Multilinear Algebra 1

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1 Multilinear Algebra 1 Tin-Yau Tam Department of Mathematics and Statistics 221 Parker Hall Auburn University AL 36849, USA tamtiny@auburn.edu November 30, Some portions are from B.Y. Wang s Foundation of Multilinear Algebra (1985 in Chinese)

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3 Chapter 1 Review of Linear Algebra 1.1 Linear extension In this course, U, V, W are finite dimensional vector spaces over C, unless specified. All bases are ordered bases. Denote by Hom (V, W ) the set of all linear maps from V to W and End V := Hom (V, V ) the set of all linear operators on V. Notice that Hom (V, W ) is a vector space under usual addition and scalar multiplication. For T Hom (V, W ), the image and kernel are Im T = {T v : v V } W, Ker T = {v : T v = 0} V which are subspaces. It is known that T is injective if and only if Ker T = 0. The rank of T is rank T := dim Im T. Denote by C m n the space of m n complex matrices. Each A C m n can be viewed as in Hom (C n, C m ) in the obvious way. So we have the concepts, like rank, inverse, image, kernel, etc for matrices. Theorem Let E = {e 1,..., e n } be a basis of V and let w 1,..., w n W. Then there exists a unique T Hom (V, W ) such that T (e i ) = w i, i = 1,..., n. Proof. For each v = n a ie i, define T v = n a it e i = n a iw i. Such T is clearly linear. If S, T Hom (V, W ) such that T (e i ) = w i, i = 1,..., n, then Sv = n a ise i = n a it e i = T v for all v = n a ie i V so that S = T. In other words, a linear map is completely determined by the images of basis elements in V. A bijective T Hom (V, W ) is said to be invertible and its inverse (T 1 T = I V and T T 1 = I W ) is linear, i.e., T 1 Hom (W, V ): T 1 (α 1 w 1 + α 2 w 2 ) = v means that T v = α 1 w 1 + α 2 w 2 = T (α 1 T 1 w 1 + α 2 T 1 w 2 ), 3

4 4 CHAPTER 1. REVIEW OF LINEAR ALGEBRA i.e., v = T 1 w 1 + α 2 T 1 w 2 for all w 1, w 2 W and v V. We will simply write ST for S T for S Hom (W, U), T Hom (V, W ). Two vector spaces V and W are said to be isomorphic if there is an invertible T Hom (V, W ). Theorem Let T Hom (V, W ) and dim V = n. 1. Then rank T = k if and only if there is a basis {v 1,..., v k, v k+1,..., v n } for V such that T v 1,..., T v k are linearly independent and T v k+1 = = T v n = dim V = dim Im T + dim Ker T. Proof. Since rank T = k, there is a basis {T v 1,..., T v k } for Im T. Let {v k+1,..., v k+l } be a basis of Ker T. Set E = {v 1,..., v k, v k+1,..., v k+l }. For each v V, T v = k a it v i since T v Im T. So T (v k a iv i ) = 0, i.e., v k a iv i Ker T. Thus v k a iv i = k+l i=k+1 a iv i, i.e., v = k+l a iv i so that E spans V. So it suffices to show that E is linearly independent. Suppose k+l a iv i = 0. Applying T on both sides, k a it v i = 0 so that a 1 = = a k = 0. Hence k+l i=k+1 a iv i = 0 and we have a k+1 = = a k+l = 0 since v k+1,..., v k+l are linear independent. Thus E is linearly independent and hence a basis of V ; so k + l = n. Theorem Let A C m n. 1. rank A A = rank A. 2. For each A C m n, rank A = rank A = rank A T, i.e., column rank and row rank of A are the same. Proof. (1) Notice Ax = 0 if and only if A Ax = 0. It is because A Ax = 0 implies that (Ax) (Ax) = x A Ax = 0. So Ax 1,..., Ax k are linearly independent if and only if A Ax 1,..., A Ax k are linearly independent. Hence rank A A = rank A. (2) Since rank A = rank A A rank A from Problem 3 and thus rank A rank A since (A ) = A. Hence rank A = rank A T (why?). Problems 1. Show that dim Hom (V, W ) = dim V dim W. 2. Let T Hom (V, W ). Prove that rank T min{dim V, dim W }. 3. Show that if T Hom (V, U), S Hom (U, W ), then rank ST min{rank S, rank T }. 4. Show that the inverse of T Hom (V, W ) is unique, if exists. Moreover T is invertible if and only if rank T = dim V = dim W. 5. Show that V and W are isomorphic if and only dim V = dim W.

5 1.2. MATRIX REPRESENTATIONS OF LINEAR MAPS 5 6. Show that if T Hom (V, U), S Hom (U, W ) are invertible, then ST is also invertible. In this case (ST ) 1 = T 1 S Show that A C m n is invertible only if m = n, i.e., A has to be square. Show that A C n n is invertible if and only if the columns of A are linearly independent. 8. Prove that if A, B C n n such that AB = I n, then BA = I n. Solutions to Problems Let {e 1,..., e n } and {f 1,..., f m } be bases of V and W respectively. Then ξ ij Hom (V, W ) defined by ξ ij (e k ) = δ ik f j, i, j, k = 1,..., n, form a basis of Hom (V, W ). Thus dim Hom (V, W ) = dim V dim W. 2. From definition rank T dim W. From Theorem rank T dim V. 3. Since Im ST Im S, rank ST rank S. By Theorem rank ST = dim V dim Ker ST. But Ker T Ker ST so that rank ST dim V dim Ker T = rank T. 4. Suppose that S and S Hom (V, W ) are inverses of T Hom (V, W ). Then S = S(T S ) = (ST )S = S ; inverse is unique. T Hom (V, W ) invertible T is bijective (check!). Now injective T Ker T = 0, and surjective T rank T = dim W 5. One implication follows from Problem 4. If dim V = dim W, let E = {e 1,..., e n } and {f 1,..., f n } be bases of V and W, define T Hom (V, W ) by T e i = f i for all i which is clearly invertible since T 1 f i = e i. 6. (T 1 S 1 )(ST ) = I V and (ST )(T 1 S 1 ) = I W. 7. From Problem 4, a matrix A is invertible only if A is square. The second statement follows from the fact the rank A is the dimension of the column space of A. 8. If AB = I, then rank AB = n so rank A = n(= rank B) by Problem 3. So Ker A = 0 and Ker B = 0 by Theorem Thus A and B is invertible. Then I = A 1 ABB 1 = A 1 B 1 (BA) 1 so that BA = I. (or, without using Theorem 1.1.2, note that B = B(AB) = (BA)B so that (I BA)B = 0. Since Im B = C n, I BA = 0, i.e., BA = I). 1.2 Matrix representations of linear maps In this section, U, V, W are finite dimensional vector spaces. Matrices A C m n can be viewed as elements in Hom (C n, C m ). On the other hand, each T Hom (V, W ) can be realized as a matrix once we fix bases for V and W.

