Linear transformation of a given coordinate

[kidsasd987]

I have a question about weights of a basis set with respect to the other basis set of one specific vector space.
It seems the weights do not covert linearly when basis sets convert linearly. I've got this question from the video on youtube "linear transformation"



Let's consider a vector space V spanned upon field K.

Then its element x∈V can be expressed in the form of a linear combination with a given basis set and its corresponding weights.

x=Σa_ix_iNow if we were to express a vector x with respect to the new basis x_hat_i,
we would express the same vector as

x=Σb_ix_hat_iWe can derive the relation

Λ(x_i)=x_hat_i
b_i=Λ^-1a_iI did this to cylindrical coordinate, and because
https://wikimedia.org/api/rest_v1/media/math/render/svg/cf553bbb290f2b6ad76c9cce12f8807d43ab09ee

and according to the equation from video
b_i=Λ^-1a_i

inverted matrix (transpose of matrix above applied to <x,y,z> should give converted weight bi but this isn't the case.
rho=sqrt(x^2+y^2) and phi=tan^-1(y/x) and they are not linear functions.could anyone tell me why is it so?

 

[fresh_42]

When people talk about a linear transformation between two basis which leads to different coordinates (weights), then it is meant to transform one basis vector linearly to another. Let's keep it simple and consider a two dimensional real vector space: a sheet of graph paper.

You can scale it by one box per centimeter on your graph paper or be two boxes or any other multiple or even different multiples for the $x$ and $y-$axis. You could even chose the $y-$axis to be the diagonal instead of the perpendicular axis. All these are linear transformations from one basis to another: $\begin{bmatrix}2&0\\0&2\end{bmatrix}\, , \,\begin{bmatrix}4&0\\0&1\end{bmatrix}\, , \,\begin{bmatrix}1&0\\1&1\end{bmatrix}$ or whatever.

This changes if (at least) one coordinate is transformed non-linearly: bent into a circle as in your example or another example would be a logarithmic scale: $\log x' \leftrightarrow x \nLeftrightarrow x'=Ax$ because $\log (x'_1+x'_2) \neq c_1\log x_1 +c_2 \log x_2$. In addition: Which angle has the origin in cylindrical coordinates? So you changed from something linear (Cartesian coordinates) to something non-linear (cylindrical coordinates) which thus cannot be done by something (function) linear. One could say you changed the nature of the coordinate system.