Connection between $S$ and $A$ when $S^{-1} A S$ is a diagonal matrix In diagonalizing a matrix $A$, we use a matrix $S$, which consists of eigenvectors of $A$.  To compute $S$, we simply take each eigenvector of $A$ and write it as a linear combination of the standard basis.  So if $(1,4,3)$ were an eigenvector of $A$, then writing it in the standard basis gives $1e_1 + 4e_2 + 3e_3$.  Taking the transpose of the coefficients gives the first column in $S$.
My question is: the matrix $S$ is not a matrix representation of the operator corresponding to matrix $A$.  It is simply one of the factors that multiplies with $A$ and $S^{-1}$, to give the diagonal matrix, which is a matrix representation of the operator corresponding to matrix $A$, but in the diagonalizing basis.  So, what is $S$ by itself?  Just something that we call a "transition matrix", or a "change of coordinate matrix"?

The most general name for it is a change of basis matrix. Since $A$ is diagonalizable, its eigenvectors span its domain, so every vector in its domain can be written as a linear combination of eigenvectors. Thus, the action of $A$ can be described by its action on each of the eigenvectors (which is multiplication by the eigenvalues). So if we choose to represent the operator corresponding to $A$ in the basis of its eigenvectors, $A$ is diagonal. The matrix $S$ transforms the diagonal matrix into the standard basis.
