NONCOMMUTATIVE POLYNOMIAL EQUATIONS. Edward S. Letzter. Introduction

NONCOMMUTATIVE POLYNOMIAL EQUATIONS Edward S Letzter Introduction My aim in these notes is twofold: First, to briefly review some linear algebra Second, to provide you with some new tools and techniques for studying equations involving matrices I don t expect you to grasp all of the material immediately, but I hope that you will have a good understanding of it by the end of the course Note: There are exercises scattered throughout; some are trivial, and a few require a substantial effort Preliminaries 0 (i) If X and Y are sets then X Y refers to the new set {(x, y) x X and y Y } (ii) The notation f: X Y will be used to denote a function whose domain is X and whose values are contained in the image set Y ; we will also describe this situation as follows: X Y x f(x) (iii) Let f: X Y If W X, then f(w ) = {f(w) w W } These notes were originally developed for the summer REU (Research Experiences for Undergraduates) programs at Texas A&M University (1998 2000) and Temple University (2001 2004), supported in part by NSF REU Site Grants 1 Typeset by AMS-TEX

2 NONCOMMUTATIVE POLYNOMIAL EQUATIONS If f(x) = Y then f is said to be surjective and we further say that f maps X onto Y Note that f always maps X surjectively onto f(x) If f(x 1 ) = f(x 2 ) if and only if x 1 = x 2, for all x 1, x 2 X, then we say that f is injective If f is both injective and surjective, then we say that f is bijective or that f is a bijection Matrices 1 (i) The set of complex numbers, equipped with the usual addition and multiplication operations, will be denoted C complex entries All matrices discussed below will be assumed to have (ii) I will assume that you are already comfortable with matrix arithmetic (ie, matrix addition, scalar multiplication, and matrix multiplication) The n n identity matrix will be denoted I; the size of I will always be clear from the context in which it is used Recall that an inverse of an n n matrix A is a matrix A 1 such that A 1 A = AA 1 = I While A may not have an inverse, if A 1 exists then it is unique (prove this!) When A 1 exists we say that A is either invertible or nonsingular If A 1 does not exist then A is said to be singular (iii) The trace of an n n matrix A is the sum of the entries along the main diagonal and is denoted trace(a) Recall (or prove for yourself), assuming A and B to be n n matrices, that trace(a + B) = trace(a) + trace(b), that trace(ab) = trace(ba), and that trace(αa) = α trace(a) for all α C However, it is not generally true that trace(ab) = trace(a) trace(b) If Q is an invertible matrix then trace(a) = trace(q 1 AQ) Also, when I is the n n identity matrix, trace(i) = n (iv) The determinant of an n n matrix A is described in almost every text on linear algebra (see, for example, Chapter 5 of Linear Algebra, 2nd Edition, by K Hoffman and R Kunze) and is denoted det(a) (Unfamiliarity with determinants should not present an obstacle for you when working on your research projects)

MATH 4096: SPRING 2009 3 Vector Spaces 2 A (complex) vector space V is a nonempty set equipped with a (vector) sum V V V (u, v) u + v, a scalar product C V V (α, v) αv, and a zero vector 0 V such that the following properties hold, for u, v, w V and α, β C: u + (v + w) = (u + v) + w, u + v = v + u, u + 0 = u, there exists an element u V such that u + ( u) = 0, α(βu) = (αβ)u, (α + β)u = αu + βu, α(u + v) = αu + βv, 1u = u The elements of V are referred to as vectors 3 (i) Using the definition in (2), you should check that the zero vector is unique, that u = ( 1)u for all u V, that α0 = 0 for all α C, and that if u V is nonzero then αu = 0 if and only if α = 0 (ii) We will use C n to designate the set of n 1 complex matrices, α 1 α 2 α n α 1, α 2,, α n C It follows immediately that C n is a vector space under the usual matrix addition and scalar multiplication When there is no possibility of confusion, we will use 0 to designate the zero n 1 matrix (equivalently, the zero vector) in C n 4 For the remainder of these notes V will denote a vector space

