In this expository paper, we present a new and easier proof of the Polar Decomposition Theorem. Unlike in classical proofs, we do not use the square root of a positive matrix. The presented proof is accessible to a broad audience

Wójcik, Paweł

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Annales Mathematicae Silesianae 31 (2017), 165–171DOI: 10.1515/amsil-2016-0013A SIMPLE PROOF OF THE POLAR DECOMPOSITIONTHEOREMPaweł WójcikAbstract. In this expository paper, we present a new and easier proof of thePolar Decomposition Theorem. Unlike in classical proofs, we do not use thesquare root of a positive matrix. The presented proof is accessible to a broadaudience.1. IntroductionThe algebra of all real (or complex) n× n matrices is denoted by Mn(K).Let us recall a well-known result.Theorem 1.1 (Polar Decomposition). Suppose that A ∈ Mn(K) is anonzero matrix. Then there are U,P ∈Mn(K) such that U is unitary, P ≥ 0,and A = UP .This result is called the Polar Decomposition, and its proof uses the squareroot of a positive matrix (or The Functional Calculus). Different proofs canbe found, e.g., in [1, 2, 3]. The aim of this article is to introduce a new proofof the Polar Decomposition. Let us point out that our proof neither uses thesquare root of a positive operator nor The Functional Calculus.It should be easier to prove Polar Decomposition Theorem, if we consideroperators instead of matrices. Using elementary techniques, Polar Decompo-sition will be proved. Throughout this paper we assume that the consideredReceived: 16.02.2016. Accepted: 27.05.2016. Online: 06.08.2016.(2010) Mathematics Subject Classification: 47A05, 47A99.Key words and phrases: Hilbert space, linear operator, polar decomposition.166 Paweł WójcikHilbert spaces are finite dimensional and their dimensions are not less than 2.Let B(H;K) denote the Banach space of all bounded linear operators (betweenHilbert spaces H and K) and we write B(H) for B(H;H). We shall identifyB(H) (where dimH = n) and Mn(K) in the natural way. Let us denote theunit sphere by S(H) := {x ∈ H : ‖x‖ = 1}. Throughout this work, all Hilbertspaces are assumed to be real or complex.If P ∈ B(H), then P is positive if 〈Px|x〉 ≥ 0 for all x ∈ H. In symbolsthis is denoted by P ≥ 0. If U ∈ B(H), then U is an isometry if ‖Ux‖ = ‖x‖for all x ∈ H, or, equivalently, 〈Ux|Uy〉 = 〈x|y〉 for all x, y ∈ H.2. A new and easy proofIn this section we present an elementary proof of the Polar Decomposition.The method of proof presented here is different from that of [1, 2, 3]. We startwith the following lemma.Lemma 2.1. Assume that dimH = 2. If A ∈ B(H;K) and A 6= 0, thenthere are vectors x1, x2 in S(H) such that x1⊥x2 and Ax1⊥Ax2.Proof. Fix x, z ∈ S(H) such that x⊥z. If 〈Ax|Az〉 = 0, we define x1 := x,x2 := z.Now, assume that 〈Ax|Az〉 = c 6= 0. Then we define a vector y := c|c|z. Itfollows that 〈Ax|Ay〉 ∈ R and y ∈ S(H). Moreover, x⊥y and 〈Ay|Ax〉 ∈ R.Then we define a mapping ϕ : [0, 1]→ K byϕ(t) :=〈A((1− t)x+ ty‖(1− t)x+ ty‖) ∣∣∣ A( t(−x) + (1− t)y‖t(−x) + (1− t)y‖)〉.Define now N1(t) := ‖(1− t)x+ ty‖ and N2(t) := ‖t(−x) + (1− t)y‖. It iseasy to check that ϕ(t) ∈ R for all t ∈ [0, 1]. Indeed, we