Why do we want the diagonalizing transformation matrix unitary?

Question

In general for a matrix $A$, we have $P^{-1}AP = D$

$P$ can be computed easily as they are formed from eigenvectors.

But sometimes we want $U^{-1}AU = D$, where $U$ is unitary. Why do we want $U$ when we have $P$ which can be computed more easily?

Answer 1

because we can use the spectral theorem:

a matrix is unitarily diagonalizable if and only if it is normal (AA*=A*A)

so it is more easy to check if a matrix is unitarily diagonalizable. and we prefer to find U, because the basis is orthonormal and we know that such U exist if the matrix normal.