I was following this example online to diagonalize a matrix. It lists the eigenvectors as $\lambda =3,2,4$ (note the order). It then arranges each eigenvalue's corresponding eigenvector (3 column vectors) in the same order and puts them together to form the invertible matrix $P$ to solve $A=PDP^{-1}$.
My question is: does it matter what order the eigenvectors are put in to form $P$? Or does it not matter because $P$ will always be invertible? Or is it purely a trial and error thing?
The order of the eigenvalues in the diagonal matrix $D$ have to be the same as the order of the eigenvectors in the in the change of basis matrix $P$; i.e. the $n-$ column of $P$ is the eigenvector of of the $n-$th eigenvalue of the diagonal.