sin(\theta) \right)$ and $\mathbf{e}_\theta = \dfrac{\partial{\mathbf{r}}}{\partial{\theta}} = \left( -r\sin(\theta), r\cos(\theta) \right)$ form a basis (except at $r = 0$).

I was wondering why it doesn't form a basis at $r = 0$? I don't understand why this is the case?

I would greatly appreciate it if people could please take the time to clarify this.

Answer 1

Because when $r=0$, you get $\mathbf{e}_\theta = (0,0)$ which makes $\{\mathbf{e}_r,\mathbf{e}_\theta\}$ linearly dependent ($0\mathbf{e}_r+1\mathbf{e}_\theta=\mathbf{0}$).

Answer 2

The second vector has a factor of r in its components, so at r=0, it is the 0 vector.