Determining Unitary Equivalence to a 3 3 Complex Symmetric Matrix from the Upper Triangular Form. Jay Daigle Advised by Stephan Garcia

Transcription

1 Determining Unitary Equivalence to a 3 3 Complex Symmetric Matrix from the Upper Triangular Form Jay Daigle Advised by Stephan Garcia April 4, 2008

2 2

3 Contents Introduction 5 2 Technical Background 9 3 The Angle Test 3 4 Breaking Down the Problem 9 4. Our Approach The One Eigenvalue Case The Two Eigenvalue Case The Three Eigenvalue Case References 37 3

4 4 CONTENTS

5 Chapter Introduction A complex symmetric matrix (CSM) is a square complex matrix T such that T = T t, where T t denotes the transpose of the matrix T. However, since any given operator has many different matrix representaions, we wish to identify those linear operators that can be represented as a complex symmetric matrix with respect to some orthonormal basis. To do this we introduce the concept of unitary equivalence. Definition. Let U be a n n matrix and let U denote the adjoint (i.e., conjugate transpose) of U. We say U is unitary if UU = U U = I. Definition. Let T and S be n n matrices. We say T and S are unitarily equivalent if there exists some unitary matrix U such that U T U = S. It turns out that two matrices are unitarily equivalent precisely when they represent the same operator with respect to two (possibly different) orthonormal bases. Thus the class of matrices we wish to study is precisely the class of matrices which are unitarily equivalent to a complex symmetric matrix (UECSM). A powerful tool for analyzing such matrices (Theorem. below) comes from Garcia and Putinar [2, 3]. We first require a few preliminaries. Definition. A function C : C n C n is a conjugation if. C is antilinear (i.e., C(ax + y) = acx + Cy for all a C, x, y C n ). 2. C is isometric (i.e., x, y = Cy, Cx for all x, y C n ). 3. C is an involution (i.e., C 2 = I). 5

6 6 CHAPTER. INTRODUCTION A simple example of a conjugation is the standard conjugation J : C n C n, which is defined by J(x,..., x n ) = (x,..., x n ). Definition. Let T be a n n complex matrix and let T denote its conjugate transpose. Let C be a conjugation on C n ; then T is C-symmetric if T = CT C. We now have the following theorem: Theorem.. Let T be a n n complex matrix. T is UECSM if and only if T is C-symmetric for some conjugation C. Proof. Suppose T is C-symmetric and let {e i } be an orthonormal basis with Ce i = e i (for a proof that such a basis exists, see [3, Lemma ]). Then T e i, e j = e i, T e j = Ce i, T Ce j = CT Ce j, e i (by the isometric property) = T e j, e i. Thus, since T e i, e j is the jith entry of T when expressed with respect to the orthonormal basis {e i }, we have represented T as a CSM with respect to this basis. Therefore the matrix T is UECSM. Conversely, suppose T is UECSM. It must be complex symmetric with respect to some orthonormal basis {u i }. Define C by extending Cu i = u i antilinearly to all of C n. Then as above, we have T u i, u j = CT Cu j, u i for all i, j, but since T is symmetric with respect to {u i } we have T u i, u j = T u j, u i. Therefore CT Cu i, u j = T u i, u j for all i, j, and thus T = CT C. Though many examples of matrices which are UECSM are known, no complete classification exists. In this paper we complete a classification of the 3 3 matrices which are UECSM. Since unitary equivalence is an equivalence relation, we may pick one representative of each equivalence class and determine whether this matrix is UECSM. To select this representative we use Schur s Theorem (see [5] for details):

7 7 Theorem.2 (Schur). Given a n n matrix A with eigenvalues λ,..., λ n in any prescribed order, there is a unitary matrix U M n such that U AU = T = [t ij ] is upper triangular, with diagonal entries t ii = λ i, i =,..., n. That is, every square matrix A is unitarily equivalent to a triangular matrix whose diagonal entries are the eigenvalues of A in a prescribed order. Thus, given a 3 3 matrix A, we know it is unitarily equivalent to a (not necessarily unique) matrix of the form λ a b 0 λ 2 c, (.) 0 0 λ 3 where λ i are the eigenvalues of A in any given order. In Chapter 2 we show that further simplifying assumptions can be made to yield a particularly easyto-work-with form. We provide a complete classification of which matrices in this form are UECSM, allowing us to determine whether any given 3 3 matrix is UECSM.