6 6 CHAPTER 1. REVIEW OF LINEAR ALGEBRA Let T Hom (V, W ) with bases E = {e 1,..., e n } for V and F = {f 1,..., f m } for W. Since T e j W, we have T e j = m a ij f i, j = 1,..., n. The matrix A = (a ij ) C m n is called the matrix representation of T with respect to the bases E and F, denoted by [T ] F E = A. From Theorem [S] F E = [T ]F E if and only if S = T, where S, T Hom (V, W ). The coordinate vector of v V with respect to the basis E = {e 1,..., e n } for V is denoted by [v] E := (a 1,..., a n ) T C n where v = n a iv i. Indeed we can view v Hom (C, V ) with 1 as a basis of C. Then v 1 = n a ie i so that [v] E is indeed [v] E 1. Theorem Let T Hom (V, W ) and S Hom (W, U) with bases E = {e 1,..., e n } for V, F = {f 1,..., f m } for W and G = {g 1,..., g l } for U. Then [ST ] G E = [S] G F [T ] F E and in particular [T v] F = [T ] F E[v] E, v V. Proof. Let A = [T ] F E and B = [S]G F, i.e., T e j = m a ijf i, j = 1,..., n. and T f i = l k=1 b kig k, i = 1,..., m. So m m l ST e j = a ij Sf i = a ij ( b ki )g k = So [ST ] G E = BA = [S]G F [T ]F E. k=1 l (BA) kj g k. Consider I := I V End V. The matrix [I] E E is called the transitive matrix from E to E since k=1 [v] E = [Iv] E = [I] E E [v] E, v V, i.e., [I] E E transforms the coordinate vector [v] E with respect to E to [v] E with respect to E. From Theorem 1.2.1, [I] E E = ([I]E E ) 1. Two operators S, T End V are said to be similar if there is an invertible P End V such that S = P 1 AP. Similarity is an equivalence relation and is denoted by. Theorem Let T End V with dim V = n and let E and E be bases of V. Then A := [T ] E E and B := [T ]E E are similar, i.e., there is an invertible P C n n such that P AP 1 = B. Conversely, similar matrices are matrix representations of the same operator with respect to different bases.

7 1.2. MATRIX REPRESENTATIONS OF LINEAR MAPS 7 Proof. Let I = I V. By Theorem 1.2.1, [I] E E [I]E E = [I]E E = I n where I n is n n identity matrix. Denote by P := [I] E E (the transitive matrix from E to E) so that P 1 = [I] E E. Thus [T ] E E = [IT I]E E = [I]E E [T ] E E[I] E E = P 1 [T ] E EP. Suppose that A and B are similar, i.e., B = R 1 BR. Let E be a basis of V. By Theorem A uniquely determines T End V such that [T ] E E = A. By Theorem 1.2.1, an invertible R C n n uniquely determines a basis E of V such that [T ] E E = R so that [T ] E E = [I]E E [T ] E E[I] E E = R 1 [T ] E ER = B. So functions ϕ : C n n C that take constant values on similarity orbits of C n n, i.e., ϕ(a) = P 1 AP for all invertible P C n n, are defined for operators, for examples, determinant and trace of a operator. Problems 1. Let E = {e 1,..., e n } and F = {f 1,..., f m } be bases for V and W. Show that the matrix representation ϕ := [ ] F E : Hom (V, W ) C m n is an isomorphism. 2. Let E and F be bases of V and W respectively. Show that if T Hom (V, W ) is invertible, then [T 1 ] E F = ([T ]F E ) Let E and F be bases of V and W respectively. Let T Hom (V, W ) and A = [T ] F E. Show that rank A = rank T. Solutions to Problems It is straightforward to show that ϕ is linear by comparing the (ij) entry of [αs+βt ] F E and α[s]f E +β[t ]F E. Since dim Hom (V, W ) = mn = dim C m n, it suffices to show that ϕ is injective. From Theorem 1.1.1, ϕ is injective. 2. From Theorem 1.2.1, [T 1 ] E F [T ]F E = [T 1 T ] E E = [I V ] E E = I n where dim V = n. Thus use Problem 1.8 to have [T 1 ] E F = ([T ]F E ) 1, or show that [T ] F E [T 1 ] E F = I n similarly. 3. (Roy) Let rank T = k and let E = {v 1,..., v n } and F = {w 1,..., w m } be bases for V and W. When we view v V as an element in Hom (C, V ), by Problem 1, the coordinate maps [ ] E : V C n and [ ] F : W C m are isomorphisms. So rank T = dim Im T = dim T v 1,..., T v n = dim [T v 1 ] F,..., [T v n ] F = dim [T ] F E[v 1 ] E,..., [T ] F E[v n ] E = dim Im A = rank A.

8 8 CHAPTER 1. REVIEW OF LINEAR ALGEBRA 1.3 Inner product spaces Let V be a vector space. An inner product on V is a function (, ) : V V C such that 1. (u, v) = (v, u) for all u, v V. 2. (α 1 v 1 + α 2 v 2, u) = α 1 (v 1, u) + α 2 (v 2, u) for all v 1, v 2, u V, α 1, α 2 C. 3. (v, v) 0 for all v V and (v, v) = 0 if and only if v = 0. The space V is then called an inner product space. The norm induced by the inner product is defined as v = (v, v), v V. Vectors v satisfying v = 1 are called unit vectors. Two vectors u, v V are said to be orthogonal if (u, v) = 0, denoted by u v. A basis E = {e 1,..., e n } is called an orthogonal basis if the vectors are orthogonal. It is said to be orthonormal if (e i, e j ) = δ ij, i, j = 1,..., n. where is the Kronecker delta notation. δ ij = { 1 if i = j 0 if i j Theorem (Cauchy-Schwarz inequality) Let V be an inner product space. Then (u, v) u v, u, v V. Equality holds if and only if u and v are linearly dependent, i.e., one is a scalar multiple of the other. Proof. It is trivial when v = 0. Suppose v 0. Let w = u (u,v) v 2 v. Clearly (w, v) = 0 so that 0 (w, w) = (w, u) = (u, u) (u, v) v 2 (v, u) = (u, u 2 v) 2 v 2. Equality holds if and only if w = 0, i.e. u and v are linearly dependent. Theorem (Triangle inequality) Let V be an inner product space. Then Proof. u + v u + v, u, v V. u + v 2 = (u + v, u + v) = u 2 + 2Re (u, v) + v 2 u (u, v) + v 2 By Theorem 1.3.1, we have u + v 2 u u v + v 2 = ( u + v ) 2.

9 1.3. INNER PRODUCT SPACES 9 Theorem Let E = {e 1,..., e n } be an orthonormal basis of V. For any u, v V, u = (u, v) = n (u, e i )e i, n (u, e i )(e i, v). Proof. Let u = n j=1 a je j. Then (u, e i ) = ( n j=1 a je j, e i ) = a i, i = 1,..., n. Now (u, v) = ( n (u, e i)e i, v) = n (u, e i)(e i, v). Denote by v 1,..., v k the span of the vectors v 1,..., v k. Theorem (Gram-Schmidt orthogonalization) Let V be an inner product space with basis {v 1,..., v n }. Then there is an orthonormal basis {e 1,..., e n } such that v 1,..., v k = e 1,..., e k, k = 1,..., n. Proof. Let e 1 = e 2 = e n = v 1 v 1, v 2 (v 2, e 1 )e 1 v 2 (v 2, e 1 )e 1... v n (v n, e 1 )e 1 (v n, e n 1 )e n 1 v n (v n, e 1 )e 1 (v n, e n 1 )e n 1 It is direct computation to check that {e 1,..., e n } is the desired orthonormal basis. The inner product on C n : for any x = (x 1,..., x n ) T and y = (y 1,..., y n ) T C n, n (x, y) := x i y i is called the standard inner product on C n. The induced norm is called the 2-norm and is denoted by x 2 := (x, x) 1/2. Problems 1. Let E = {e 1,..., e n } be a basis of V and v V. Prove that v = 0 if and only if (v, e i ) = 0 for all i = 1,..., n.

