There is another useful consequence of the theory of square roots, namely, the analogue of the polar representation \zeta=\rho e^{i \theta} of a complex number.
Theorem 1. If A is an arbitrary linear transformation on a finite-dimensional inner product space, then there is a (uniquely determined) positive transformation P , and there is an isometry U , such that A=U P . If A is invertible, then U also is uniquely determined by A .
Proof. Although it is not logically necessary to do so, we shall first give the proof in case A 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 A .
Since the transformation A^{*} A is positive, we may find its (unique) positive square root, P=\sqrt{A^{*} A} . We write V=P A^{-1} ; since V A=P , the theorem will be proved if we can prove that V is an isometry, for then we may write U=V^{-1} . Since V^{*}=(A^{-1})^{*} P^{*}=(A^{*})^{-1} P, we see that V^{*} V=(A^{*})^{-1} P P A^{-1}=(A^{*})^{-1} A^{*} A A^{-1}=1, so that V is an isometry, and we are done.
To prove uniqueness we observe that U P=U_{0} P_{0} implies P U^{*}=P_{0} U_{0}^* and therefore P^{2}=P U^{*} U P=P_{0} U_{0}^{*} U_{0} P_{0}=P_{0}^{2}. Since the positive transformation P^{2}=P_{0}^{2} has only one positive square root, it follows that P=P_{0} . (In this part of the proof we did not use the invertibility of A .) If A is invertible, then so is P (since P=U^{-1} A ), and from this we obtain (multiplying the relation U P=U_{0} P_{0} on the right by P^{-1}=P_{0}^{-1} ) that U=U_{0} .
We turn now to the general case, where we do not assume that A is invertible. We form P exactly the same way as above, so that P^{2}=A^{*} A , and then we observe that for every vector x we have \begin{align} \|P x\|^{2} &= (P x, P x)\\ &= (P^{2} x, x)\\ &= (A^{*} A x, x)\\ &= \|A x\|^{2}. \end{align}If for each vector y=P x in the range \mathcal{R}(P) of P we write U y=A x , then the transformation U is length-preserving wherever it is defined. We must show that U is unambiguously determined, that is, that P x_{1}=P x_{2} implies A x_{1}=A x_{2} . This is true since P(x_{1}-x_{2})=0 is equivalent to \|P(x_{1}-x_{2})\|=0 and this latter condition implies \|A(x_{1}-x_{2})\|=0 . The range of the transformation U , defined so far on the subspace \mathcal{R}(P) only, is \mathcal{R}(A) . Since U is linear, \mathcal{R}(A) and \mathcal{R}(P) have the same dimension, and therefore (\mathcal{R}(A))^{\perp} and (\mathcal{R}(P))^{\perp} have the same dimension. If we define U on (\mathcal{R}(P))^{\perp} to be any linear and isometric transformation of (\mathcal{R}(P))^{\perp} onto (\mathcal{R}(A))^{\perp} , then U , thereby determined on all \mathcal{V} , is an isometry with the property that U P x=A x for all x . This completes the proof. ◻
Applying the theorem just proved to A^{*} in place of A , and then taking adjoints, we obtain also the dual fact that every A may be written in the form A=P U with an isometric U and a positive P . In contrast with the Cartesian decomposition ( Section: Self-adjoint transformations ), we call the representation A=U P a polar decomposition of A .
In terms of polar decompositions we obtain a new characterization of normality.
Theorem 2. If A=U P is a polar decomposition of the linear transformation A , then a necessary and sufficient condition that A be normal is that P U=U P .
Proof. Since U is not necessarily uniquely determined by A , the statement is to be interpreted as follows: if A is normal, then P commutes with every U , and if P commutes with some U , then A is normal. Since A A^{*}=U P^{2} U^{*}=U P^{2} U^{-1} and A^{*} A=P^{2} , it is clear that A is normal if and only if U commutes with P^{2} . Since, however, P^{2} is a function of P and vice versa P is a function of P^{2} ( P=\sqrt{P^{2}} ), it follows that commuting with P^{2} is equivalent to commuting with P . ◻
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 A is an arbitrary linear transformation on a finite-dimensional inner product space, then there is a partial isometry U , and there is a positive transformation P , such that \mathcal{N}(U)=\mathcal{N}(P) and such that A=U P . The transformations U and P are uniquely determined by these conditions.
- The transformation A is normal if and only if the transformations U and P described in (a) commute with each other.