8 8 CHAPTER. INTRODUCTION

9 Chapter 2 Technical Background In this section we present several lemmas that allow us to conduct our classification. First we describe some simple categories of matrices that are known to be UECSM. The results in this section are known and can be found in [2, 3, 4]. We begin with a definition: Definition. We say a matrix is algebraic of degree n if its minimal polynomial is a degree n polynomial. In other words, T is of degree n if there is some polynomial f of degree n with f(t ) = 0, but no polynomial g of degree less than n such that g(t ) = 0. We now present a theorem due to Garcia and Wogen [4]. Theorem 2.. If T is a square matrix that is algebraic of degree two, then T is UECSM. This theorem has a few useful corollaries: Corollary 2.2. All and 2 2 matrices are UECSM. Proof. The degree of the minimum polynomial of a n n matrix is always n. Thus every and 2 2 matrix is algebraic of degree 2, and by Theorem 2. is UECSM. Corollary 2.3. Every rank matrix is UECSM. Proof. Every rank one operator has the form T x = x, v u for some u, v C n. But then T 2 = ( x, v u), v u = x, v u, v u = u, v T, so we have T 2 u, v T = 0 and T is algebraic of degree two. Thus by Theorem 2., T is UECSM. 9

10 0 CHAPTER 2. TECHNICAL BACKGROUND In our attempt to classify the 3 3 matrices which are UECSM, it will be useful to be able to decompose our matrices into simpler matrices. The next lemma shows that we can, in some instances, do this: Lemma 2.4. Suppose T = n i= T i, where T i is C i symmetric for some C i. Then T is C-symmetric for C = n i= C i. Proof. We prove the result for n = 2; the theorem then follows by induction. Let T and T 2 be square complex matrices and C, C 2 conjugations such that C T C = T and C 2 T 2 C 2 = T2. Then we claim C = C C 2 is a conjugation. Antilinearity is trivially preserved by direct summation, and C 2 = I since ( ) ( ) ( ) C 0 C 0 C C 0 0 C 2 0 C 2 = 0 C 2 C 2 = ( I 0 0 I To show C is isometric, let x = (x, x 2 ) be in the domain of T. Since (x, 0) and (0, x 2 ) are orthogonal we have (x, 0) 2 + (0, x 2 ) 2 = (x, x 2 ) 2. Thus ( ) ( ) Cx 2 = C 0 x 2 0 C 2 x 2 ( ) = C x 2 C 2 x 2 ( ) = C x 2 ( ) C 2 x 2 = x 2 + x 2 2 = x 2. Thus C is a conjugation. Now we show that T is C-symmetric: ( ) ( ) ( ) C 0 T 0 C 0 CT C = 0 C 2 0 T 2 0 C 2 ( ) ( ) C T = C 0 T = 0 0 C 2 T 2 C 2 0 T2 = T. Thus we have the desired result. Our final lemma allows us to make certain simplifying assumptions about the structure of our matrix: Lemma 2.5. If T is a C-symmetric matrix, then so is at +bi for all a, b C. ).

11 Proof. Consider the matrix C(aT + bi)c. We have C(aT + bi)c = C(aT )C + C(bI)C = act C + bc 2 = at + bi = (at + bi), which is the desired result.

12 2 CHAPTER 2. TECHNICAL BACKGROUND

13 Chapter 3 The Angle Test Suppose that T is a C-symmetric matrix and that u is a generalized eigenvector with eigenvalue λ of order n (i.e., (T λi) n u = 0 but (T λi) n u 0). We have (T λi) k (Cu) = C((T λi) k u) for all k, and thus Cu is a generalized eigenvector of T with eigenvalue λ of order n. Furthermore, since C is an isometry we have u = Cu. Note that the case k = implies that if u is an eigenvector of T, then Cu is an eigenvector of T. Now suppose the eigenspace of T corresponding to the eigenvalue λ is one-dimensional, and u is a unit eigenvector with eigenvalue λ. Let v be a unit eigenvector of T with eigenvalue λ. Then since each eigenspace is one-dimensional, these eigenvectors are unique up to multiplication by a unimodular constant, and we have Cu = αv for some unimodular α. A similar argument will hold for generalized eigenvectors as long as we have a canonical method for choosing a basis for the generalized eigenspace which is respected by C. Our next lemma provides such a canonical basis: Lemma 3.. Let T be a n n matrix with eigenvalues λ,..., λ r. Suppose that T has precisely one Jordan block for each eigenvalue. Let K = ker(t λ i I) k ker(t λ i I) k 3