10 10 CHAPTER 1. REVIEW OF LINEAR ALGEBRA 2. Show that each orthogonal set of nonzero vectors is linearly independent. 3. Let E = {e 1,..., e n } be an orthonormal basis of V. If u = n a ie i and n = n b ie i, then (u, v) = n a ib i. 4. Show that an inner product is completely determined by an orthonormal basis, i.e., if the inner products (, ) and, have the same orthonormal basis, then (, ) and, are the same. 5. Show that (A, B) = tr (B A) defines an inner product on C m n, where B denotes the complex conjugate transpose of B. 6. Prove tr (A B) tr (A A)tr (B B) for all A, B C m n. Solutions to Problems v = 0 (v, u) = 0 for all u V (v, e i ) = 0 for all i. Conversely each u V is of the form n a ie i so that (v, e i ) = 0 for all i implies (v, u) = 0 for all u V. 2. Suppose that S = {v 1,..., v n } is an orthogonal set of nonzero vectors. With n a iv i = 0, a j (v j, v j ) = ( n a iv i, v j ) = 0 so that a j = 0 for all j since (v j, v j ) 0. Thus S is linearly independent. 3. Follows from Theorem Follows from Problem Straightforward computation. 6. Apply Cauchy-Schwarz inequality on the inner product defined in Problem Adjoints Let V, W be inner product spaces. For each T Hom (V, W ), the adjoint of T is S Hom (W, V ) such that (T v, w) W = (v, Sw) V for all v V, w W and is denoted by T. Clearly (T ) = T. Theorem Let W, V be inner product spaces. Each T Hom (V, W ) has a unique adjoint. Proof. By Theorem 1.3.4, let E = {e 1,..., e n } be an orthonormal basis of V. For any w W, define S Hom (W, V ) by Sw := n (w, T e i ) W e i.

11 1.4. ADJOINTS 11 For any v V, by Theorem 1.3.3, v = n (v, e i) V e i so that (v, Sw) V = (v, = (T n (w, T e i ) W e i ) V = n (T e i, w) W (v, e i ) V n (v, e i ) V e i, w) W = (T v, w) W. Uniqueness follows from Problem 1. Theorem Let E = {e 1,..., e n } and F = {f 1,..., f m } be orthonormal bases of the inner product spaces V and W respectively. Let T Hom (V, W ). Then [T ] E F = ([T ]F E ), where the second denotes the complex conjugate transpose. Proof. Let A := [T ] F E, i.e., T e j = m k=1 a kjf k, j = 1,..., n. So (T e j, f i ) W = a ij. By the proof of Theorem T f j = n (f j, T e i ) W e i = So [T ] E F = A = ([T ] F E ). n (T e i, f j ) W e i = n a ji e i. Notice that if E and F are not orthonormal, then [T ] E F = ([T ]F E ) may not hold. Problems 1. Let S, T Hom (V, W ) where V, W are inner product spaces. Prove that (a) (T v, w) = 0 for all v V and w W if and only if T = 0. (b) (Sv, w) = (T v, w) for all v V and w W if and only if S = T. 2. Show that (ST ) = T S where T Hom (V, W ) and S L(W, U) and U, V, W are inner product spaces. 3. Let E and F be orthonormal bases of inner product space V. Prove that ([I] F E ) = ([I] F E ) 1 = [I] E F. 4. Let G be a basis of the inner product space V. Let T End V. Prove that [T ] G G and ([T ]G G ) are similar. 5. Let V, W be inner product spaces. Prove that if T Hom (V, W ), then rank T = rank T. 6. The adjoint : Hom (V, W ) Hom (W, V ) is an isomorphism.

12 12 CHAPTER 1. REVIEW OF LINEAR ALGEBRA Solutions to Problems (a) (T v, w) = 0 for all v V and w W T v = 0 for all v V, i.e., T = 0. (b) Consider S T. 2. Since (v, T S w) V = (T v, S w) V = (ST v, w) V, by the uniqueness of adjoint, (ST ) = T S. 3. Notice that I = I where I := I V. So from Theorem ([I] F E ) = [I ] E F = [I]F E. Then by Problem 2.2, ([I]F E ) 1 = [I] E F. 4. (Roy and Alex) Let E be an orthonormal basis of V. Let P = [I V ] E G (the transition matrix from G to E). Then [T ] G G = P 1 [T ] E E P and [T ] E E = P [T ]G G P 1. By Theorem [T ] E E = ([T ]E E ). Together with Problem 2 [T ] G G = P 1 [T ] E EP = P 1 ([T ] E E) P = P 1 (P [T ] G GP 1 ) P = (P P ) 1 ([T ] G G) (P P ), i.e. ([T ] G G ) and ([T ] G G ) are similar. Remark: If G is an orthonormal basis, then P is unitary and thus ([T ] G G ) = ([T ] G G ), a special case of Theorem Follows from Theorem 1.4.2, Problem 2.3 and rank A = rank A where A is a matrix. 6. It is straightforward to show that : Hom (V, W ) Hom (W, V ) is a linear map. Since dim(v, W ) = dim(w, V ) it remains to show that is injective. First show that (T ) = T : for all v V, w W, (T w, v) = (v, T w) V = (T v, w) V = (w, T v) Then for any S, T Hom (V, W ), S = T S = T. 1.5 Normal operators and matrices An operator T End V on the inner product space V is normal if T T = T T ; Hermitian if T = T, positive semi-definite, abbreviated as psd or T 0 if (T x, x) 0 for all x V ; positive definite, abbreviated as pd or T > 0 if (T x, x) > 0 for all 0 x V ; unitary if T T = I. Unitary operators form a group. When V = C n is equipped with the standard inner product and orthonormal basis, linear operators are viewed as matrices in C n n and the adjoint is simply the complex conjugate transpose. Thus a matrix A C n n is said to be normal if A A = AA ; Hermitian if A = A, psd if (Ax, x) 0 for all x C n ; pd if (Ax, x) > 0 for all 0 x C n ; unitary if A A = I. Unitary matrices in

13 1.5. NORMAL OPERATORS AND MATRICES 13 C n n form a group and is denoted by U n (C). One can see immediately that A is unitary if A has orthonormal columns (rows since AA = I as well from Problem 1.8). Theorem (Schur triangularization theorem) Let A C n n. There is U U n (C) such that U AU is upper triangular. Proof. Let λ 1 be an eigenvalue of A with unit eigenvector x 1, i.e., Ax 1 = λ 1 x 1 with x 1 2 = 1. Extend via Theorem to an orthonormal basis {x 1,..., x n } for C n and set Q = [x 1 x n ] which is unitary. Since Q Ax 1 = λ 1 Q x 1 = λ 1 (1,..., 0) T, λ 1 0 Â := Q AQ =. A 1 0 where A 1 C (n 1) (n 1). By induction, there is U 1 U n 1 (C) such that U1 A 1 U 1 is upper triangular. Set P :=. U 1 0 which is unitary. So is upper triangular and set U = QP. λ 1 P 0 ÂP =. U1 A 1 U 1 0 Theorem Let T End V, where V is an inner product space. Then there is an orthonormal basis E of V such that [T ] E E is upper triangular. Proof. For any orthonormal basis E = {e 1,..., e n} of V let A := [T ] E E. By Theorem there is U U n (C), where n = dim V, such that U AU is upper triangular. Since U is unitary, E = {e 1,..., e n } is also an orthonormal basis, where e j = n u ije i, j = 1,..., n (check!) and [I]E E = U. Hence [T ] E E = [I]E E [T ]E E [I]E E = U AU since U = U 1. Lemma Let A C n n be normal and let r be fixed. Then a rj = 0 for all j r if and only if a ir = 0 for all i r. Proof. From a rj = 0 for all j r and AA = A A, n a rr 2 = a rj 2 = (AA ) rr = (A A) rr = a rr 2 + a ir 2. j=1 i r So a ir = 0 for all i r. Then apply it on the normal A.

