Can a basis always be chosen to diagonalise the matrix representation of an operator with the eigenvalues ordered arbitrarily?

Question

In other words, if I can find a basis to diagonalise a matrix, $M$, such that its diagonal elements are its eigenvalues, $$M=\text{diag}(a,b,c,d),$$ can I always find a basis such that $$M=\text{diag}(c,b,a,d),$$ for example, or any other ordering of the eigenvalues.

Answer 1

Yes. If the matrix with respect to the basis $\{v_1,v_2,v_3,v_4\}$ is $\operatorname{diag}(a,b,c,d)$, then the matrix with respect to the basis $\{v_3,v_2,v_1,v_4\}$ is $\operatorname{diag}(c,b,a,d)$. That is, you apply to the vectors the same permutation that you applied to de eigenvalues.