14 4 CHAPTER 3. THE ANGLE TEST and ˆK = ker(t λ i I) k ker(t λ i I) k such that K {0}, and let u K and v ˆK be unit vectors. Then Cu = αv for some unimodular constant α. Proof. It is clear from our discussion that Cu is a generalized eigenvector of T with eigenvalue λ of order k, and thus is an element of ˆK. Since there is only one Jordan block for λ i in the Jordan form of T we have that dim ker(t λ i I) k = k, and thus dim ˆK =. Then since v is non-zero it must span K, and so there is some scalar with αv = Cu. But v = Cu, so α =. We are now prepared to state and prove a necessary condition for a matrix T to be C-symmetric. Lemma 3.2 (The Angle Test). Let T be a n n matrix with r distinct eigenvalues λ,..., λ r which is UECSM, and suppose T has precisely one Jordan block for each eigenvalue. Let and K = ker(t λ i I) k ker(t λ i I) k L = ker(t λ j I) l ker(t λ j I) l, and let ˆK and ˆL be the analogous spaces for T. Let u K, u 2 L, v ˆK, v 2 ˆL. Then u, u 2 = v, v 2. (3.) Proof. If K (or L) is {0}, then so is ˆK (or ˆL). We can see that u = v = 0 (or u 2 = v 2 = 0) and so u, u 2 = v, v 2 = 0. So assume K, L {0}. By the isometric property of C, u, u 2 = Cu 2, Cu. By Lemma 3. there exist unimodular α, α 2 with Cu i = α i v i. Thus we have u, u 2 = Cu 2, Cu = α 2 v 2, α v

15 5 = α α 2 v, v 2. Since α i are unimodular, taking the absolute value of both sides yields Equation (3.). Lemma 3.2 provides a useful technique to prove that a given matrix with a single Jordan block for each eigenvalue is not UECSM: find a canonical basis for the generalized eigenspaces of T and T as described in Lemma 3., take the pairwise inner products of the basis eigenvectors, and check whether the norms are all the same. In particular, if a n n matrix T has n distinct eigenvalues then the pairwise inner products of the eigenvectors must have matching norms. If equation (3.) holds for all pairs of vectors in our canonical basis, we say that T passes the Angle Test. While the condition in Lemma 3.2 is clearly a necessary condition for T to be UECSM, it is not clear that it is sufficient. However, Garcia showed [] that the following stronger condition is sufficient when T has distinct eigenvalues: Theorem 3.3. Let T be a 3 3 matrix with distinct eigenvalues λ, λ 2, λ 3. Let u i be a unit eigenvector of T with eigenvalue λ i, and let v i be a unit eigenvector of T with eigenvalue λ i. Then T is UECSM if and only if T passes the Angle Test and there exist unimodular constants b,2, b,3, b 2,3 such that b i,j = u i, u j v j, v i whenever v j, v i 0, and the matrix has rank one. B = b,2 b,3 b,2 b 2,3 b,3 b 2,3 In fact, a similar condition holds for any n n matrix with n distinct eigenvalues. We omit the proof for brevity. However, we will prove the analogous (and somewhat more difficult) result for the case where a 3 3 matrix T has only two distinct eigenvalues. Let T be a 3 3 C-symmetric matrix with two distinct eigenvalues. Without loss of generality we may, by Lemma 2.5, assume that the eigenvalues

16 6 CHAPTER 3. THE ANGLE TEST are 0 and, and that 0 has multiplicity two. Furthermore, we assume that dim(ker T ) =, since whenever dim(ker T ) = 2 then T is a rank one matrix and the case is trivial due to Corollary 2.3. Let u 0 be a unit eigenvector of T corresponding to the eigenvalue 0, u 00 be a unit generalized eigenvector of T corresponding to the eigenvalue 0 that is orthogonal to u 0, u be a unit eigenvector of T corresponding to the eigenvalue. Furthermore, T also has eigenvalues 0 and with the same multiplicity. Let v 0 be a unit eigenvector of T corresponding to the eigenvalue 0, v 00 be a unit generalized eigenvector of T corresponding to the eigenvalue 0 that is orthogonal to v 0, v be a unit eigenvector of T corresponding to the eigenvalue. Lemma 3. tells us that C determines a triple of unimodular constants α 0, α 00, α with Thus for any i, j we have Cu 0 = α 0 v 0, Cu 00 = α 00 v 0, Cu = α v. u i, u j = Cu j, Cu i = α j v j, α i v i = α i α j v j, v i. Let B be the matrix given by α 0 B = α 00 ( ) α 0 α 00 α. α Whenever v j, v i 0 we have b i,j = α i α j = u i, u j v j, v i.

