FACTORING A QUADRATIC OPERATOR AS A PRODUCT OF TWO POSITIVE CONTRACTIONS

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1 FACTORING A QUADRATIC OPERATOR AS A PRODUCT OF TWO POSITIVE CONTRACTIONS CHI-KWONG LI AND MING-CHENG TSAI Abstract. Let T be a quadratic operator on a complex Hilbert space H. We show that T can be written as a product of two positive contractions if and only if T is of the form ai bi on H H 2 (H 3 H 3 for some a, b [0, ] and strictly positive operator P with P a b ( a( b. Also, T T 3 we give a necessary condition for a bounded linear operator T with operator matrix on 0 T 2 H K that can be written as a product of two positive contractions.. Introduction There has been considerable interest in studying factorization of bounded linear operators (see [2, 3, 4, 5, 5]. For example, a 2 2 matrix C can be written as a product of two orthogonal ( a a( a projections if and only if C is the identity operator or C is unitarily similar to 0 0 for some a [0, ]. For more results about products of orthogonal projections, one may consult [, 7, 8, ]. Note that one can write an n n matrix C as a product of two positive (semi-definite operators exactly when C is similar to a positive operator (see [4, Theorem 2.2]. However, in the infinite dimensional case, the product of two positive operators may not be similar to a positive operator (see [2], [5, Example 2.]. For more development in this direction, one may consult [2, 4, 5]. In this paper, we study the problem when a bounded linear operator T on a complex Hilbert space H can be written as a product of two positive contractions. contraction, and we have that I/8 Re T and I/4 Im T I/4 In this case, T must be a (see [0, Theorem. and Corollary 4.3]. In Proposition 2.4, we give a necessary condition for this problem when T has operator matrix T T 3 0 T 2 on H K. 99 Mathematics Subject Classification. 47A60, 47A68, 47A63. Key words and phrases. quadratic operator, positive contraction, spectral theorem.

2 In such a case, T and T 2 must also be products of two positive contractions. This is an extension of the result of Wu in [4, Corollary 2.3] concerning the finite dimensional case. However, even for a 2 2 matrix C, it is not easy to determine when it is the product of two positive contractions. For example, consider C The diagonalizable contraction C is similar to a positive operator. Thus it is a product of two positive operators. Moreover, C satisfies I/8 Re C and I/4 Im C I/4. However, we will see that C cannot be written as a product of two positive contractions by Lemma 2.. Let B(H be the algebra of bounded linear operators acting on a complex Hilbert space H. We identify B(H with M n, the algebra of n n complex matrices, if H has finite dimension n. Recall that a bounded linear operator T B(H is positive (resp., strictly positive if T h, h 0 (resp., T h, h > 0 for every h 0 in H. We write as usual T 0 (resp., T > 0 when T is positive (resp., strictly positive. We call T B(H a quadratic operator if (T ai(t bi 0 for some scalars a, b C. Every quadratic operator T B(H is unitarily similar to ai bi on H H 2 (H 3 H 3 for some a, b C, P > 0 (see [3]. In this paper, we prove the following. Theorem.. A quadratic operator T B(H with operator matrix ai bi on H H 2 (H 3 H 3 for some a, b C and P > 0, can be written as a product of two positive contractions if and only if a, b [0, ], and P a b ( a( b. 2. Proof First we consider the 2 2 case so that we can identify B(H M 2 and H C 2. Lemma 2.. Suppose C if and only if a, b [0, ], a z with z 0. Then C is a product of two positive contractions z S {c : 0 c a b ( a( b}. If the above equivalent conditions hold, then there are continuous maps a ij (z, b ij (z for i, j 2 with (2. 0 a ii (z, b ii (z, a 2 (z a 2 (z 0, b 2 (z b 2 (z 0, 0 (a ij (z I, 0 (b ij (z I. 2

