The following question is taken from Pg. no. 156 of "Linear Algebra Done Right (3rd Edition)" by Sheldon Axler.
Define $T\in L(R^2)$ by \begin{equation} T(x,y)= (41x + 7y, -20x + 74y) \end{equation} The matrix of $T$ with respect to the standard basis of $R^2$ is \begin{equation} \begin{pmatrix} 41 & 7 \\ -20 & 74 \end{pmatrix} \end{equation} which is not a diagonal matrix. However, $T$ is diagonalizable, because the matrix of $T$ with respect to the basis $(1, 4), (7, 5)$ is \begin{equation} \begin{pmatrix} 69 & 0 \\ 0 & 46 \end{pmatrix} \end{equation}
Now I understand the first part that the matrix formed by standard basis is \begin{equation} \begin{pmatrix} 41 & 7 \\ -20 & 74 \end{pmatrix} \end{equation} since for $(1,0)$ we have $(41+0, -20+0)$ and for $(0,1)$ we have $(0+7, 0+74)$ and so the matrix is of the form given above.
However, in the second part for the basis $(1, 4), (7, 5)$, should it not be something like this \begin{equation} \begin{pmatrix} 41(1)+7(4) & 41(7)+7(5) \\ -20(1)+74(4) & -20(7)+74(5) \end{pmatrix} \end{equation}
Also, how can we guess a basis for which a matrix is diagonalizable?
You have $T(1,4) = (69,276) = 69(1,4)$, and similarly $T(7,5) = (322,230) = 46(7,5)$, which is why the resulting matrix is diagonal.
You don't have to guess: the basis consists of eigenvectors for the linear transformation.