$T:\Bbb R^2 \to \Bbb R^2$, linear, diagonal with respect to any basis. Is there a linear transformation from $\Bbb R^2$ to $\Bbb R^2$ which is represented by a diagonal matrix when written with respect to any fixed basis?
If such linear transformation $T$ exists, then its eigenvector should be the identity matrix for any fixed basis $\beta$ of $\Bbb R^2$.
Then, I don't see, if this is possible or not.

Yes:
More generally:
let $R$ be a unital ring, then the for a matrix $Z\in M_{n\times n}(R)$ the following are equivalent:


*

*is invariant with respect to any chosen basis of $R^n$ 

*$Z$ is representable as a diagonal matrix in some basis of $R^n$.  

*$Z$ is a scalar matrix, ie: $\left( (\exists r \in R) Z=rI_n \right)$

*$Z$ is in the center of $GL_n(R)$.  


Proof (assuming 4 $\Leftrightarrow 2$ which can be found in a text on algebraic groups, or affine group schemes):
Any element $Z$ of the center of $M_{n\times n}(R)$ then satisfies:
\begin{equation}
(\forall P \in GL_n(R)) \mbox{ } Z= IZ=PP^{-1}Z = PZ P^{-1}.  
\end{equation}
However, since the center of $M_{n\times n}(R)$ is the diagonal subgroup of $GL_n(R)$ and since any change of basis of $R^n$ is the same as conjugating by some $P\in GL_n(R)$.
Therefore a matrix is the same under any basis of $R^n$ if and only if it is invariant with respect to any conjugation by some $P\in GL_n(R)$ if and only if it is in the center of $GL_n(R)$ if and only if it is diagonal is some basis if and only if its is a scalar matrix.  