4 NONCOMMUTATIVE POLYNOMIAL EQUATIONS Subspaces and Bases 5 A nonempty subset U of V is a (vector) subspace of V provided that U is closed under vector sums and scalar products: if u, v U and α C then u + v U and αu U (It is easy to check that a subspace U of V is itself a vector space under the sum and scalar product defined for V ) The subspace of V consisting only of the zero vector will be denoted (0) 6 Let X be a nonempty (but not necessarily finite) subset of V (i) The span of X, denoted span X, is the set of linear combinations α 1 x 1 + α 2 x 2 + + α t x t, for all α 1,, α t C, all x 1,, x t X, and all positive integers t It is standard to say that the span of the empty subset of V is (0), and it is straightforward to check that span X is a subspace of V If U is the subspace of V spanned by X, then we say that X is a spanning set for U Every subspace U of V has a spanning set, since span U = U If u V, then span{u} is also denoted Cu (ii) The nonempty set X is linearly dependent if there exist α 1, α 2, α t C, at least one of which is not equal to zero, and pairwise distinct x 1, x 2,, x t X such that α 1 x 1 + α 2 x 2 + + α t x t = 0 Note that any subset containing the zero vector is immediately linearly dependent If X is not linearly dependent then it is said to be linearly independent Check that a subset of V consisting of exactly one nonzero vector is necessarily linearly independent, and more generally that every nonempty subset of a linearly indedpendent set is linearly independent (iii) Exercise: Let X be a finite set spanning the subspace U, and suppose that X contains at least one nonzero vector Prove that X contains a linearly independent subset also spanning U (The conclusion remains true when X is infinite, but the proof is somewhat more complicated) (iv) A linearly independent (and so necessarily nonempty) spanning set of V is termed a basis for V It follows from the preceding that every vector space has a (possibly infinite) basis If V has a basis comprised of exactly n (pairwise distinct) vectors, then every basis of V consists of precisely n vectors (see, for example, Section 3, Chapter 2, of Linear Algebra,

MATH 4096: SPRING 2009 5 2nd Edition, by K Hoffman and R Kunze, for a proof), and we say that the dimension of V is n; if no such n exists we say that V is infinite dimensional In general we will write dim V to denote the dimension of V If dim V = n, and if X is linearly independent, then X has no more than n distinct elements (again see, for example, Section 3, Chapter 2, of Hoffman s and Kunze s text) Therefore, if U is a subspace of V, then dim U dim V Exercise: If u is a nonzero vector in V, then Cu is a one-dimensional subspace of V (v) Suppose that X is linearly independent, that x 1,, x t are pairwise distinct vectors in X, and that α 1, α 2,, α t, β 1, β 2,, β t C Then α 1 x 1 + α 2 x 2 + + α t x = β 1 x 1 + β 2 x 2 + + β t x t if and only if α 1 = β 1, α 2 = β 2,, α t = β t (Prove this yourself) Now suppose that X is a basis for V It follows from the last if and only if that every vector u V can be uniquely written in the form u = α 1 x 1 + α 2 x 2 + + α t x t for x 1, x 2,, x t X and α 1, α 2,, α t C (Check this last statement Don t assume that X is finite) (vi) If X is linearly independent, and u V is not contained in span X, then X {u} is linearly independent (Prove this yourself) 7 The set e 1 = 1 0 0, e 2 = 0 1 0,, e n 1 = 0 1, e n = 0 0 0 1 is a basis for C n called the standard basis 8 If X is a linearly independent subset of V, then we say that X can be extended to a basis of V provided there exists a (necessarily linearly independent) subset Y of V such that X Y is a basis for V