haveϕ(t) =(1− t)(−t)N1(t)N2(t)〈Ax|Ax〉+ (1− t)2N1(t)N2(t)〈Ax|Ay〉+−t2N1(t)N2(t)〈Ay|Ax〉+ t(1− t)N1(t)N2(t)〈Ay|Ay〉 ∈ R.A simple proof of the Polar Decomposition Theorem 167In fact, we can write ϕ : [0, 1] → R. It is easy to see that ϕ is continuous.Moreover, we haveϕ(0) =〈A(x‖x‖) ∣∣∣A( y‖y‖)〉andϕ(1) = −〈A(y‖y‖) ∣∣∣A( x‖x‖)〉= −〈A(x‖x‖) ∣∣∣A( y‖y‖)〉,which means ϕ(0) = −ϕ(1). Thus we get ϕ(0) ≤ 0 ≤ ϕ(1) or ϕ(1) ≤ 0 ≤ ϕ(0).Without loss of generality, we may assume that ϕ(0) ≤ 0 ≤ ϕ(1). Using theDarboux property we get ϕ(to) = 0 for some to ∈ [0, 1]. Thus for the vectorsx1 :=(1− to)x+ toy‖(1− to)x+ toy‖, x2 :=to(−x) + (1− to)y‖to(−x) + (1− to)y‖we have x1⊥x2 and 0 = ϕ(to) = 〈A(x1) |A(x2)〉, therefore Ax1⊥Ax2. Theproof is complete. The next result is a consequence of the above lemma.Theorem 2.2. Assume that dimH = n. If A ∈ B(H;K) and A 6= 0, thenthere are vectors x1, . . . , xn in S(H) such thatxj⊥xk and Axj⊥Axk, j, k ∈ {1, . . . , n}, j 6= k.Proof. We proceed by induction (with respect to the dimension of H).For n = 2 we have proved that it is true (see Lemma 2.1).Assume the statement holds for n. We will prove it for n + 1. Supposethat dimH = n + 1. Obviously S(H) is compact. Therefore there is a yoin S(H) such that ‖A‖ = ‖Ayo‖. It is clear that dim{yo}⊥ = n. Then, byinductive assumption, there are the vectors x1, . . . , xn ∈ S({yo}⊥)⊂ S(H)such that xj⊥xk and Axj⊥Axk, for j, k ∈ {1, . . . , n}, j 6= k. We define avector xn+1 := yo. It is easy to observe that xj⊥xn+1 for all j ∈ {1, . . . , n}.We will show that Axj⊥Axn+1 for all j ∈ {1, . . . , n}. Assume, contrary toour claim, that 〈Axjo |Axn+1〉 = c 6= 0, for some xjo ∈ {x1, . . . , xn}. We definea vector u := c|c|xjo . It follows that u⊥xn+1, ‖u‖ = 1 and(1) 〈Au|Axn+1〉 = |c| ∈ R.168 Paweł WójcikLet α ∈ (0, 1). It is easy to check that αu+√1− α2xn+1 ∈ S(H). Therefore‖A‖2 ≥∥∥∥A(αu+√1− α2xn+1)∥∥∥2= α2‖Au‖2 + 2Re(α√1− α2 〈Au|Axn+1〉)+ (1− α2)‖Axn+1‖2and making use of (1), we obtain‖A‖2 ≥ α2‖Au‖2 + 2α√1− α2 〈Au|Axn+1〉+ (1− α2)‖Ayo‖2= −α2‖Au‖2 + 2α√1− α2|c|+ (1− α2)‖A‖2.It follows from the above inequality thatα2‖A‖2 ≥ α2‖Au‖2 + 2α√1− α2|c|andα2(‖A‖2 − ‖Au‖2)≥ 2α√1− α2|c|.Thus we haveα(‖A‖2 − ‖Au‖2)≥ 2√1− α2|c|.By letting α tend to 0, we get 0 ≥ 2|c|, which is a contradiction. As an illustration of the applications of this theorem we prove here thepolar decomposition of an operator. The main result of this paper is thefollowing.Theorem 2.3 (Polar Decomposition). Let H be a Hilbert space such thatdimH = n. If A ∈ B(H), then there are U,P ∈ B(H) such that U is unitary,P ≥ 0, and A = UP .Proof. Assume that dim(kerA)⊥ = p. Thus we obtain dimkerA = n−p.It is clear that A|(kerA)⊥ : (kerA)⊥ → H is injective. We choose {x1, . . . , xp}⊂S(H) ∩ (kerA)⊥ such thatxj⊥xk and Axj⊥Axk, j, k ∈ {1, . . . , p}, j 6= k,A simple proof of the Polar Decomposition Theorem 169by Theorem 2.2. By the injectivity of A|(kerA)⊥ , we obtain Axk 6= 0 for all k ∈{1, . . . , p}. It is easy to see that {x1, . . . , xp} and{1‖Ax1‖Ax1, . . . ,1‖Axp‖Axp}are two orthonormal bases for (kerA)⊥ and A((kerA)⊥), respectively.Let {e1, . . . , en−p} be an orthonormal basis for kerA and let {y1, . . . , yn−p}be an orthonormal basis for A((kerA)⊥)⊥. Then we define a positive operatorP ∈ B(H) byPxk := ‖Axk‖xk, k ∈ {1, . . . , p}; Pet := 0, t ∈ {1, . . . , n− p}.We can now define an isometry U ∈ B(H) byUxk :=1‖Axk‖Axk, k ∈ {1, . . . , p}; Uet := yt, t ∈ {1, . . . , n− p}.We haveUPxk = U (‖Axk‖xk) = ‖Axk‖U (xk) = ‖Axk‖1‖Axk‖Axk = Axk,and UPet = U (0) = 0 = Aet. We have shown that UP and A coincide on thebasis, thus they are equal: UP = A. This completes the proof. 