3 such that (2.2 (a ij (z(b ij (z a z, z S. Proof. We first prove the sufficiency. Without loss of generality, we may assume 0 a b. If a b or b, then z 0 and C diag (a, diag (, b. In the following, we may assume 0 a < b <, and consider two cases. Case. 0 a < b <. For z S, we have that z 2 b( b and hence (z 2 /b+b ( b+b. Consider ( a (z a A 2 (z z 2 /b z a 2 (z a 22 (z z b b (z b and B 2 (z b 2 (z b 22 (z 0 0, 0 Then A is rank with eigenvalue (z 2 /b + b, and C AB. Evidently, a ij (z, b ij (z are continuous maps for i, j 2 and satisfy (2., (2.2. Case 2. 0 < a < b <. For z S, we have a + b Let λ (z λ 2 (z be roots of the equation z 2 ( a( b a + b ( a b 2 2 ab. λ 2 (a + b z 2 λ + ab 0. ( a( b Then, a λ 2 (z λ (z b and λ (z, λ 2 (z are continuous maps on z S. Note that λ (zλ 2 (z ab, λ (z + λ 2 (z a + b z 2 ( a( b. We have (2.3 z ( a( b(λ j a(b λ j λ j, j, 2. We will construct a (z a A 2 (z a 2 (z a 22 (z a a 2 a 2 a 3 b (z b and B 2 (z b 2 (z b 22 (z a3 a γ 2 a 2 a 4 such that A has eigenvalues, λ, B has eigenvalues, λ 2, and C AB. First, we set (2.4 γ λ 2 b a λ <. Because b γ + bγ ( b( γ > 0, we can let a 3 b a + bγ γ a < b a b a 3

4 so that by (2.4, Then we can let a 3 λ (b a ( + bγ γ a a γ γb γa a γab + γa + a2 γ( + bγ γ a γ (γb a( a ( + bγ γ a (b λ ( a ( + bγ γ a 0. a + λ a 3 > 0 so that a + a 3 + λ and a 2 a a 3 λ ( + λ a 3 a 3 λ ( a 3 (a 3 λ so that a a 3 a 2 2 λ. As a result, a + a 3 + λ, det(a λ, and hence A has eigenvalues, λ. Further, let Then by (2.4, a 4 a 3 ( λ 2 γ 2 + a2 2 so that γ 2 (a 3 a 4 a 2 2 λ 2. γ(a 3 + a 4 γa 3 + γ λ2 a 3 γ 2 + a2 2 γ λ2 a 3 γ 2 + (a 3 λ + λ a 3 γ a 3 ( λ2 γ 2 λ + γ( + λ γ a 3 (b a γ + bγ γ a + γ + γλ + λ 2. + γ( + λ As a result, trb + λ 2 and det(b λ 2. Therefore, B has eigenvalues, λ 2. Denote by (AB ij the (i, j entry of AB. By (2.4, (AB γ(a a 3 a 2 2 γλ a, (AB 22 γ(a 3 a 4 a 2 2 γ(λ 2 /γ 2 b. Clearly, (AB 2 γ(a 2 a 3 a 3 a 2 0. By (2.4 and (2.3, (AB 2 γa 2 (a 4 a γ ( a 3 (a 3 λ ((a 3 + a 4 (a 3 + a γ ( b γ + bγ(b λ ( a ( + λ2 ( + λ ( + bγ γ a γ ( b( γ( a(b λ ( + λ 2 γ γλ ( + bγ γ γλ ( b( a(λ a(b λ ( b( a( γ(b λ z. λ For the converse, since A, B are positive contractions with σ(c σ(ab σ(b /2 AB /2 [0,, we have 0 a, b. Without loss of generality, we may assume a b. First, consider A B. Then the assumption C AB implies C is unitarily similar to α b 2 α b α b 2, 0 b 2 b 4 b 2 b 4 4