6 NONCOMMUTATIVE POLYNOMIAL EQUATIONS Lemma Every linearly independent subset of V can be extended to a basis for V Proof (The following inductive proof assumes that V is finite dimensional) Suppose that X is a (necessarily finite, by (6iv)) linearly independent subset of V Set X 1 = X For i 1, we may assume that a series X 1 X 2 X i of linearly independent subsets of V has been chosen If span X i = V, then X can be extended to a basis for V, and the proof follows Now suppose that span X i V We can therefore choose a vector y V \ span X i, and by (6vi), we can set X i+1 = X i {y} to obtain a linearly independent subset of V properly containing X i cannot be extended to a basis for V there exists an infinite series Therefore, if X X 1 X 2 X i X i+1 of linearly independent subsets of V a contradiction to the finite dimensionality of V, by (6iv) 9 It follows from (8) that if U is a subspace of V then any basis for U can be extended to a basis for V 10 Corollary If U is a subspace of V, and if dim U = dim V, then U = V Proof Exercise Linear Transformations 11 For the remainder of these notes W will denote a vector space 12 (i) A function T : V W is a linear transformation provided that T (u + v) = T (u) + T (v), and T (αu) = αt (u) for all u, v V and α C (ii) A linear transformation T : V V is also called a linear operator on V (iii) A bijective linear transformation is termed a (linear) isomorphism, and a bijective linear operator is called a (linear) automorphism

MATH 4096: SPRING 2009 7 13 Let T : V W be a linear transformation The following facts are routine to verify (i) Let U be a vector space, and let S: U V be a linear transformation The composition T S: U W is also a linear transformation If S and T are isomorphisms then so is T S (v) If T is an isomorphism then T 1 : W V is also an isomorphism We thus say that V and W are isomorphic to each other when there exists any isomorphism from V onto W 14 Let T : V W be a linear transformation (i) Set K = {u V T (u) = 0} It is easy to verify (and you should!) that K is a subspace of V, called the kernel of T (ii) Exercise: Prove that T is injective if and only if its kernel is (0) (Hint: T (x 1 ) = T (x 2 ) if and only if T (x 1 x 2 ) = 0) 15 Suppose that X = {x 1, x 2,, x n } is a basis for V, and let T : V W be a linear transformation (i) Let u be an arbitrary vector in V Set u = α 1 x 1 + α 2 x 2 + + α n x n for suitable α 1, α 2,, α n C Then T (u) = α 1 T (x 1 ) + α 2 T (x 2 ) + + α n T (x n ) Roughly speaking, the preceding equality asserts that T is completely determined by its action on X (ii) Exercise, using (i): T is injective if and only if T (X) is a linearly independent subset of W Consequently, if T is injective then dim V dim W (iii) Exercise, also using (i): T is surjective if and only if T (X) spans W Consequently, if T is surjective then dim V dim W (iv) We deduce from (ii) and (iii) that if T is an isomorphism then dim V = dim W

8 NONCOMMUTATIVE POLYNOMIAL EQUATIONS 16 Assume that {x 1, x 2,, x n } is a basis for V Assume that {y 1, y 2,, y n } is a basis for W ; in particular dim V = dim W (i) It is routine to check that the assignment α 1 x 1 + α 2 x 2 + + α n x n α 1 y 1 + α 2 y 2 + + α n y n produces an isomorphism from V onto W (ii) It follows from (ii) that any two vector spaces of dimension n are isomorphic In particular, if {x 1, x 2,, x 3 } is a basis for V, then it follows from (7) and (i) that there is an isomorphism from V onto C n given by for α 1,, α n C α 1 x 1 + α 2 x 2 + + α n x n α 1 α 2 α n, Matrices Associated to Linear Transformations 17 (i) Let A be an m n matrix (with complex entries) Then the assignment x Ax, for x C n produces a linear transformation from C n to C m We will let A denote both the matrix and the corresponding linear transformation (ii) An n n matrix A is invertible (or, equivalently, nonsingular) if and only Ax 0 for all nonzero x C n, if and only if the determinant of A is nonzero (See, for example, Theorem 13, Chapter 1, and Theorem 4, Chapter 5, of Linear Algebra, 2nd Edition, by K Hoffman and R Kunze) (iii) Exercise: Let A be an n n matrix Then the linear transformation A: C n C n is an isomorphism if and only if A is invertible (iv) Let A be an m n matrix, and let C 1,, C n denote the columns of A, ordered appropriately It is useful to note, for α 1 α 2 that x = α n, and α 1, α 2,, α n C, Ax = α 1 C 1 + α 2 C 2 + + α n C n