17 Furthermore, we see that b i,i = for all i, and that b i,j = b j,i for all i, j. Thus the matrix B is self-adjoint. Also, note that rank(b) = since α 0 colspace(b) = span α 00. α Finally, note that since T u 00 ker(t ), we have T u 00 = au 0 for some a C. But T v 00 = CT Cv 00 = CT α 00 u 00 = α 00 Cau 0 = (aα 00 α 0 )v 0. Thus the coefficients of u 0 in T u 00 and v 0 in T v 00 are related by a unimodular a factor of α 00 α 0. a We are now prepared to state our main theorem. Theorem 3.4. Let T be a 3 3 matrix under the above hypotheses, with T u 00 = au 0. We have that T is C-symmetric if and only if there exists a set of unimodular constants α 0, α 00, α such that. u i, u j = α i α j v j, v i. 2. The function C defined by letting Cu i = α i v i and extending antilinearly to C 3 is an involution. 3. T v 00 = (aα 00 α 0 )v 0. Proof. We wish to show that C is a conjugation and that CT C = T. We see that C is antilinear by construction and an involution by condition (2). We wish to show that it is an isometry. The u i form a basis for our space, so let x = a i u i. We have x 2 = x, x = = = i {0,00,} i,j {0,00,} i,j {0,00,} a i u i, j {0,00,} a i u i, a j u j a i aj u i, u j a j u j 7

18 8 CHAPTER 3. THE ANGLE TEST = = = = = i,j {0,00,} i,j {0,00,} i,j {0,00,} i,j {0,00,} j {0,00,} = Cx, Cx = Cx 2. a i a j u i, u j a i a j α i α j v j, v i a i a j α j v j, α i v i a i a j Cu j, Cu i a j Cu j, i {0,00,} a i Cu i Thus C is indeed a conjugation. We now wish to show that CT C = T. Since both maps are linear we only need to show that they agree on the basis {u 0, u 00, u }. Note that since C is an involution we have Cv i = α i u i. For u 0 and u we have Now consider CT Cu 00. We get CT Cu i = CT α i v i = α i Cλ i v i = α i λ i α i u i = λ i u i = T u i. CT Cu 00 = α 00 CT v 00 = α 00 C(aα 00 α 0 )v 0 = α 00 aα 00 α 0 Cv 0 = aα 0 α 0 u 0 = au 0 = T u 00. Thus CT C = T and we have that T is C-symmetric.

19 Chapter 4 Breaking Down the Problem 4. Our Approach By Schur s Theorem (Theorem.2), any 3 3 matrix T is unitarily equivalent to a matrix of the form shown in (.). Thus a matrix is UECSM if and only if its upper triangular form is UECSM. By Lemma 2.5 we may, without loss of generality, subtract a multiple of the identity to ensure one eigenvalue is 0. If there is more than one eigenvalue we may further divide by a scalar to obtain a matrix with an eigenvalue of. Based on the number of distinct eigenvalues we have the following three cases: Case : One distinct eigenvalue. We may assume T has the form T = 0 a b 0 0 c (4.) Case 2: Two distinct eigenvalues. We may assume T has the form T = 0 a b 0 0 c 0 0. (4.2) 9

20 20 CHAPTER 4. BREAKING DOWN THE PROBLEM Case 3: Three distinct eigenvalues. We may assume T has the form for some λ {0, }. T = 0 a b 0 c 0 0 λ (4.3) 4.2 The One Eigenvalue Case Proposition 4.. A 3 3 matrix T in the upper triangular form given in (4.) is UECSM if and only if one of the following holds:. a = c = a = c, a, c 0. Proof. The proof breaks down into three cases. Case : a = 0 or c = 0. We have T = 0 0 b 0 0 c or T = 0 a b In either case, T is a rank one matrix and thus UECSM by Corollary 2.3. Case 2: a, c 0, a = c.. Define an antilinear map C : C 3 C 3 by x 0 0 u C y = ā 0 0 cu z u 0 0 x y z

