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.