MATH 4096: SPRING 2009 9 18 Let T : C n C m be a linear transformation, and let e 1, e 2,, e n be the standard basis vectors of C n Let A denote the m n matrix whose jth column is equal to T (e j ), for 1 j n For it follows from (17iv) that x = α 1 α 2 α n, and α 1, α 2,, α n C, T (x) = T (α 1 e 1 + α 2 e 2 + + α n e n ) = α 1 T (e 1 ) + α 2 T (e 2 ) + + α n T (e n ) = Ax Therefore, every linear transformation from C n to C m corresponds to an m n matrix Change of Basis 19 (i) Suppose that T : V W is a linear transformation, that {x 1, x 2,, x n } is a basis for V, and that {y 1, y 2,, y m } is a basis for W Let ϕ: V C n denote the isomorphism sending α 1 x 1 + α 2 x 2 + α n x n for α 1, α 2,, α n C, and let ψ: C m W be the isomorphism mapping β 1 β 2 β m for β 1, β 2,, β m C Next, set α 1 α 2 α n, β 1y 1 + β 2 y 2 + β m y m, T (x 1 ) = a 11 y 1 + a 21 y 2 + + a m1 y m T (x 2 ) = a 12 y 1 + a 22 y 2 + + a m2 y m T (x n ) = a 1n y 1 + a 2n y 2 + + a mn y m,

10 NONCOMMUTATIVE POLYNOMIAL EQUATIONS for a 11, a 12, a 21,, a mn C, and set A equal to the m n matrix (a ij ) Then T = ψ A ϕ To illustrate: ϕ V C n T W ψ A C m (ii) Note that the matrix A in (i) depends on the choice of bases for V and W (iii) Part (i) can be used to justify the informal assertion that the study of linear transformations of finite dimensional vector spaces is identical to matrix theory 20 (i) Let T : V V and T : V V be linear operators, respectively, on the vector spaces V and V We say that T and T are equivalent if there exists an isomorphism ϕ: V V such that T = ϕ T ϕ 1 The following diagram may help to visualize the involved compositions of functions: ϕ V T V ϕ V T V Observe that T = ϕ T ϕ 1 if and only if T = ϕ 1 T ϕ (ii) Let T : V V be a linear operator, and suppose that we are given a vector space isomorphism ϕ: V V Set T = ϕ T ϕ 1 Then T : V V is a linear operator that is trivially equivalent to T (iii) Exercise: Suppose that A and A are n n matrices Then A, A : C n C n are equivalent linear transformations if and only if there exists an invertible n n matrix M such that A = MAM 1 In other words, A and A are equivalent if and only if they are conjugate (iv) Let T : V V be a linear operator, and assume that X = {x 1, x 2,, x n } is a basis for V Let ϕ: V C n be the isomorphism α 1 x 1 + α 2 x 2 + + α n x n α 1 α 2 α n,

