 Solution

The $j$th column of the matrix representation of a linear map is the coordinate vector of the image of the $j$th basis vector:Define $C$ by $w_j={\sum }_i{C}_{ij}{u}_i$, so its columns express the new basis vectors in the old basis. For every vector, $[v{]}_u=C[v{]}_w$. Applying $\varphi$ in either coordinate system gives $AC[v{]}_w=C{A}^{\prime }[v{]}_w$, proving the change of basis formulaTo make $A^{\prime }$ diagonal, choose the columns of $C$ to be linearly independent eigenvectors of $A$, expressed in the $u$ basis, and put their corresponding eigenvalues on the diagonal in the same order. This is possible precisely when $A$ is a diagonalizable matrix over the stated field; an eigenbasis need not exist over an arbitrary field.