14 14 CHAPTER 1. REVIEW OF LINEAR ALGEBRA Theorem Let A C n n. Then 1. A is normal if and only if A is unitarily similar to a diagonal matrix. 2. A is Hermitian (psd, pd) if and only if A is unitarily similar to a real (nonnegative, positive) diagonal matrix. 3. A is unitary if and only if A is unitarily similar to a diagonal unitary matrix. Proof. Notice that if A is normal, Hermitian, psd, pd, unitary, so is U AU for any U U n (C). By Theorem 1.5.1, there is U U n (C) such that U AU is upper triangular and also normal. By Lemma 1.5.3, the result follows. The rest follows similarly. From Gram-Schmidt orthgonalization (Theorem 1.3.4), one has the QR decomposition of an invertible matrix. Theorem (QR decomposition) Each invertible A C n n can be decomposed as A = QR, where Q U n (C) and R is upper triangular. The diagonal entries of R may be chosen positive; in this case the decomposition is unique. Problems 1. Show that upper triangular in Theorem may be replaced by lower triangular. 2. Prove that A End V is unitary if and only if Av = v for all v V. 3. Show that A C n n is psd (pd) if and only if A = B B for some (invertible) matrix B. In particular B may be chosen lower or upper triangular. 4. Let V be an inner product space. Prove that (i) (Av, v) = 0 for all v V if and only if A = 0, (ii) (Av, v) R for all v V if and only if A is Hermitian. What happens if the underlying field C is replaced by R? 5. Show that if A is psd, then A is Hermitian. What happens if the underlying field C is replaced by R? 6. Prove Theorem Prove that if T Hom (V, W ) where V and W are inner product spaces, then T T 0 and T T 0. If T is invertible, then T T > 0 and T T > 0. Solutions to Problems Apply Theorem on A and take complex conjugate transpose back.

15 1.5. NORMAL OPERATORS AND MATRICES A End V is unitary A A = I V So unitary A implies Av 2 = (Av, Av) = (A Av, v) = (v, v) = v 2. Conversely, Av = v for all v V implies ((A A I)v, v) = 0 where A A I is Hermitian. Apply Problem If A = B B, then x Ax = xb Bx = Bx for all x C n. Conversely if A is psd, we can define a psd A 1/2 via Theorem Apply QR on A 1/2 to have A 1/2 = QR where Q is unitary and R is upper triangular. Hence A = A 1/2 A 1/2 = (A 1/2 ) A 1/2 = R Q QR = R R. Set B := R. Let A 1/2 = P L where P is unitary and L is lower triangular (apply QR decomposition on (A ) 1 ). Then A = L L. 4. (i) It suffices to show for Hermitian A because of Hermitian decomposition A = H +ik where H := (A+A )/2 and K = (A A )/(2i) are Hermitian and (Av, v) = (Hv, v) + i(kv, v) and (Hv, v), (Kv, v) R. So (Av, v) = 0 for all v if and only if (Hv, v) = (Kv, v) = 0 for all v V. Suppose A is Hermitian and (Av, v) = 0 for all v V. Then there is an orthonormal basis of eigenvectors {v 1,..., v n } of A with corresponding eigenvalues λ 1,..., λ n, but all eigenvalues are zero because λ i (v i, v i ) = (Av i, v i ) = 0. When C is replaced by R, we cannot conclude that A = 0 since for any real skew symmetric matrix A R n n, x T Ax = 0 (why?). (Brice) Notice that A is psd (because (Av, v) = 0). From Problem 3 via matrix-operator approach, A = B B for some B End V. Thus from Problem 4.2, A = A. (ii) 2[(Av, w) + (Aw, v)] = (A(v + w), v + w) (A(v w), v w) R so that Im [(Av, w) + (Aw, v)] = 0 for all v, w. Similarly 2i[ (Av, w) + (Aw, v)] = (A(v + iw), v + iw) (A(v iw), v iw) R so that Re [ (Av, w) + (Aw, v)] = 0 for all v, w. So (Av, w) = (Aw, v) = (v, Aw), i.e., A = A. When C is replaced by R, we cannot conclude that A is real symmetric since for any real skew symmetric matrix A R n n, x T Ax = Follow from Problem Express the columns (basis of C n ) of A in terms of the (orthonormal) columns of Q. 7. (T T v, v) = (T v, T v) 0 for all v V. So T T 0 and similar for T T 0. When T is invertible, T v 0 for all v 0 so (T T v, v) = (T v, T v) > 0.

16 16 CHAPTER 1. REVIEW OF LINEAR ALGEBRA 1.6 Inner product and positive operators Theorem Let V be an inner product space with inner product (, ) and let T End V. Then u, v := (T u, v), u, v V, defines an inner product if and only if T is pd with respect to (, ). Proof. Suppose u, v := (T u, v), u, v V, defines an inner product, where (, ) is an inner product for V. So (T u, v) = u, v = v, u = (T v, u) = (T u, v) for all u, v V. So T = T, i.e., self-adjoint. For any v 0, 0 < v, v = (T v, v) so that T is pd with respect to (, ). The other implication is trivial. The next theorem shows that in the above manner inner products are in one-one correspondence with pd matrices. Theorem Let (, ) and, be inner products of V. Then there exists a unique T End V such that u, v := (T u, v), u, v V. Moreover, T is positive definite with respect to both inner products. Proof. Let E = {e 1,..., e n } be an orthonormal basis of V with respect to (, ). So each v V can be written as v = n (v, e i)e i. Now for each v V defines T End V by n T v := v, e i e i Clearly (T u, v) = n n u, e i (e i, v) = u, (v, e i )e i = u, v. Since, is an inner product, from Theorem T is pd with respect to (, ). When v 0, T v, v = (T 2 v, v) = (T v, T v) > 0) so that T is pd with respect to,. The uniqueness follows from Problem 4.1. Theorem Let F = {f 1,..., f n } be a basis of V. There exists a unique inner product (, ) on V such that F is an orthonormal basis. Proof. Let (, ) be an inner product with orthonormal basis E = {e 1,..., e n }. By Problem 1.4 S End V defined by Sf i = e i is invertible. Set T := S S > 0 ( and T > 0 from Problem 5.7 are with respect to (, )). So u, v = (T u, v) is an inner product by Theorem and f 1,..., f n are orthonormal with respect to,. It is straightforward to show the uniqueness. Problems

17 1.7. INVARIANT SUBSPACES Let E = {e 1,..., e n } be a basis of V. For any u = n a ie i and v = n b ie i, show that (u, v) := n a ib i is the unique inner product on V so that E is an orthonormal basis. 2. Find an example that there are inner products (, ) on, on V and T End V so that T is pd with respect to one but not to the other. 3. Let V be an inner product space. Show that for each A C n n there are u 1,..., u n ; v 1,..., v n C n such that a ij = (u i, v j ) 1 i, j n. 4. Suppose that (, ) on, are inner products on V, T End V and let T ( ) and T be the corresponding adjoints. Show that T ( ) and T are similar. Solutions to Problems Uniqueness follows from Theorem By Theorem there is a pd S End V with respect to both inner products such that u, v = (Su, v) for all u, v V and thus S 1 u, v = (u, v). So u, T v = T u, v = (ST u, v) = (u, T ( ) S ( ) v) = S 1 u, T ( ) S ( ) v Since S ( ) = S = S, say, and (A 1 ) = (A ) 1 for any A End V ((A 1 ) A = (AA 1 ) = I by Problem 4.2). Then u, T v = u, (S 1 ) T ( ) S v = u, (S ) 1 T ( ) S v. Hence T = (S ) 1 T ( ) S. 1.7 Invariant subspaces Let W be a subspace of V. For any T Hom (V, U), the restriction of T, denote by T W, is the unique T 1 Hom (W, U) such that T 1 (w) = T (w) for all w W. Let W be a subspace of V and T End V. Then W is said to be invariant under T if T w W for all w W, i.e., T (W ) W. The trivial invariant subspaces are 0 and V. Other invariant subspaces are called nontrivial or proper invariant subspaces. If W is invariant under T, then the restriction T W Hom (W, V ) of T induces T W End W (we use the same notation T W ).

