The polar decomposition as stated in Horn and Johnson's Matrix Analysis (corollary 7.3.3):
"If A is an n\times n complex matrix, then it may be written in the form
A = PU,
where P is a positive semidefinite matrix and U is unitary. The matrix P is always uniquely determined as P = ( AA^*)^{1/2}; if A is nonsingular, then U is uniquely determined as U = P^{-1}A."
So, we need to know how to take square roots of matrices. Let M be any positive semidefinite matrix. It is therefore hermitian. We know by the spectral theorem for hermitian matrices (theorem 4.1.5, Horn and Johnson) that it is unitarily equivalent to a diagonal matrix, a.k.a., it is diagonalizable via a unitary:
M = U \Lambda U^*
where \Lambda is the usual matrix of eigenvalues. Define M^{1/2} = U \Lambda^{1/2} U^*, where
\Lambda^{1/2} = diag ( \lambda_1^{1/2}, \ldots, \lambda_n^{1/2})
and the unique NONNEGATIVE square root is taken in each case. This is possible since the spectrum of a positive definite matrix is nonnegative. Then M^{1/2} is hermitian (this is the converse of the spectral theorem) and it is positive semidefinite since its spectrum is nonnegative (theorem 7.2.1, Horn and Johnson).
OK, so now we are to compute (AA^*)^{1/2} = (A^2)^{1/2}. Note that this is in general NOT equal to A itself. Why? Because the eigenvalues of A may be negative, and we take nonnegative roots when computing the square root. First, we had better verify that AA^*=A^2 is positive semidefinite: let x be any column vector, then
x^*AA^*x = (A^*x)^*(A^*x) = ||A^*x||^2 \geq 0,
and (AA^*)^* = AA^*, as required.
Now, A is hermitian and therefore we may invoke the spectral theorem one more time to write
A = W D W^*,
where D is the diagonal matrix of eigenvalues and W is unitary. Then
AA^* = A^2 = W D^2 W^*
and thus
(AA^*)^{1/2} = (A^2)^{1/2} = W (D^2)^{1/2} W^*.
Remember to take the nonnegative roots! And we've found our P! Now, if A is nonsingular, then we can find U via the formula in the original theorem. Verifying that U is unitary (EDIT: and uniqueness of P and U!) can be found in Horn and Johnson.
If you have any more questions or want to see an example, write back.
PS, buy Horn and Johnson.