21 4.2. THE ONE EIGENVALUE CASE 2 where u is a unimodular constant such that uc R. Then we have CT C x y z u z = CT ay cu u x = = = C a ȳ cu 0 āx cu = cu u bx + a yū cu x ā 0 0 y. b c 0 z + bu x cu x 0 0 āx bx + c y c Therefore CT C = T and thus T is UECSM. Case 3: a, c 0, a c. Let e, e 2, e 3 be the standard basis vectors for C 3. We can see that T has precisely one one-dimensional eigenspace for the eigenvalue 0, spanned by e. Similarly, T has precisely one one-dimensional eigenspace, spanned by e 3. By Lemma 3., if T is C-symmetric then Ce = αe 3 for some α =. Since C is an involution, this gives us e = Cαe 3 = αce 3 and thus Ce 3 = αe. We see that e 2 is orthogonal to e and e 3, so since C is an isometry we must have Ce 2 orthogonal to Ce = αe 3 and Ce 3 = αe. Thus Ce 2 must be some scalar multiple of e 2, and in fact we have Ce 2 = βe 2 where β = since C is norm-preserving. Now T e 2 = ae. But T e 2 = CT Ce 2 = CT (βe 2 ) = βc(ce 3 ) = αβce and thus a = αβc. Since α = β = this gives us a = c, a contradiction. Thus in this case T is not UECSM.

22 22 CHAPTER 4. BREAKING DOWN THE PROBLEM 4.3 The Two Eigenvalue Case Proposition 4.2. A 3 3 matrix T in the upper triangular form given in (4.2) is UECSM if and only if one of the following holds:. a = b = c = a, c 0 and b + ac 2 = c 2 + c 4. Proof. The proof breaks down into five cases. Case : a = 0. We have T = 0 0 b 0 0 c 0 0. Thus T is a rank one matrix, and is UECSM by Corollary 2.3. Case 2: b = c = 0 It is clear that T is the direct sum of a 2 2 matrix and a matrix: 0 a 0 T = By Corollary 2.2, every 2 2 matrix and every matrix is UECSM, and the direct sum of matrices which are UECSM is also UECSM by Lemma 2.4. Thus T is UECSM. Case 3: c = 0 and a, b 0. We have T = 0 a b We see that T has a one-dimensional eigenspace with eigenvalue, and some eigenspace with eigenvalue 0. Since T has -dimensional nullspace,.

23 4.3. THE TWO EIGENVALUE CASE 23 the eigenspace with eigenvalue 0 cannot be 2-dimensional. Thus T has one one-dimensional eigenspace with eigenvalue 0, and one one-dimensional eigenspace with eigenvalue. These spaces are spanned by the following unit vectors: u 0 = 0 with eigenvalue 0, 0 b b 2 + u = 0 with eigenvalue. b 2 + Note that u 0, u = b b. 2 + Similarly, T has two one-dimensional eigenspaces: 0 v 0 = with eigenvalue 0, v = with eigenvalue. Note that v 0, v = 0. But since b 0 we have u 0, u v 0, v, and T fails the Angle Test of Lemma 3.2. Thus T is not UECSM. Case 4: a, c 0. We have T = 0 a b 0 0 c 0 0 We use the conditions of Theorem 3.4: T is UECSM if and only if there exist unimodular α 0, α 00, α such that. u i, u j = α i α j v j, v i. 2. The function C defined by letting Cu i = α i v i and extending antilinearly is an involution. 3. T v 00 = (aα 00 α 0 )v 0, where a is as in T..

24 24 CHAPTER 4. BREAKING DOWN THE PROBLEM It is clear that if such α i exist then the matrix α 0 B = α 00 ( ) α 0 α 00 α α has rank one. Conversely, if there exists some matrix B such that b i,i =, b i,j =, B is self-adjoint, and whenever v j, v i 0 we have b i,j = u i, u j v j, v i, then this matrix will factor as above and we will have a triple (α 0, α 00, α ) that satisfies condition (). Looking now at the normalized eigenvectors of T and T, we get that u 0 = (, 0, 0), u 00 = (0,, 0), u = (b + ac, c, ), b + ac 2 + c 2 + and v 0 = v 00 = (0,, c), + c 2 ( + c 2 )( ac + b 2 + c 2 + ) ( + c 2, c(ac + b), (ac + b)), v = (0, 0, ). Thus u 0, u 00 = v 00, v 0 = 0, but the other pairs of eigenvectors are not orthogonal. Assuming that T is UECSM, the Angle Test gives us u 0, u = v 0, v, b + ac c = b + ac 2 + c c 2, b + ac 2 ( + c 2 ) = c 2 ( b + ac 2 + c 2 + ), b + ac 2 = c 2 + c 4, (4.4)