5 where b 2 0. b b 2 b 2 b 4 is unitarily similar to α2 0 0 for some 0 α, α 2, α 2 b, b 4 and Thus we have + α 2 b + b 4, a + b α b + b 4, ab α α 2 α (b b 4 b 2 2, and a 2 + b 2 + z 2 α 2(b2 + b2 2 + b2 2 + b2 4. These imply that z 2 [α 2 (b 2 + b b b 2 4] [(α b + b 4 2 2α α 2 ] ( α 2 b 2 2. Hence we may assume α <. In addition, we also obtain that a + b α b + b 4 α b + + α 2 b + α 2 ( α b and hence b α ( + α 2 a b α [( a( b ab + α 2 ] α [α 2 ( α + ( a( b], where the last equality follows from ab α α 2. Let c ( a( b/( α. Then b α 2 + c and b 4 c. By a direct computation, we see that z 2 ( α 2 b 2 2 ( α 2 (b b 4 α 2 ( α 2 [(α 2 + c( c α 2 ] (because α 2 b b 4 b 2 2 c( α [( α ( α 2 c( α ] ( a( b[(a + b (α + α 2 ], where the last equality follows from c ( a( b/( α and ab α α 2. Since ab α α 2, we have α + α 2 2 α α 2 2 ab. This implies that In general, since C α ( a α z α b 0 α z a b ( a( b. α( A A ( B B, where 0 < α A B, the scalars a, b, z in the above can be replaced by a/α, b/α, z/α, respectively, to get 0 a/α, b/α and z α ( a α ( b a b α α α. This shows that 0 a, b α and z a b (α a( b α a b ( a( b. This proves the necessity. In order to prove Theorem., we need the following fact; see, for example, [9, p. 547]. 5

6 Lemma 2.2. Let A be a bounded linear operator of the form ( A A 2 A 2 A 22 on H K, where H and K are Hilbert spaces. Then A is positive if and only if A and A 22 are both positive and there exists a contraction D mapping K into H satisfying A 2 A /2 DA/2 22. Lemma 2.3. Suppose a (z, a 22 (z, a 2 (z a 2 (z are continuous real-valued functions defined a (z a on S [0, such that A 2 (z a (P a 0 for all z S. Then 2 (P 0 on a 2 (z a 22 (z a 2 (P a 22 (P H H for all positive operators P B(H with spectrum in S. Proof. Since A 0, we have a (z, a 22 (z 0 and Define h(z by 0 a 2 (za 2 (z a (za 22 (z, z S. h(z : { a 2 (z a /2 (za/2 if a 2 (z > 0, 22 (z 0 if a 2 (z 0. Then h(z is a bounded Borel function on S with h(z, which satisfies a 2 (z a /2 (zh(za/2 22 (z. By the spectral theorem, for all positive operators P B(H with spectrum in S, we have a (P 0, a 22 (P 0, a 2 (P a 2 (P 0 and a 2 (P a /2 (P h(p a/2 22 (P for the contraction h(p B(H. Our assertion follows from Lemma 2.2. In the finite dimensional case, Wu [4, Corollary 2.3] has shown that if C product of two positive operators, then so are C and C 2. C C 3 0 C 2 is a Proposition 2.4 gives another proof which holds for both finite and infinite dimensional Hilbert spaces. In fact, it is also true that positive operators are replaced by positive contractions. Proposition 2.4. Let T be a bounded linear operator of the form T T 3 on H K, 0 T 2 where H and K are both Hilbert spaces. If T is a product of two positive contractions, then so are T and T 2. Proof. By our assumption and Lemma 2.2, we may assume that T AB, where A and B are of the form ( A A /2 D A /2 2 A /2 2 D A/2 A 2 and ( B B /2 D 2 B /2 2 B /2 2 D2 B/2 B 2 6 on H K,

