When we do this diagonalizing, note that there is flexibility in defining an eigenvector. Eigenvectors represent a "direction", and if we scale them, they are still an eigenvector.
Example: if eigenvector x= [ 1/3 1] corresponds to eigenvalue λ and we scale x by 3 to get x' = [1 3] , it is still an eigenvector associated with λ
M(3x)=λ(3x)
still works.
We would scale x to get rid of fractions.
Also, when diagonalizing, there is no forced order on the eigenvalues in matrix \(\Lambda\) The only requirement is that the eigenvalue in position n,n of matrix \(\Lambda\) corresponds to the eigenvector in column n of matrix P
Often, though, people put the eigenvalues in ascending (or descending ) order.
The point is we can get different variations of P and Λ, and all are correct, unless the order of the eigenvalues in Λ are stipulated to be in some order.