18 18 CHAPTER 1. REVIEW OF LINEAR ALGEBRA Theorem Let V be an inner product space over C and T End V. If W is an invariant subspace under T and T, then (a) (T W ) = T W. (b) If T is normal, Hermitian, psd, pd, unitary, so is T W. Proof. (a) For any x, y W, T W x = T x, T W y = T y. By the assumption, (T W x, y) W = (T x, y) V = (x, T y) V = (x, T W y) W. So (T W ) = T W. (b) Follows from Problem 7.3. Problems 1. Prove that if W V is a subspace and T 1 Hom (W, V ), then there is T Hom (V, U) so that T W = T 1. Is T unique? 2. Show that the restriction on W V of the sum of S, T Hom (V, U) is the sum of their restrictions. How about restriction of scalar multiple of T on W? 3. Show that if S, T End V and the subspace W is invariant under S and T, then (ST ) W = (S W )(T W ). 4. Let T End V and S End W and H Hom (V, W ) satisfy SH = HT. Prove that Im H is invariant under S and Ker H is invariant under T. Solutions to Problems Let {e 1,..., e m } be a basis of W and extend it to E = {e 1,..., e m, e m+1,..., e n } a basis of V. Define T Hom (V, U) by T e i = T 1 e i, i = 1,..., m and T e j = w j where j = m + 1,..., n and w m+1,..., w n W are arbitrary. Clearly T W = T 1 but T is not unique. 2. (S + T ) W v = (S + T )v = Sv + T v = S W v + T W v for all v V. So (S + T ) W = S W + T W. 3. (S W )(T W )v = S W (T v) = ST v since T v W and thus (S W )(T W )v = ST W v for all v. 4. For any v Ker H, H(T v) = SHv = 0, i.e., T v Ker H so that Ker H is an invariant under T. For any Hv Im H (v V ), S(Hv) = HT (v) Im H.

19 1.8. PROJECTIONS AND DIRECT SUMS Projections and direct sums The vector space V is said to be a direct sum of the subspaces W 1,..., W m if each v V can be uniquely expressed as v = n w i where w i W i, i = 1,..., n and is denoted by V = W 1 W m. In other words, V = W W m and if w w m = w w m where w i W i, then w i = w i for all i. Theorem V = W 1 W m if and only if V = W W m and W i (W 1 + +Ŵi+ +W m ) = 0 for all i = 1,..., m. Here Ŵi denotes deletion. In particular, V = W 1 W 2 if and only if W 1 + W 2 = V and W 1 W 2 = 0. Proof. Problem 7. In addition, if V is an inner product space, and W i W j if i j, i.e., w i w j whenever w i W i, w j W j, we call V an orthogonal sum of W 1,..., W m, denoted by V = W 1 + +W m. Suppose V = W 1 W m. Each P i End V defined by P i v = w i, i = 1,..., m satisfies Pi 2 = P i and Im P i = W i. In general P End V is called a projection if P 2 = P. Notice that P is a projection if and only if I V P is a projection; in this case Im P = Ker (I V P ). Theorem Let V be a vector space and let P End V be a projection. Then the eigenvalues of P are either 0 or 1 and rank P = tr P. Proof. Let rank P = k. By Theorem 1.1.2, there are v 1,..., v n V such that P v 1,..., P v k is a basis of Im P and {v k+1,..., v n } is a basis of Ker P. From P 2 = P, E = {P v 1,..., P v k, v k+1,..., v n } is linearly independent (check) and thus a basis of V. So [P ] E E = diag (1,..., 1, 0,..., 0) in which there are k ones. So we have the desired results. The notions direct sum and projections are closely related. Theorem Let V be a vector space and let P 1,..., P m be projections such that P P m = I V. Then V = Im P 1 Im P m. Conversely, if V = W 1 W m, then there are unique projections P 1,..., P m such that P P m = I V and Im P i = W i, i = 1,..., m. Proof. From P 1 + +P m = I V, each v V can be written as v = P 1 v+ +P m v so that V = Im P Im P m. By Theorem m m m dim V = tr I V = tr P i = rank P i = dim Im P i. So V = Im P 1 Im P m (Problem 8.1). Conversely if V = W 1 W m, then for any v V, there is unique factorization v = w w m, w i W i, i = 1,..., m. Define P i End V by P i v = w i. It is easy to see each P i is a projection and Im P i = W i and P P m = I V. For the uniqueness, if there are projections Q 1,..., Q m such that Q Q m = I V and Im Q i = W i for all i, then for each v V, from the unique factorization Q i v = P i v so that P i = Q i for all i.

20 20 CHAPTER 1. REVIEW OF LINEAR ALGEBRA Corollary If P End V is a projection, then V = Im P Ker P. Conversely if V = W W, then there is a projection P End V such that Im P = W and Ker P = W. We call P End V an orthogonal projection if P 2 = P = P. Notice that P is an orthogonal projection if and only if I V P is an orthogonal projection. The notions orthogonal sum and orthogonal projections are closely related. Theorem Let V be an inner product space and let P 1,..., P m be orthogonal projections such that P P m = I V. Then V = Im P 1 + +Im P m. Conversely, if V = W 1 + +W m, then there are unique orthogonal projections P 1,..., P m such that P P m = I V and Im P i = W i, i = 1,..., m. Proof. From Theorem 1.8.3, if P 1,..., P m are orthogonal projections such that P P m = I V, then V = Im P 1 Im P m. It suffices to show that the sum is indeed orthogonal. From the proof of Theorem 1.8.3, P i v = v i Im P i for all i, where v V is expressed as v = m v i. Thus for all u, v V, if i j, then (P i u, P j v) = (P i u, v j ) = (u, P i v j ) = (u, P i v j ) = (u, 0) = 0. Conversely if V = W 1 + +W m, then from Theorem there are unique projections P 1,..., P m such that P P m = I V and Im P i = W i, for all i. It remains to show that each projection P i is an orthogonal projection, i.e., P i = Pi. For u, v V, write u = m u i, v = m v i where u i, v i W i for all i. Then for all i m m (P i u, v) = (u i, v) = (u i, v i ) = (u i, v i ) = ( u i, v i ) = (u, P i v). Corollary Let V be an inner product space. If P End V is an orthogonal projection, then V = Im P +Ker P. Conversely if V = W +W, then there is an orthogonal projection P End V such that Im P = W and Ker P = W. Proof. Apply Theorem on P and I V P. Problems 1. Show that if V = W W m, then V = W 1 W m if and only if dim V = dim W dim W m. 2. Let P 1,..., P m End V be projections on V and P P m = I V. Show that P i P j = 0 whenever i j. 3. Let P 1,..., P m End V be projections on V. Show that P P m is a projection if and only if P i P j = 0 whenever i j.