7 respectively, such that 0 A I H, 0 A 2 I K, 0 B I H, 0 B 2 I K, D and D 2 are contractions from K into H. From T AB, we obtain that (2.5 (2.6 T A B + A /2 D (A /2 2 B /2 2 D 2B /2, A /2 2 (D A /2 B /2 B /2 A /2 2 (A /2 2 B /2 2 D 2B /2, T 2 (A /2 2 D A /2 B /2 D 2 B /2 2 + A 2 B 2. Let E be the restriction of A /2 2 to (ker A /2 2, then E is injective. Since 0 A /2 2 I K, so we can consider the (possibly unbounded inverse E : E : ran A/2 2 (ker A /2 2 such that EA /2 2 P 0, where P 0 is the orthogonal projection from K onto rana /2 2. Hence by (2.6, we derive that A /2 2 B /2 2 D 2B /2 P 0 (A /2 2 B /2 2 D 2B /2 P 0 (D A /2 B. Moreover, substitute this into (2.5 to get Note that P 0 D implies that T A B A /2 D (P 0 (D A /2 B [A /2 (I H D P 0 D A /2 ]B [A /2 (I H (P 0 D (P 0 D A /2 ]B. 0 (I H (P 0 D (P 0 D I H. Therefore, T [(A /2 P P A /2 ]B, where P P I H (P 0 D (P 0 D for some positive contraction P on H. This shows that T is a product of two positive contractions. Similarly, we can show that T 2 proof. is a product of two positive contractions, and hence so is T 2. This completes our Now we are ready to give the proof of Theorem.. Proof of Theorem.. We first prove the necessity. By assumption, we can focus on the part B(H 3 H 3 a z for some P > 0. Now, consider a 2 2 matrix with a, b [0, ] and z S : {c : 0 c a b ( a( b}. Then by Lemma 2., there are continuous maps a ij (z, b ij (z for i, j 2 with a 2 (z a 2 (z 0, b 2 (z b 2 (z 0 and satisfy 0 (a ij (z I 2, 0 (b ij (z I 2, (a ij (z(b ij (z 7 a z, z S.

8 By Lemma 2.3, By the spectral theorem on positive operators, 0 (a ij (P I and 0 (b ij (P I. (a ij (P (b ij (P. To prove the converse, suppose there is a factorization of the quadratic operator T B(H with operator matrix ai bi for some P 0 as the product of two positive contractions. By Proposition 2.4, we know that T AB for some 0 A, B I, A, B B(H 3 H 3. We may use the Berberian construction (see [6] to embed H 3 into a larger Hilbert space K 3, B(H 3 into B(K 3. Suppose A (A ij i,j 2, B (B ij i,j 2 B(H 3 H 3. Then P, A, and B are extended to P B(K 3, Ã (Ãij i,j 2 B(K 3 K 3, and B ( B ij i,j 2 B(K 3 K 3, respectively, such that the following conditions hold. (a P 0 with P P such that all the elements in σ( P are eigenvalues of P. (b 0 Ã, B I such that T ai P Ã B. Since P 0 and σ( P are eigenvalues of P, the quadratic operator T is unitarily similar to a P T 2 that admits a factorization as the product of two positive contractions. By Proposition 2.4, we see that is a product of two positive contractions. Thus, a P P a b ( a( b. Remark 2.5. Inspired by a comment of the referee, we see that if one considers the set of operators of the form with respect to a fixed orthonormal basis, then our proof of Theorem. shows that the decomposition depends continuously on P, and therefore continuous on T. acknowledgment Li is an honorary professor of the University of Hong Kong and the Shanghai University. His research was supported by US NSF and HK RCG. The Research of Tsai was supported by the National Science Council of the Republic of China under the project NSC M Tsai would like to thank Pei Yuan Wu and Ngai-Ching Wong for their helpful suggestions and comments. Some results in this paper are contained in the doctorial thesis of Ming-Cheng Tsai under the supervisor of Pei Yuan Wu to whom Tsai would express his heartfelt thanks. The authors would give their thanks to the referee for useful comment. 8

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