MATH 4096: SPRING 2009 11 for α 1, α 2,, α n C Set T (x 1 ) = a 11 x 1 + a 21 x 2 + + a n1 x n T (x 2 ) = a 12 x 1 + a 22 x 2 + + a n2 x n T (x n ) = a 1n x 1 + a 2n x 2 + + a nn x n, for a 11, a 12, a 21,, a nn C, and set A equal to the n n matrix (a ij ) Then T = ϕ 1 A ϕ, and T is equivalent to A As before, A depends on the choice of basis X, and we describe the above scenario by writing [T ] X = A (v) Retain the notation of (iv) To better understand how A depends on X, let {y 1, y 2,, y n } be another basis for V, and let ψ be the isomorphism for β 1, β 2,, β n C As before, set β 1 y 1 + β 2 y 2 + + β n y n β 1 β 2 β n, T (y 1 ) = b 11 y 1 + b 21 y 2 + + b n1 y n T (y 2 ) = b 12 y 1 + b 22 y 2 + + b n2 y n T (y n ) = b 1n y 1 + b 2n y 2 + + b nn y n, for b 11, b 12, b 21,, b nn C, and set B equal to the n n matrix (b ij ) We now have T = ψ 1 B ψ, and so T is equivalent to B Now combine the preceding two situations: C n ϕ ψ V C n A C n ϕ T V ψ B C n Note that ϕ ψ 1 : C n C n is an isomorphism, and so there exists an invertible n n matrix Q whose associated linear transformation Q: C n C n is the map ϕ ψ 1 We

12 NONCOMMUTATIVE POLYNOMIAL EQUATIONS therefore obtain a new diagram: Q C n C n A C n Q B C n Consequently, B = Q 1 AQ, and in particular, A and B are conjugate Exercise: Suppose that T is equivalent to the matrix A, as above, and that C is an n n matrix conjugate to A Then T is equivalent to C (vi) To summarize, if T is a linear operator on a finite dimensional vector space V, then different bases for V only affect the matrix depiction of T up to conjugation More generally, the study of linear operators on n-dimensional vector spaces, up to equivalence, is identical to the study of n n matrices, up to conjugation Eigenvectors and Eigenvalues 21 Let T : V V be a linear operator (i) A nonzero vector v V is an eigenvector for T, or more briefly a T -eigenvector, provided there exists a λ C such that T (v) = λv In the preceding setting, λ is termed the T -eigenvalue associated to v Note that an eigenvector is necessarily nonzero but that an eigenvalue may equal zero (ii) Let ϕ: V V be an isomorphism of vector spaces Further suppose T : V V is a linear operator such that T = ϕ T ϕ 1 Let v be a nonzero vector in V Exercise: v is an eigenvector for T with eigenvalue λ if and only if ϕ(v) is an eigenvector for T with eigenvalue λ (iii) Exercise: Let v 1, v 2,, v n be eigenvectors for T with pairwise distinct eigenvalues λ 1, λ 2,, λ n Then v 1, v 2,, v n are linearly independent (iv) Suppose that the dimension of V is n, and let v be an eigenvector for T with eigenvalue λ Since {v} is a linearly independent set, we can extend it to a basis X =

MATH 4096: SPRING 2009 13 {x 1, x 2,, x n } with x 1 = v Set A = (a ij ) = [T ] X, as in (20iv) Exercise: a 11 = λ and a 21 = a 31 = = a n1 = 0 In other words, λ a 12 a 1n 0 a 22 a 2n A = 0 a 32 a 3n 0 a n2 a nn 22 We now consider the existence of eigenvectors and eigenvalues, first studying the case for n n matrices This is the one part of our discussion where some knowledge of the properties of determinants is required To start, let A be an n n matrix, also viewed as a linear transformation A: C n C n (i) Recall that I denotes the n n identity matrix, and let λ C The scalar product λi is the matrix with entries λ along the main diagonal and zero elsewhere Such matrices are referred to as scalar matrices More generally, the linear operator on V sending each u V to λv is referred to as a scalar operator and is often simply denoted λ: V V (ii) Note that a nonzero vector v C n is an A-eigenvector with eigenvalue λ if and only if Av = λv, if and only if (A λi)v = 0 But there exists a nonzero v such that (A λi)v = 0 if and only if A is singular, if and only if det(a λiv) = 0 (iii) Set f(x) = det(a xi) Then f(x) is a polynomial of degree n 2 and is called the characteristic polynomial of A (Moreover, the coefficient of x n2, the so-called leading coefficient, is 1 Polynomials whose leading coefficient is 1 are termed monic) From (ii) we see that A has an eigenvector if and only if f(x) has a root To be more specific, the eigenvalues of A are exactly the roots of f(x) (iv) The Fundamental Theorem of Algebra asserts that every complex polynomial in one variable can be factored into a product of polynomials all of degree one Therefore, for suitably chosen λ 1,, λ t C, f(x) = (x λ 1 ) i 1 (x λ 2 ) i2 (x λ t ) i t, for t n 2 The λ 1, λ 2,, λ t are precisely the eigenvalues of A (That all of the degreeone factors in the above product can be chosen to be monic follows from the fact that f(x) is itself monic) In particular, there exists at least one A-eigenvector in C n (v) We have been working with complex vector spaces and matrices precisely because they allow the above application of the Fundamental Theorem of Algebra

