has determinant 1 then both and have determinant 1. The positive-semidefinite matrix P is always unique, even if A is singular, and is denoted aswhere denotes the conjugate transpose of . The uniqueness of P ensures that this expression is well-defined. The uniqueness is guaranteed by the fact that is a positive-semidefinite Hermitian matrix and, therefore, has a unique positive-semidefinite Hermitian square root. If A is invertible, then P is positive-definite, thus also invertible and the matrix U is uniquely determined by Relation to the SVD In terms of the singular value decomposition (SVD) of , , one haswhere , , and are unitary matrices (called orthogonal matrices if the field is the reals ). This confirms that is positive-definite and is unitary. Thus, the existence of the SVD is equivalent to the existence of polar decomposition. One can also decompose in the formHere is the same as before and is given byThis is known as the left polar decomposition, whereas the previous decomposition is known as the right polar decomposition. Left polar decomposition is also known as reverse polar decomposition. The polar decomposition of a square invertible real matrix is of the form where is a positive-definite matrix and is an orthogonal matrix. Relation to Normal matrices The matrix with polar decomposition is normal if and only and commute: , or equivalently, they are simultaneously diagonalizable. Construction and proofs of existence The core idea behind the construction of the polar decomposition is similar to that used to compute the singular-value decomposition. Derivation for normal matrices If is normal, then it is unitarily equivalent to a diagonal matrix: for some unitary matrix and some diagonal matrix . This makes the derivation of its polar decomposition particularly straightforward, as we can then write where is a diagonal matrix containing the phases of the elements of , that is, when , and when . The polar decomposition is thus , with and diagonal in the eigenbasis of and having eigenvalues equal to the phases and absolute values of those of , respectively. Derivation for invertible matrices From the singular-value decomposition, it can be shown that a matrix is invertible if and only if (equivalently, ) is. Moreover, this is true if and only if the eigenvalues of are all not zero. In this case, the polar decomposition is directly obtained by writing and observing that is unitary. To see this, we can