21 1.8. PROJECTIONS AND DIRECT SUMS Show that if P End V is an orthogonal projection, then P v v for all v V. 5. Prove that if P 1,..., P m End V are projections on V with P P m = I V, then there is an inner product so that P 1,..., P n are orthogonal projections. 6. Show that if P End V is an orthogonal projection on the inner product space and if W is P -invariant subspace of V, then P W End W is an orthogonal projection. 7. Prove Theorem Solutions to Problems By Theorem 1.8.3, V = Im P 1 Im P m so that for any v V, P i P j v = P i v j = 0 if i j. 3. P P m is a projection means P i + P i P j = i i j i P 2 i + i j P i P j = (P P m ) 2 = P P m. So if P i P j = 0 for i j, then P P m is a projection. Conversely, if P P m is a projection, so is I V (P P m ). (Roy) Suppose that P := P 1 + +P m is a projection. Then the restriction of P on its image P (V ) is the identity operator and P i P (V ) is still a projection for each i = 1,..., m. According to Theorem 1.8.3, P (V ) = m Im P i P (V ) = m Im P i, which implies that P i P j = 0 whenever i j. 4. I P is also an orthogonal projection. Write v = P v + (I P )v so that v 2 = P v 2 + (I P )v 2 P v By Theorem 1.8.3, V = Im P 1 Im P m. Let {e i1,..., e ini } be a basis of Im P i for all i. Then there is an inner product, by Theorem so that the basis {e 11,..., e 1n1,..., e m1,..., e mnm } is orthonormal. Thus P 1,..., P m are orthogonal projections by Theorem Notice that P 2 W = P 2 W = P W since P 2 = P and from Theorem (P W ) = P W = P W since P = P. 7.

22 22 CHAPTER 1. REVIEW OF LINEAR ALGEBRA 1.9 Dual spaces and Cartesian products The space V = Hom (V, C) is called the dual space of V. Elements in V are called (linear) functionals. Theorem Let E = {e 1,..., e n } be a basis of V. Define e 1,..., e n V by e i (e j ) = δ ij, i, j = 1,..., n. Then E = {e 1,..., e n} is a basis of V. Proof. Notice that each f V can be written as f = n f(e i)e i since both sides take the same values on e 1,..., e n. Now if n a ie i = 0, then 0 = n a ie i (e j) = a j, j = 1,..., n. So e 1,..., e n are linearly independent. The basis E is called a basis of V dual to E, or simply dual basis. If T Hom (V, W ), then its dual map (or transpose) T Hom (W, V ) is defined by T (ϕ) = ϕ T, ϕ W. The functional T (ϕ) is in V, and is called the pullback of ϕ along T. Immediately the following identity holds for all ϕ W and v V : T (ϕ), v = ϕ, T (v) where the bracket f, v := f(v) is the duality pairing of V with its dual space, and that on the right is the duality pairing of W with its dual. This identity characterizes the dual map T, and is formally similar to the definition of the adjoint (if V and W are inner product spaces). We remark that we use the same notation for adjoint and dual map of T Hom (V, W ). But it should be clear in the context since adjoint requires inner products but dual map does not. Theorem The map : Hom (V, W ) Hom (W, V ) defined by (T ) = T is an isomorphism. Proof. Problem 10. When V has an inner product, linear functionals have nice representation. Theorem (Riesz) Let V be an inner product space. For each f V, there is a unique u V such that f(v) = (v, u) for all v V. Hence the map ξ : V V defined by ξ(u) = (, u) is an isomorphism. Proof. Let E = {e 1,..., e n } be an orthonormal basis of V. Then for all v V, by Theorem f(v) = f( n (v, e i )e i ) = n n f(e i )(v, e i ) = (v, f(e i )e i )

23 1.9. DUAL SPACES AND CARTESIAN PRODUCTS 23 so that u := n f(e i)e i. Uniqueness follows from the positive definiteness of the inner product: If (v, u ) = (v, u) for all v V, then (v, u u ) = 0 for all v. Pick v = u u to have u = u. The map ξ : V V defined by ξ(u) = (, u) is clearly a linear map bijective from the previous statement. Let V 1,..., V m be m vector spaces over C. m i V i = V 1 V m = {(v 1,..., v m ) : v i V i, i = 1,..., m} is called the Cartesian product of V 1,..., V m. It is a vector space under the natural addition and scalar multiplication. Remark: If V is not finite-dimensional but has a basis {e α : α A} (axiom of choice is needed) where A is the (infinite) index set, then the same construction as in the finite-dimensional case (Theorem 1.9.1) yields linearly independent elements {e α : α A} in V, but they will not form a basis. Example (for Alex s question): The space R, whose elements are those sequences of real numbers which have only finitely many non-zero entries, has a basis {e i : i = 1, 2,..., } where e i = (0,..., 0, 1, 0,... ) in which the only nonzero entry 1 is at the ith position. The dual space of R is R ℵ, the space of all sequences of real numbers and such a sequence (a n ) is applied to an element (x n ) R to give n a nx n, which is a finite sum because there are only finitely many nonzero x n. The dimension of R is countably infinite but R ℵ does not have a countable basis. Problems 1. Show that Theorem remains true if the inner product is replaced by a nondegenerate bilinear form B(, ) on a real vector space. Nondegeneracy means that B(u, v) = 0 for all v V implies that u = Let {e 1,..., e n } be an orthonormal basis of the inner product space V. Show that {f 1,..., f n } is a dual basis if and only if f j (v) = (v, e i ) for all v V and j = 1,..., n. 3. Let {f 1,..., f n } be a basis of V. Show that if v V such that f j (v) = 0 for all j = 1,..., n, then v = Let {f 1,..., f} be a basis of V. Prove that there is a basis E = {e 1,..., e n } of V such that f i (e j ) = δ ij, i, j = 1,..., n. 5. Let E = {e 1,..., e n } and F = {f 1,..., f n } be two bases of V and let E and F be their dual bases. If [I V ] F E = P and [I V ]E F = Q, prove that Q = (P 1 ) T. 6. Show that if S Hom (W, U) and T Hom (V, W ), then (ST ) = T S.

24 24 CHAPTER 1. REVIEW OF LINEAR ALGEBRA 7. Show that ξ : V V defined by ξ(v)(ϕ) = ϕ(v) for all v V and V V, is an (canonical) isomorphism. 8. Suppose E and F are bases for V and W and let E and F be their dual bases. For any T Hom (V, W ), what is the relation between [T ] F E and [T ] E F? 9. Show that dim(v 1 V m ) = dim V dim V m. 10. Prove Theorem Solutions to Problems Let B(, ) be a nondegenerate bilinear form on a real vector space V. Define ϕ : V V by ϕ(v) = B(v, ). Clearly ϕ is linear and we need to show that ϕ is surjective. Since dim V = dim V, it suffices to show that ϕ is injective (and thus bijective). Let v Ker ϕ, i.e., B(v, w) = 0 for all w V. By the nondegeneracy of B(, ), v = Notice that (e i, e j ) = δ ij. If F is a dual basis, then for each v = n (v, e i)e i, f j (v) = f j ( n (v, e i)e i ) = n (v, e i)f j (e i ) = (v, e j ). On the other hand, if f j (v) = (v, e j ) for all v, then f j (e i ) = (e i, e j ) = δ ij, i.e., F is a dual basis. 3. The assumption implies that f(v) = 0 for all f V. If v 0, extends it to basis E = {v, v 2,..., v n } of V. Define g V by g(v) = 1 and g(v i ) = 0 for all i = 2,..., n. 4. Introduce an inner product on V so that by Theorem we determine u 1,..., u n via f j (v) = (v, u j ). By Theorem and Theorem there is a pd T End V such that (T u i, u j ) = δ ij. Set e i = T u i for all i. Uniqueness is clear. Another approach: Since {f 1,..., f n } is a basis, each f i 0 so that dim Ker f i = n 1. So pick v i j i Ker f j 0 (why?) such that f i (v i ) = 1. Then f j (v i ) = 0 for all j For any ϕ U, (T S )ϕ = T (ϕ S) = (ϕ S) T = ϕ (ST ) = (ST ) ϕ. So (ST ) = T S. 7. Similar to Problem 10. Notice that ξ(e i ) = e i because of e e. for all i. It is canonical 8. Let A = [T ] F E, i.e., T e j = n a ijf i. Write v = i α ie i so that (T f j )v = f j (T v) = f j ( i α i T e i ) = f j ( i,k α i a ki f k ) = i,k α i a ki f j (f k ) = i α i a ji.