25 4.3. THE TWO EIGENVALUE CASE 25 and u 00, u = v 00, v, c b + ac 2 + c 2 + = b + ac ( + c 2 )( ac + b 2 + c 2 + ), c 2 ( + c 2 ) = b + ac 2, c 2 + c 4 = b + ac 2, which is the same as (4.4). Thus the conditions of the Angle Test are satisfied if and only if (4.4) holds. Now consider the matrix B = x x u,u 0 v 0,v u 0,u v,v 0 u 00,u v,v 00 u,u 00 v 00,v, where the x is a free unimodular constant that we get since v 00, v 0 = 0. It is clear that this matrix has rank one only if But as long as we also have x = u 0,u u,u 00. v,v 0 v 00,v u,u 0 v 0,v = u 00,u v,v 00 = then we will have rank(b) =. Thus as long as (4.4) holds there exists a triple that satisfies condition (). In particular, if we set ( ) ( ) α0 α 00 α = u 0,u u,u 00 v,v 0 v 00,v u 0,u v,v 0 then the conditions in () are satisfied. Assuming that (4.4) holds, we get α 0 =, α 00 = b + ac + c + c 2 2 c α = (b + ac) c + c 23, c + c + c (b + ac) =,

26 26 CHAPTER 4. BREAKING DOWN THE PROBLEM and further we see that u = v 00 = (b + ac, c, ), + c 2 + c 2 3 ( + c 2, c(ac + b), (ac + b)). Now assume the preceding conditions are met and let C be defined as in (2), by extending the map Cu i = α i v i antilinearly. To see if C is an involution we calculate Ce i where e i are the standard basis vectors. We see that e = u 0 and e 2 = u 00, so Ce = α v and Ce 2 = α 00 v 00. To compute Ce 3 we use which implies that e 3 = (0, 0, ) = b + ac 2 + c 2 + u (b + ac)u 0 cu 00 = ( + c 2 )u (b + ac)u 0 cu 00 by (4.4), Ce 3 = ( c 2 + )Cu (b + ac)cu 0 ccu 00 = ( c 2 + )α v (b + ac)α 0 v 0 cα 00 v 00 = ( + c 2 ) (c( + 3 c 2 ), b + ac, b+ac ) c We know that C is an involution if Cv i = α i u i for all i. We can see that v = e 3 so we need But we have ( + c 2 ) 3 (c( + c 2 ), b + ac, b+ac c ) = α u. α u = u 0, u v, v 0 = b + ac + c 2 = (b + ac, c, ) + c 2 + c 2 (b + ac, c, ) c + c 2 ), b + ac, b+ac c ( b+ac 2 ( + c 2 ) 3 c

27 4.3. THE TWO EIGENVALUE CASE 27 = ) (( + c 2 )c, b + ac, b+ac ( + c 2 ) 3 c = Ce 3 For v 0 we have v 0 = + c 2 (e 2 ce 3 ). Using our definition of C on the basis vectors we get ( ) Cv 0 = α 00 v 00 c + c 2 ( + c 2 ) (c( + 3 c 2 ), b + ac, b+ac ) c ( ) ( + c 2 )( + c 2 ), c(ac + b) + c(b + ac), (b + ac) (ac + b) = ( + c 2 ) 2 = (, 0, 0) and so = u 0. Finally, we consider v 00. We see that Cv 00 = v 00 = + c 2 3 ( + c 2 )e c(ac + b)e 2 (ac + b)e 3, + c 2 3 ( + c 2 )α 0 v 0 c(ac + b)α 00 v 00 (ac + b)ce 3. Taking each of these terms in turn, we have ( + c 2 )α 0 v 0 = + c 2 3 (0, ( + c 2 ) 2, c( + c 2 ) 2 ), c(ac + b)α 00 v 00 = (ac + b)ce 3 = + c 2 3 (c(ac + b)( + c 2 ), c 2 ac + b 2, c ac + b 2 ), + c 2 3 (c(ac + b)( + c 2 ), ac + b 2, ac+b 2 c ). So we see that Cv 00 = ( + c 2 ) 3 (0, ( c 2 ) 3, c( + 2 c 2 + c 4 ) + c( + c 2 ) 2 )

28 28 CHAPTER 4. BREAKING DOWN THE PROBLEM = (0,, 0) = α 00 u 00. Thus for any u i, we get C 2 u i = Cα i v i = α i α i u i = u i, and we see that C 2 is the identity on the basis formed by {u 0, u 00, u }. Thus C is an involution, and condition (2) of Theorem 3.4 is satisfied. Finally, we wish to show that condition (3) of Theorem 3.4 holds. We have T u 00 = (a, 0, 0) = au 0, so we wish to show that But T v 00 = T v 00 = av 0. + c 2 3 (0, ( + c 2 )a, ( + c 2 )ac = a (0,, c) + c 2 = av 0. Thus T meets the conditions of Theorem 3.4 if and only if b+ac 2 = c 2 + c 4, and we have the desired result. 4.4 The Three Eigenvalue Case Proposition 4.3. Let T M 3 3 (C) be in the upper triangular form given in (4.3). Then T is UECSM if and only if one of the following holds:. a = b = b = c = a = c = a, c, b ac 0, (ac + b(λ ))a cλ + c a 2 + ab(λ ) = a( c 2 + λ 2 λ) bc λ ( + b ac 2 ( + a 2 )) 2 (a c 2 bc),

