Polar decomposition

There is another useful consequence of the theory of square roots, namely, the analogue of the polar representation $ζ = ρ e^{i θ}$ 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 $ℛ (P)$ 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 $ℛ (P)$ only, is $ℛ (A)$ . Since $U$ is linear, $ℛ (A)$ and $ℛ (P)$ have the same dimension, and therefore $(ℛ (A))^{⟂}$ and $(ℛ (P))^{⟂}$ have the same dimension. If we define $U$ on $(ℛ (P))^{⟂}$ to be any linear and isometric transformation of $(ℛ (P))^{⟂}$ onto $(ℛ (A))^{⟂}$ , then $U$ , thereby determined on all $𝒱$ , 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 $𝒩 (U) = 𝒩 (P)$ 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.

EXERCISES

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