25 1.10. NOTATIONS 25 On the other hand ( i a ji e i )(v) = i a ji e i ( k α k e k ) = i α i a ji. So T fj = i a jie i, i.e., [T ] E F = AT, i.e., ([T ] F E )T = [T ] E F. explains why T is also called the transpose of T. This (Roy) Suppose E = {e 1,..., e n } and F = {f 1,..., f m }. Let [T ] F E = (a ij ) C m n and [T ] E F = (b pq) C n m. By definition, T e j = m a ijf i and T fq = n p=1 b pqe p. Since e j (e i) = δ ij and fj (f i) = δ ij, the definition (T fj )e i = fj (T e i) implies that b ij = a ji. Thus [T ] E F = ([T ]F E )T. 9. If E i = {e i1,..., e ini } is a basis of V i, i = 1,..., m, then E = {(e 11,..., e m1 ),..., (e 1n1,..., e mnm )} is a basis of V 1 V m (check!). 10. is linear since for any scalars α, β, (αs + βt ) ϕ = ϕ(αs + βt ) = αϕs + βϕt = αs ϕ + βt ϕ = (αs + βt )ϕ. It is injective since if T = 0, then ϕ T = T ϕ = 0 for all ϕ V. By Problem 3, T = 0. Thus is an isomorphism Notations Denote by S m the symmetric group on {1,..., m} which is the group of all bijections on {1,..., m}. Each σ S m is represented by ( ) 1,..., m σ =. σ(1),..., σ(m) The sign function ε on S m is ε(σ) = { 1 if σ is even 1 if σ is odd where σ S m. Let α, β, γ denote finite sequences whose components are natural numbers Γ(n 1,..., n m ) := {α : α = (α(1),..., α(m)), 1 α(i) n i, i = 1,..., m} Γ m,n := {α : α = (α(1),..., α(m)), 1 α(i) n, i = 1,..., m} G m,n := {α Γ m,n : α(1) α(m)} D m,n := {α Γ m,n : α(i) α(j) whenever i j} Q m,n := {α Γ m,n : α(1) < < α(m)}.

26 26 CHAPTER 1. REVIEW OF LINEAR ALGEBRA For α Q m,n and σ S m, define Then ασ D m,n and ασ := (α(σ(1)),..., ασ(m))). D m,n = {ασ : α Q m.n, σ S m } = Q m,n S m. (1.1) Suppose n > m. For each ω Q m,n, denote by ω the complementary sequence of ω, i.e., ω Q n m,n, the components of ω and ω are 1,..., n, and ( ) 1,..., m, m + 1,..., n σ = ω(1),..., ω(m), ω (1),..., ω S (n m) n. It is known that ε(σ) = ( 1) s(ω)+m(m+1)/2, (1.2) where s(ω) := ω(1)+ +ω(m). Similarly for ω Q m,n, θ S m and π S n m, ( ) 1,..., m, m + 1,..., n σ = ωθ(1),..., ωθ(m), ω π(1),..., ω S π(n m) n and Moreover S n = {σ = ε(σ) = ε(θ)ε(π)( 1) s(ω)+m(m+1)/2, (1.3) ( ) 1,..., m, m + 1,..., n ωθ(1),..., ωθ(m), ω π(1),..., ω : π(n m) ω Q m,n, θ S m, π S n m }. (1.4) Let A C n k. For any 1 m n, 1 l k, α Q m,n and β Q l,k. Let A[α β] denote the submatrix of A by taking the rows α(1),..., α(m) of A and the columns β(1),..., β(l). So the (i, j) entry of A[α β] is a α(i)β(j). The submatrix of A complementary to A[α β] is denoted by A(α β) := A[α β ] C (n m) (k l). Similarly we define A(α β] := A[α β] C (n m) k and A[α β) := A[α β ] C n (k l). Recall that if A C n n, the determinant function is given by It is easy to deduce that det A = det A = σ S n ε(σ) σ S n ε(π)ε(σ) n a iσ(i). (1.5) n a π(i),σ(i). (1.6) Let A C n k. For any 1 m min{n, k}, α Q m,n and β Q m,k, det A[α β] = σ S m ε(σ) a α(i),βσ(i). (1.7)

27 1.10. NOTATIONS 27 For any θ S n, we have det A[αθ β] = det A[α βθ] = ε(θ) det A[α β]. (1.8) When α Γ m,n, β Γ m,k, A[α β] C m m whose (i, j) entry is a α(i)β(j). However if α Γ m,n and α D m,n, then A[α β] has two identical rows so that det A[α β] = 0, α Γ m,n \ D m,n. (1.9) Theorem (Cauchy-Binet) Let A C r n, B C n l and C = AB. Then for any 1 m min{r, n, l} and α Q m,r and β Q m,l, det C[α β] = det A[α ω] det B[ω β]. ω Q m,n Proof. det C[α β] = σ S m ε(σ) = σ S m ε(σ) = = = = = = = σ S m ε(σ) j=1 c α(i),βσ(i) n a α(i),j b j,βσ(i) τ Γ m,n a α(i),γ(i) b γ(i),βσ(i) a α(i),γ(i) ε(σ)b γ(i),βσ(i) τ Γ m,n σ S m τ Γ m,n ω D m,n m a α(i),γ(i) det B[γ β] a α(i),γ(i) det B[γ β] ω Q m,n σ S m ω Q m,n ( σ S m ε(σ) a α(i),ωσ(i) det B[ωσ β] a α(i),ωσ(i) ) det B[ω β] ω Q m,n det A[α ω] det B[ω β]. Theorem (Laplace) Let A C n n, 1 m n and α Q m,n. Then det A = det A[α ω]( 1) s(α)+s(ω) det A(α ω). ω Q m,n

28 28 CHAPTER 1. REVIEW OF LINEAR ALGEBRA Proof. The right side of the formula is ( 1) s(α) = ( 1) s(α) = ( 1) s(α) = n ε(π)ε(σ) σ S n = det A. ω Q m,n ( 1) s(ω) det A[α ω] det A[α ω ] π S n m ε(θ)ε(π)( 1) s(ω) n m a α(i)ωθ(i) ω Q m,n θ S m j=1 n ε(σ)( 1) m(m+1)/2 a α(i)σ(i) a α (i m)σ(i) σ S n i=m+1 a π(i)σ(i) a α (j)ω π(j) Theorem Let U U n (C). Then for α, β Q m,n with 1 m n, det U det U[α β] = ( 1) s(α)+s(β) det U(α β). (1.10) In particular det U[α β] = det U(α β). (1.11) Proof. Because det I[β ω] = δ βω and U [β γ] = U[γ β] T, apply Cauchy-Binet theorem on I n = U U to yield δ βω = det U[γ β] det U[γ ω]. (1.12) τ Q m,n Using Laplace theorem, we have δ αγ det U = det U[γ ω]( 1) s(α)+s(ω) det U(α ω). ω Q m,n Multiplying both sides by U[γ β] and sum over τ S m, the left side becomes γ Q m,n δ αγ det U det U[γ β] = det U det U[α β] and the right side becomes = γ Q m,n ω Q m,n det U[γ β] det U[γ ω]( 1) s(α)+s(ω) det U(α ω) ω Q m,n δ βω ( 1) s(α)+s(ω) det U(α ω) = ( 1) s(α)+s(β) det U(α β) Since det U = 1, we have the desired result.