29 4.4. THE THREE EIGENVALUE CASE 29 a( a 2 c + ab(λ ) + cλ) ( + a 2 )(ac + b(λ )) a c 2 bc a( c 2 + λ 2 λ) bc = a( c 2 + λ 2 λ) bc ( + λ c 2 )(a c 2 bc), (ac + b(λ ))( a 2 c + ab(λ ) + cλ) = c aλ(λ )( + λ 2 + ac+b(λ ) λ 2 λ 2 )(λ 2 λ). Proof. The proof breaks down into seven cases. Case : a = b = 0 It is clear that T is the direct sum of a matrix and a 2 2 matrix, and thus UECSM: T = 0 c. 0 0 λ Case 2: b = c = 0. Similarly, T is the direct sum of a 2 2 matrix and a matrix, and thus UECSM: 0 a 0 T = λ Case 3: a = c = 0, b 0. We have If wee define U by T = U = 0 0 b λ , we see that U is a unitary matrix. Furthermore, 0 b 0 U T U = 0 λ 0, 0 0 and thus T is unitarily equivalent to the direct sum of a 2 2 matrix and a matrix, and UECSM.

30 30 CHAPTER 4. BREAKING DOWN THE PROBLEM Case 4: c = 0 and a, b 0. We have T = 0 a b λ We can see that T has three one-dimensional eigenspaces, spanned by the following unit eigenvectors:. u 0 = (, 0, 0) with eigenvalue 0, ( ) u = a, a a 2 + a a, 0 with eigenvalue, 2 + ( ) u λ = b, 0, λ b with eigenvalue λ. λ 2 + b 2 b λ 2 + b 2 Similarly, T has three corresponding eigenspaces with corresponding unit eigenvectors: ( ) λ v 0 = ( a, λā 2 +)λ 2 + b ( a, b, 2 2 +)λ 2 + b 2 v = (0,, 0), v λ = (0, 0, ). ( a 2 +)λ 2 + b 2 By Lemma 2.4 we know that if T is C-symmetric, then u 0, u λ = v 0, v λ, and computing these inner products gives us which implies that either b = 0, or b λ 2 + b 2 = b ( a 2 +)λ 2 + b 2, λ 2 + b 2 = ( a 2 + )λ 2 + b 2 and thus a = 0, which is a contradiction. Thus T is not UECSM.

31 4.4. THE THREE EIGENVALUE CASE 3 Case 5: a = 0, b, c 0. By an argument similar to the argument in Case 4, T is not UECSM. We have 0 0 b T = 0 c. 0 0 λ Now T has three one-dimensional eigenspaces, spanned by the following unit eigenvectors: u 0 = (, 0, 0) with eigenvalue 0, u = (0,, 0) with eigenvalue, ( u λ = + c 2 + b b, c 2 λ λ with eigenvalue λ. λ λ Similarly, T has three corresponding one-dimensional eigenspaces spanned by corresponding unit eigenvectors: v 0 = + λ 2 b v = + λ c v λ = (0, 0, ). ( λb, 0, ), 2 ( 0, λ c, ), By Lemma 2.4 we know that if T is C-symmetric, then u 0, u = v 0, v, and thus we have 0 = ( + λ 2) ( + > 0, λ 2) b c which is a contradiction. Thus T is not UECSM.

32 32 CHAPTER 4. BREAKING DOWN THE PROBLEM Case 6: a, c 0, b = ac. It is clear that T has three one-dimensional eigenspaces, spanned by the following unit eigenvectors: u 0 = (, 0, 0) with eigenvalue 0, u = (a,, 0) + a 2 with eigenvalue, u λ = (ac, c, λ ) with eigenvalue λ. ac 2 + c 2 + λ 2 Similarly, T has three corresponding one-dimensional eigenspaces spanned by corresponding unit eigenvectors: but v 0 = v = (, a, 0), + a 2 (0, λ, c), λ 2 + c 2 v λ = (0, 0, ). We can compute that ac u 0, u λ = 0 ac 2 + c 2 + λ 2 v 0, v λ = 0, and thus T fails the Angle Test and is not UECSM. Case 7: a, c, b ac 0. We use the rank one condition given in Theorem 3.3. Simple computation gives us that T has eigenspaces spanned by the normalized eigenvectors u 0 = (, 0, 0), u = (a,, 0), + a 2 u λ = c + λ 2 + b( λ) ac λ 2 λ 2 ( ac+b(λ ) λ 2 λ, c λ, ).