14 NONCOMMUTATIVE POLYNOMIAL EQUATIONS 23 Theorem Let T : V V be a linear operator, and assume that V is finite dimensional Then V contains a (nonzero) T -eigenvector Proof Assuming that dim V = n, we have already seen that T will be equivalent to A: C n C n, for some n n matrix A Furthermore, V contains a T -eigenvector if and only C n contains an A-eigenvector, as noted in (21ii) However, in (22iv) we proved that A has at least one eigenvector Invariant Subspaces 24 Let T : V V be a linear operator, and let U be a subspace of V (i) We say that U is T -invariant if T (U) U Note that (0) and V are T -invariant subspaces of V (ii) If U is T -invariant, then the restriction U T U U u T (u) is a linear operator on U If S is a set of linear operators then we will say that U is S-invariant when U is F -invariant for each F S 25 (i) A nonzero vector v V is a T -eigenvector if and only if Cv is a T -invariant subspace of V In particular, the set of T -eigenvectors in V corresponds exactly to the set of one-dimensional T -invariant subspaces of V (ii) Choose λ C, and set V T λ = {v V T (v) = λv} Note that 0 V T λ Also, V T λ (0) if and only if V has a T -eigenvector We call V T λ the (T, λ)-eigenspace of V Exercise: V T λ is a T -invariant subspace of V

MATH 4096: SPRING 2009 15 Upper Triangular, Nilpotent, and Semisimple Matrices 27 (i) Problem: Use the results obtained so far in these notes to prove that every n n matrix is conjugate to an upper triangular matrix (An n n matrix A is termed upper triangular if every entry below the main diagonal is equal to zero) (ii) Exercise: An n n matrix A is nilpotent if A m = 0 for some positive integer m Prove that a matrix is nilpotent if and only if it is conjugate to a strictly upper triangular matrix (An upper triangular n n matrix is said to be strictly upper triangular when the entries along the main diagonal are also equal to zero) (iii) Exercise: An n n matrix A is semisimple, or diagonalizable, if C n has a basis of A-eigenvectors Prove that a matrix is semisimple if and only if it is conjugate to a diagonal matrix (An n n matrix is diagonal if all of the entries not on the main diagonal are zero A scalar matrix is diagonal, but a diagonal matrix need not be be scalar) (iv) Exercise: Every n n matrix is the sum of a nilpotent and semisimple matrix 28 A more precise decomposition of a square matrix can be obtained as follows: A Jordan Block is an m m matrix of the form λ 1 λ 1 λ 1, λ 1 λ where the unlisted entries are understood, by standard convention, to equal zero Theorem Every (complex) n n matrix is conjugate to a matrix J 1 J 2, Jt where each J i is a Jordan Block Proof See, for example, Chapter 7, Section 3 of Linear Algebra, 2nd Edition, by K Hoffman and R Kunze We say that the matrix listed in the theorem is in Jordan Canonical Form

