The main purpose of this post is to prove the polar decomposition theorem for invertible matrices. As an application, we extract some information about the topology of , namely that . Along the way, we recall a few facts and also define real powers of a positive definite Hermitian matrix. We assume that the matrices below are non singular, even though some of the results are true without this assumption. We first prove that every Hermitian matrix has only real eigenvalues. Notice that with respect to the standard Hermitian inner product for a column vector . In fact, more generally, .
Proposition 1 Let be a Hermitian matrix i.e. , then has only real eigenvalues.
Proof: Over complex numbers every non constant polynomial of degree has solutions. Thus, it is clear that has eigenvalues. Let be an eigenvalue of with an eigenvector . That is Taking conjugate transpose of both sides, we get
Since , and we get which is only possible if is real.
Recall that a matrix is called positive definite if for all . Clearly, such a matrix must be nonsingular.
Proposition 2 If is a positive definite matrix, then it has only positive real eigenvalues.
Proof: As above let be an eigenvalue of with an eigenvector . Then,
Thus, must be positive as well.
If is Hermitian, it has an eigenspace decomposition. Here is the sketch of the proof. We apply induction on the dimension . It is clear for . Now, we may assume . Let and be as above. We consider the one dimensional space generated by and the complementary space . By definition, for , , or equivalently . Thus,
or in a more familiar form . This means that preserves which has dimension . So, by induction, has an eigenspace decomposition and we are basically done. The reason that this argument is not very precise is that in this post, we are using a concrete definition of being Hermitian. So, we also have to argue that somehow the matrix is still Hermitian when restricted to . Of course, using the abstract definition, this is trivial. In fact, it is pretty easy to translate every argument presented here to abstract one.
Lemma 3 If is Hermitian, then it is diagonalizable by a unitary matrix.
Proof: Since has an eigenspace decomposition, we can choose a basis of consisting of eigenvectors only. Furthermore, we may choose those vectors to be unit. Consider the matrix that takes the standard basis to this eigenbasis. Then, it is clear that is a diagonal matrix. It is also clear that .
Next, we prove the polar decomposition for invertible matrices. In this proof, we also define the square root of a matrix.
Theorem 4 Given an invertible matrix , there is a positive definite Hermitian matrix and a unitary matrix such that .
Proof: Let . Clearly, i.e. is Hermitian. Also, for nonzero , is nonzero, thus
So, is also positive definite. By the above lemma, as is Hermitian, there is a unitary matrix which diagonalizes i.e. . Since is unitary, and hence, . Also, since is positive definite, all the eigenvalues of and hence of are positive, by the above proposition. So, we define to be where is defined by taking square root of each entry on the diagonal. In fact, using this idea, we can define any power of by . Note that is also diagonal with positive diagonal entries. Hence, in particular, it is Hermitian. Clearly, a diagonal matrix with positive diagonal entries is positive definite. So, is positive definite.
We set . It is easy to check that is positive definite.
as is positive definite.
Finally, we let . Of course, here is invertible because its a product of nonsingular matrices. Now, we just need to check that .
which was to be shown.
Now, we will apply our knowledge to understand the topology of . Given , it is clear from our proof that we can choose positive definite Hermitian part so that . Hence, , in other words, is an element of . Again, in our proof, we have explained that in fact you may take any power of a positive definite Hermitian matrix. So we can define a path of matrices by . We see that and . This defines a deformation retract of onto the . It is easy to see that the space of positive definite Hermitian matrices of determinant 1 is homeomorphic to . More concretely, to write down any such matrix, we need ; . Also, we set . Then,
is positive definite Hermitian of determinant 1.
It is also not very hard to check that only the identity matrix is the only matrix that lies in which is also positive definite Hermitian of determinant 1. Thus, . We leave it as an exercise to prove that .