33 4.4. THE THREE EIGENVALUE CASE 33 Similarly, T has eigenspaces spanned by the normalized eigenvectors ( ) λ v 0 =, aλ,, λ + b ac 2 + aλ ac b b ac b ac 2 ( ) v = 0, λ,, + λ c c 2 us v λ = (0, 0, ). Computing the pairwise inner products of these eigenvectors in turn gives u 0, u = u 0, u λ = u, u λ = v 0, v = v 0, v λ = v, v λ = a + a 2, ac + b(λ ), (λ 2 c λ) + λ 2 + b( λ) ac λ 2 λ 2 a 2 c + ab(λ ) + cλ, (4.5) ( + a 2 c )( + λ 2 + b( λ) ac λ 2 λ 2 )(λ 2 + λ) + λ c a(λ λ 2 c 2 ) + bc 2 λ + b ac 2 + aλ + λ b ac 2 + aλ b ac 2,. + λ c 2, b ac 2 (bc a c 2 ) Given our hypotheses that λ 0, and b ac 0, these are all well-defined. So now consider the matrix u 0,u u 0,u λ v,v 0 v λ,v 0 u B =,u 0. v 0,v u λ,u 0 v 0,v λ u,u λ v λ,v u λ,u v,v λ The conjugate symmetry of the inner product gives us that u i, u j v j, v i = u j, u i v i, v j,

34 34 CHAPTER 4. BREAKING DOWN THE PROBLEM and so this matrix has the form of the matrix B from Theorem 3.3. The matrix B is rank one if and only if the columnspace of B is onedimensional, if and only if each column is a scalar multiple of each other column. Let B i be the ith column of B; then if B is rank one we must have B = u 0, u v, v 0 B 2, B 2 = u, u λ v λ, v B 3, (4.6) B 3 = u 0, u λ v λ, v 0 B. The system in (4.6) implies nine equations, three of which are trivially true. The other six are given by u 0, u v, v 0 =, (4.7) u, u λ v λ, v =, (4.8) u 0, u λ v λ, v =, (4.9) u 0, u u λ u 0 = u λ, u v, v 0 v 0, v λ v, v λ, (4.0) u 0, u u, u λ = u 0, u λ v, v 0 v λ, v v λ, v 0, (4.) u, u λ u λ, u 0 = u, u 0 v λ, v v 0, v λ v 0, v. (4.2) Note first that satisfying the rank-one condition implies satisfying the Angle Test because of equations (4.7), (4.8), and (4.9). Further, note that if we assume (4.0), (4.), and (4.2), then substituting (4.) into (4.0) gives u λ, u v, v λ = u 0, u u 0, u u, u λ v, v 0 v, v 0 v λ, v = u λ, u u 0, u v, v λ v, v 0 and thus (4.7) holds. A similar argument gives that (4.8) and (4.9) hold as well, so the conditions of Theorem 3.3 are met if and only if we have (4.0),

35 4.4. THE THREE EIGENVALUE CASE 35 (4.), and (4.2). Rearranging these equations gives u 0, u u 0, u λ u, u λ u 0, u u, u λ u 0, u λ u 0, u λ u, u λ u 0, u = v 0, v v 0, v λ v, v λ = v 0, v v, v λ v 0, v λ = v 0, v λ v, v λ v 0, v Plugging in values from Equation (4.5) gives the desired condition. This completes our classification of the 3 3 matrices from the upper triangular form. Given a 3 3 upper triangular matrix, these results allows us to easily determine whether it is UECSM. Since every matrix is equivalent to an upper triangular matrix, this is in effect a classification of all 3 3 matrices which are UECSM.

36 36 CHAPTER 4. BREAKING DOWN THE PROBLEM

37 Bibliography [] Stephan Garcia, Personal Communication. [2] Stephan Ramon Garcia, Conjugation and Clark operators, Contemporary Mathematics 393 (2006), [3] Stephan Ramon Garcia and Mihai Putinar, Complex symmetric operators and applications, Transactions of the American Mathematical Society 358 (2005), [4] Stephan Ramon Garcia and Warren R. Wogen, Some new classes of complex symmetric operators, preprint. [5] Roger A. Horn and Charles R. Johnson, Matrix analysis, Cambridge University Press,