3. RemarkNow, we are going to present one more application of Theorem 2.2. Namely,we will prove that any injective operator can restrict to a similarity (a scalarmultiple of an isometry).Theorem 3.1. Assume that dimH = n = 2m ≥ 4. Let A ∈ B(H) beinjective. Then there is a subspace M⊂H such that dimM = 12n = m andA|M is a similarity (a scalar multiple of an isometry).Proof. We choose {x1, x2, . . . , x2m} ⊂ S(H) such thatxj⊥xk and Axj⊥Axk, j, k ∈ {1, 2, . . . , 2m}, j 6= k;see Theorem 2.2. Without loss of generality, we may assume that‖Ax1‖ ≤ ‖Ax2‖ ≤ . . . ≤ ‖Ax2m‖.170 Paweł WójcikChoose γ ∈ R such that‖Ax1‖ ≤ . . . ≤ ‖Axm‖ ≤ γ ≤ ‖Axm+1‖ ≤ . . . ≤ ‖Ax2m‖.We consider the following collection of subspaces:X1 : = span{x1, x2m},X2 : = span{x2, x2m−1},...Xm : = span{xm, xm+1}.It is easy to observe that Xj⊥Xk for j, k ∈ {1, . . . ,m}, j 6= k. SinceS(X1) = X1 ∩ S(H), the unit sphere S(X1) is an arcwise connected subsetof H. Moreover, we have ‖Ax1‖ ≤ γ ≤ ‖Ax2m‖. Hence there is a vectorw1 ∈ S(X1) such that γ = ‖Aw1‖.In a similar way we obtain a vector w2 ∈ S(X2) such that γ = ‖Aw2‖.Indeed, since S(X2) = X2 ∩ S(H), the unit sphere S(X2) is an arcwise con-nected subset of H. Moreover, we have ‖Ax2‖ ≤ γ ≤ ‖Ax2m−1‖. Hence thereis a vector w2 ∈ S(X2) such that γ = ‖Aw2‖.This and similar reasoning shows that there are vectors w1, . . . , wm suchthatwj ∈ S(Xj), γ = ‖Awj‖, where j ∈ {1, . . . ,m}.It is easy to check that {w1, . . . , wm} is an orthonormal set in H.It is not hard to see that A(Xj)⊥A(Xk) for j, k ∈ {1, . . . ,m}, j 6= k.Therefore { 1γAw1, . . . ,1γAwm} is also an orthogonal set in H. We define asubspaceM := span{w1, . . . , wm}. Thus we have dimM = m = 12n. Now, wedefine an operator T ∈ B(M;H) as follows:Twj :=1γAwj , j ∈ {1, 2, . . . ,m}.It follows that T is an isometry. Finally, we get A|M = γT . The proof iscomplete. Theorem 3.2. Assume that dimH = n = 2m + 1 ≥ 3. Let A ∈ B(H) beinjective. Then there is a subspaceM⊂H such that dimM = 12(n+1) = m+1and A|M is a similarity.The proof of Theorem 3.2 runs similarly.A simple proof of the Polar Decomposition Theorem 171References[1] Conway J.B., A course in functional analysis, Springer-Verlag, New York, 1985.[2] Gantmacher F.R., Brenner J.L., Applications of the theory of matrices, Dover Publica-tions, New York, 2005.[3] Pedersen G.K., Analysis now, Springer-Verlag, New York, 1989.Institute of MathematicsPedagogical University of CracowPodchorążych 230-084 KrakówPolande-mail: pwojcik@up.krakow.pl

A simple proof of the Polar Decomposition Theorem

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A simple proof of the Polar Decomposition Theorem

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