I want to compute $a\otimes a$ if $a\in \mathbb{R}^3$ is given in cylindrical coordinates, i.e. $a=a_{\rho}e_{\rho}+a_{\varphi}e_{\varphi}+a_z e_z$, where $e_{\rho}=(\cos(\varphi),\sin(\varphi),0),e_{\varphi}=(-\sin(\varphi),\cos(\varphi),0),e_z=(0,0,1)$.
My first attempt: $a\otimes a$ is independent of the basis, i.e. \begin{align} a\otimes a = \begin{pmatrix}a_{\rho}^2 & a_{\rho}a_{\varphi} & a_{\rho}a_z \\\\ a_{\rho}a_{\varphi} & a_{\varphi}^2 & a_{z}a_{\varphi}\\\\ a_{z}a_{\rho}& a_{z}a_{\varphi} & a_z^2 \end{pmatrix} \end{align} But if we repeat this with the euclidean basis we get \begin{align} a\otimes a = \begin{pmatrix}a_{x}^2 & a_{x}a_{y} & a_{y}a_z \\\\ a_{x}a_{y} & a_{y}^2 & a_{z}a_{y}\\\\ a_{z}a_{x}& a_{z}a_{y} & a_z^2 \end{pmatrix} \end{align} If we compare these matrices it must hold $a_x^2=a_{\rho}^2$ which is wrong in general. By change of coordinates we know that for $e_x=(1,0,0), e_y=(0,1,0), e_z=(0,0,1)$ \begin{align} &a_x=a\cdot e_x = (a_{\rho}e_{\rho}+a_{\varphi}e_{\varphi}+a_z e_z)e_x= a_{\rho}\cos(\varphi)-a_{\varphi}\sin(\varphi)\\\\ &a_y=a\cdot e_y = (a_{\rho}e_{\rho}+a_{\varphi}e_{\varphi}+a_z e_z)e_y= a_{\rho}\sin(\varphi)+a_{\varphi}\cos(\varphi)\\\\ &a_z=a_z \end{align} If we plug these in into $a\otimes a$ we should get the right matrix.
Question What is wrong with my first attempt? It seems that $a\otimes a$ is not independent of the basis, am I at least right in this point?