I read in a lecture note that a positive-definite hermitian n-by-n matrix A has a unique positive-definite square root B and get stuck in its proof of the uniqueness.Part of the proof goes as follows.
Let B be a positive-definite square root of A. Then B commutes with A. Thus, B stabilizes each eigenspace of A. Since B is diagonalizable on V , it is diagonalizable on each eigenspace of A...
The "V" mentioned in the proof is the n-dimensional complex vector space where the matrix A acts as an operator.
I cannot figure out why B is diagonalizable on V. Any help is welcomed!