Polar decomposition

There is another useful consequence of the theory of square roots, namely, the analogue of the polar representation of a complex number.

Theorem 1. If is an arbitrary linear transformation on a finite-dimensional inner product space, then there is a (uniquely determined) positive transformation , and there is an isometry , such that . If is invertible, then also is uniquely determined by .

Proof. Although it is not logically necessary to do so, we shall first give the proof in case is invertible; the general proof is an obvious modification of this special one, and the special proof gives greater insight into the geometric structure of the transformation .

Since the transformation is positive, we may find its (unique) positive square root, . We write ; since , the theorem will be proved if we can prove that is an isometry, for then we may write . Since we see that so that is an isometry, and we are done.

To prove uniqueness we observe that implies and therefore Since the positive transformation has only one positive square root, it follows that . (In this part of the proof we did not use the invertibility of .) If is invertible, then so is (since ), and from this we obtain (multiplying the relation on the right by ) that .

We turn now to the general case, where we do not assume that is invertible. We form exactly the same way as above, so that , and then we observe that for every vector we have If for each vector in the range of we write , then the transformation is length-preserving wherever it is defined. We must show that is unambiguously determined, that is, that implies . This is true since is equivalent to and this latter condition implies . The range of the transformation , defined so far on the subspace only, is . Since is linear, and have the same dimension, and therefore and have the same dimension. If we define on to be any linear and isometric transformation of onto , then , thereby determined on all , is an isometry with the property that for all . This completes the proof. ◻

Applying the theorem just proved to in place of , and then taking adjoints, we obtain also the dual fact that every may be written in the form with an isometric and a positive . In contrast with the Cartesian decomposition ( Section: Self-adjoint transformations ), we call the representation a polar decomposition of .

In terms of polar decompositions we obtain a new characterization of normality.

Theorem 2. If is a polar decomposition of the linear transformation , then a necessary and sufficient condition that be normal is that .

Proof. Since is not necessarily uniquely determined by , the statement is to be interpreted as follows: if is normal, then commutes with every , and if commutes with some , then is normal. Since and , it is clear that is normal if and only if commutes with . Since, however, is a function of and vice versa is a function of ( ), it follows that commuting with is equivalent to commuting with . ◻

EXERCISES

Exercise 1. If a linear transformation on a finite-dimensional inner product space has only one polar decomposition, then it is invertible.

Exercise 2. Use the functional calculus to derive the polar decomposition of a normal operator.

Exercise 3.

If is an arbitrary linear transformation on a finite-dimensional inner product space, then there is a partial isometry , and there is a positive transformation , such that and such that . The transformations and are uniquely determined by these conditions.
The transformation is normal if and only if the transformations and described in (a) commute with each other.

EXERCISES

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