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If $e_{\rho},e_{\varphi},e_z$ are the unit vectors in cylindrical coordinates, one can check \begin{align} (e_{\rho}\otimes e_{\rho})\cdot (e_{\rho}\otimes e_{\rho})= e_{\rho}\otimes e_{\rho}, (e_{\varphi}\otimes e_{\varphi})\cdot (e_{\varphi}\otimes e_{\varphi})= e_{\varphi}\otimes e_{\varphi}, (e_{z}\otimes e_{z})\cdot (e_{z}\otimes e_{z})= e_{z}\otimes e_{z}. \end{align} which means that the matrices are orthogonal. I suspect that $I=e_{\rho}\otimes e_{\rho}+e_{\varphi}\otimes e_{\varphi}+e_z\otimes e_z$. At least if one multiplies by a vector one gets back the vector. Is my conjecture true?

homersimpson
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    You are confusing me multiple times: If $e_1=\left(\begin{smallmatrix}1\0\0\end{smallmatrix}\right),,,e_2=\left(\begin{smallmatrix}0\1\0\end{smallmatrix}\right),,,e_3=\left(\begin{smallmatrix}0\0\1\end{smallmatrix}\right),,$ then $e_1\otimes e_1=\left(\begin{smallmatrix}1&0&0\0&0&0\0&0&0\end{smallmatrix}\right),,,e_2\otimes e_2=\left(\begin{smallmatrix}0&0&0\0&1&0\0&0&0\end{smallmatrix}\right),,,e_3\otimes e_3=\left(\begin{smallmatrix}0&0&0\0&0&0\0&0&1\end{smallmatrix}\right),.$ These matrices are certainly not orthogonal. Idempotent I would say. – Kurt G. Mar 27 '23 at 13:27
  • Lastly, what have these properties which hold for all orthogonal vectors to do with cylindrical coordinates? – Kurt G. Mar 27 '23 at 13:27
  • @KurtG. I'm sorry, the $^{-1}$ is missing. What I meant that these matrices are orthogonal matrices because $A^{-1}=A^t$. But now that you mention it, it does not have to do anything with cylindircal coordinates. It was my observation so far... – homersimpson Mar 27 '23 at 14:00
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    Please look at the matrices I displayed in the first comment. They are not invertible. The only property that holds in your post is $I=e_1\otimes e_1+e_2\otimes e_2+e_3\otimes e_3,.$ Without the $^{-1}$ your matrix equations were correct BTW. But your wording was wrong. – Kurt G. Mar 27 '23 at 14:37
  • @KurtG. thanks you're right! So what I wrote is also true for cylindrical coordinate basis vectors, i.e. $I=e_{\rho}\otimes e_{\rho}+e_{\varphi}\otimes e_{\varphi}+e_z\otimes e_z$? – homersimpson Mar 27 '23 at 15:11
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    Yes. It is true in every orthonormal basis when we represent that basis as I did in my first comment. In 2d polar (and likewise in 3d cylindrical) coordinates you can sometimes come across a non-orthonormal basis (called holonomic or coordinate basis). See here for a detailed discussion. – Kurt G. Mar 27 '23 at 15:26
  • @KurtG. I tried to do this by hand now, $A=(a_{\rho\rho}e_{\rho}\otimes e_{\rho}+a_{\rho\varphi}e_{\rho}\otimes e_{\varphi}+a_{\rho z}e_{\rho}\otimes e_{z}+a_{\varphi\rho}e_{\varphi}\otimes e_{\rho}+a_{\varphi\varphi}e_{\varphi}\otimes e_{\varphi}+a_{\varphi z}e_{\varphi}\otimes e_{z}+ a_{z\rho}e_z\otimes e_{\rho}+a_{z\varphi} e_{z}\otimes e_{\varphi} + a_{zz} e_{z}\otimes e_{z})$. I computed $(e_{\rho}\otimes e_{\rho}+e_{\varphi}\otimes e_{\varphi} + e_{z}\otimes e_z)\cdot A= A$, but $A\cdot (e_{\rho}\otimes e_{\rho}+e_{\varphi}\otimes e_{\varphi} + e_{z}\otimes e_z)\neq A$ – homersimpson Mar 29 '23 at 09:16
  • The reason is that $(e_{\rho}\otimes e_{\varphi})\cdot (e_{\rho}\otimes e_{\rho})=0$ so we lose the coefficients $a_{\rho\varphi}$ and $a_{\rho z}$ and the same happens if you go through the second and third line. However, if the matrix is symmetric it is true that $A\cdot (e_{\rho}\otimes e_{\rho}+e_{\varphi}\otimes e_{\varphi} + e_{z}\otimes e_z)= A$. This is not really satisfying... – homersimpson Mar 29 '23 at 09:19

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