29 1.10. NOTATIONS 29 The permanent function of A is per A = n σ S n which is also known as positive determinant. a iσ(i) Problems 1. Prove equation (1.2). 2. Prove that m n j=1 a ij = γ Γ m,n m a iγ(i). 3. Prove a general Laplace expansion theorem: Let A C n n, α, β Q m,n, and 1 m n. δ αβ det A = det A[α ω]( 1) s(α)+s(ω) det A(β ω). ω Q m,n 4. Show that Γ(n 1,..., n m ) = D m,n = ( ) n + m 1 n i, Γ m,n = n m, G m,n =, m ( ) ( ) n n m!, Q m,n =. m m Solutions to Problems (Roy) If A = (a ij ) C m n, the left side of m n j=1 a ij = m γ Γ m,n a iγ(i) is the product of row sums of A. Fix n and use induction on m. If m = 1, then both sides are equal to the sum of all entries of A. Suppose that the identity holds for m 1. We have γ Γ m,n a iγ(i) = = = m 1 γ Γ m 1,n m 1 n j=1 n j=1 a ij a ij a iγ(i) n j=1 n a mj j=1 a mj by induction hypothesis

30 30 CHAPTER 1. REVIEW OF LINEAR ALGEBRA The ith slot of each sequence in Γ(n 1,..., n m ) has n i choices. So Γ(n 1,..., n m ) = m n i and thus Γ m,n = n m follows from Γ(n 1,..., n m ) = m n i with n i = n for all i. ). Now Qm,n is the number of picking m distinct num- G m,n = ( n+m 1 m bers from 1,..., n. So Q m,n = ( n ) m) is clear and Dm,n = Q m,n m! = m! ( n m

31 Chapter 2 Group representation theory 2.1 Symmetric groups Let S n be the symmetric group on {1,..., n}. A matrix P C n n is said to be a permutation matrix if p i,σ(j) = δ i,σ(j), i, j = 1,..., n where σ S n and we denote by P (σ) for P because of its association with σ. Denote by P n the set of all permutation matrices in C n n. Notice that ϕ : S n P n (δ P (σ)) is an isomorphism. Theorem (Cayley) Each finite group G with n elements is isomorphic to a subgroup of S n. Proof. Let Sym (G) denotes the group of all bijections of G. For each σ G, define the left translation l σ : G G by l σ (x) = σx for all x G. It is easy to show that l στ = l σ l τ for all σ, τ G so that l : G Sym (G) is a group homomorphism. The map l is injective since l σ = l τ implies that σx = τx for any x G and thus σ = τ. So G is isomorphic to some subgroup of Sym (G) and thus to some subgroup of S n. The following is another proof in matrix terms. Let G = {σ 1,..., σ n }. Define the regular representation Q : G GL n (C): Q(σ) ij := (δ σi,σσ j ) GL n (C). It is easy to see that Q(σ) is a permutation matrix and Q : σ Q(σ) is injective. So it suffices to show that Q is a homomorphism. Now (Q(σ)Q(π)) ij = n Q(σ) ik Q(π) kj = k=1 n δ σi,σσ k δ σk,πσ j = δ σi,σπσ j = Q(σπ) ij. k=1 31

32 32 CHAPTER 2. GROUP REPRESENTATION THEORY So we view each finite group as a subgroup of S n for some n. The element σ S n is called a cycle of length k (1 k n), if there exist 1 i 1,..., i k n such that σ(i t ) = i t+1, t = 1,..., k 1 σ(i k ) = i 1 σ(i) = i, i {i 1,..., i + k} and we write σ = (i 1,..., i k ). Two cycles (i 1,..., i k ) and (j 1,..., j m ) are said to be disjoint if there is no intersection between the two set {i 1,..., i k } and {j 1,..., j m }. From the definition, we may express the cycle σ as (i 1, σ(i 1 ),..., σ k 1 (i 1 )). For any σ S n there is 1 k n such that σ k (i) = i and let k be the smallest such integer. Then (i σ(i) σ k 1 (i)) is a cycle but is not necessarily equal to σ. The following theorem asserts that σ can be reconstructed from these cycles. Theorem Each element σ S n can be written as a product of disjoint cycles. The decomposition is unique up to permutation of the cycles. Proof. Let i n be a positive integer. Then there is a smallest positive integer r such that σ r (i) = i. If r = n, then the cycle (i σ(i) σ r 1 (i)) is σ. If r < n, then there is positive integer j n such that j {i, σ(i),..., σ r 1 (i)}. Then there is a smallest positive integer s such that σ s (j) = j. Clearly the two cycles (i σ(i) σ r 1 (i)) and (j σ(j) σ s 1 (j)) are disjoint. Continue the process we have σ = (i σ(i) σ r 1 (i))(j σ(j) σ s 1 (j)) (k σ(k) σ t 1 (k)) For example, ( ) = ( ) ( 4 7 ) ( 6 ) = ( ) ( 4 7 ). So c(σ) = 3 (which include the cycle of length 1) ( ) = ( ) ( 3 4 ) = ( 3 4 ) ( ) = ( 3 4 ) ( ) So c(σ) = 2. Cycles of length 2, i.e., (i j), are called transpositions. Theorem Each element σ S n can be written (not unique) as a product of transposition.

33 2.1. SYMMETRIC GROUPS 33 Proof. Use Theorem (i 1 i k ) = (i 1 i 2 )(i 2 i 3 ) (i k 1 i k ) = (i 1 i k )(i 1 i k 1 ) (i 1 i 2 ). (2.1) Though the decomposition is not unique, the parity (even number or odd number) of transpositions in each σ S n is unique (see Merris p.56-57). Problems 1. Prove that each element of S n is a product of transpositions. (Hint: (i 1 i 2 i k ) = (i 1 i 2 )(i 2 i 3 ) (i k 1 i k ) = (i 1 i k )(i 1 i k 1 ) (i 1 i 2 )). 2. Show that S n is generated by the transposition (12), (23),..., (n 1, n) (Hint: express each tranposition as a product of (12), (23),..., (n 1, n)). 3. Express the following permutations as a product of nonintersecting cycles: (a) (a b)(a i 1... i r b j 1... j r ). (b) (a b)(a i 1... i r b j 1... j r )(a b). (c) (a b)(a i 1... i r ). (d) (a b)(a i 1... i r )(a b). 4. Express σ = (24)(1234)(34)(356)(56) S 7 as a product of disjoint cycles and find c(σ). Solutions to Problems Use Theorem and the hint. 2. (i, i + 2) = (i + 1, i + 2)(i, i + 1)(i + 1, i + 2) and each (ij) is obtained by conjugating (i, i + 1) by the product (j, j + 1) (i + 1, i + 2) where i < j. So S n is generated by (12), (23),..., (n 1, n). (Zach) (ij) = (1i)(1j)(1i). By Problem 1, every cycle is a product of transpositions. So S n is generated by (12), (13),..., (1n). Now (1k) = (1, k 1)(k 1, k)(1, k 1) for all k > (a) (a b)(a i 1... i r b j 1... j r ) = (a i 1... i r )(b j 1... j s ). (b) (a b)(a i 1... i r b j 1... j s )(a b) = (a i 1... i r )(b j 1... j s )(a b) = (a j 1... j s b i 1... i r ). (c) (a b)(a i 1... i r ) = (a i 1... i r b). (d) (a b)(a i 1... i r )(a b) = (b i 1... i r ). ( ) σ = (24)(1234)(34)(356)(56) = = (14235)(6)(7) So c(σ) = 3.