16 NONCOMMUTATIVE POLYNOMIAL EQUATIONS Noncommutative Polynomial Equations 29 (i) Consider the equation XY = Y X We say that [ X = 1 1 ] [, Ỹ = 1 ] 1 is a (2 2) matrix solution to this equation, since XỸ = Ỹ X (ii) In the equation XY + Y X = 2, we think of the 2 as a scalar matrix 2 2, 2 but with indeterminate size Observe then that [ 1 X = 1 is a 2 2 solution to Y X + XY = 2, since ], Ỹ = [ ] 1 1 1 [ XỸ + Ỹ X 1 = 1 ] [ ] 1 1 + 1 [ ] [ 1 1 1 1 ] [ 2 = 1 2 ] (iii) Both XY = Y X and XY +Y X = 2 are examples of noncommutative polynomial equations We think of expressions such as XY, X 2 Y 3Y X, and Y 2 X 3 Y 4 + 1 as noncommutative polynomials, or more precisely as polynomials in the noncommuting variables X and Y let (iv) We need to briefly explain the use of exponents in (iii) For all positive integers m, X m = m times {}}{ XX X

Hence, for all positive integers m, m, MATH 4096: SPRING 2009 17 X m+m = X m X m We will follow the usual convention that X 0 = 1 However, we will only allow non-negative exponents to appear in noncommutative polynomials (v) Negative exponents are permitted in noncommutative Laurent polynomials such as X 4 Y X 4 + 7Y 1 + 3, X 1 Y + Y X 1 = 5, where we interpret the 1 to mean that X 1 X = XX 1 = 1 and Y 1 Y = Y 1 Y = 1 We then set X m = m times {}}{ X 1 X 1 X 1 30 We can also study systems of noncommutative polynomial equations, such as HE EH = 2E, HF F H = 2F, EF F E = H, which has the following (r + 1) (r + 1) solution, for each positive integer r, r r 2 H = r 4, r 0 1 0 F = 1 0, 1 0 0 µ 1 0 µ 2 Ẽ = 0 µ 3 0 µ r Exercise: Verify that HẼ Ẽ H = 2Ẽ, H F F H = 2 F, Ẽ F F Ẽ = H

18 NONCOMMUTATIVE POLYNOMIAL EQUATIONS 31 More generally, let S be a system of noncommutative polynomial equations f 1 (X 1,, X s ) = 0, f 2 (X 1,, X s ) = 0,, f t (X 1,, X s ) = 0 in the (noncommuting) variables X 1, X 2,, X s The n n matrices X 1, X 2,, X s form an n n matrix solution to the system when the equations all remain true after setting X i = X i, for 1 i s We will also refer to X 1,, X s as, more simply, either a matrix solution or an n n solution (More formally, the set { X 1, X 2,, X s } is the solution) 32 Exercise: Prove that there are no matrix solutions to Y X XY = 1 33 As in (31), let S be a system of noncommutative polynomial equations in the variables X 1,, X s Further suppose that X 1, X 2,, X s and X 1, X 2,, X s are two n n solutions to S (in particular, the dimensions of the two solutions are the same) We say that these solutions are equivalent if there exists an invertible n n matrix Q such that X i = Q X iq 1, for 1 i s 34 A matrix solution to a system of noncommutative polynomial equations is (upper) triangularizable if it is equivalent to a solution only consisting of upper triangularizable matrices In (27) we noted that every n n matrix is individually similar to an upper triangular matrix, but this fact does not imply that matrix solutions consisting of more than one matrix must be triangularizable For example, other than E = 0, F = 0, H = 0, there are no triangularizable solutions to the equation in (30) (Don t worry, for the moment, about the proof to this assertion) 35 Exercise: Prove that every matrix solution to XY Y X = Y is upper triangularizable Do the same for XY Y X = Y 2

MATH 4096: SPRING 2009 19 36 Let S be a system of noncommutative polynomial equations, as in (31) (i) Suppose that X 1, X 2,, X s is a 1 1 solution to S Note that X i Xj = X j Xi for all 1 i, j s Therefore, to find the 1-dimensional solutions to S we can add to it the equations X i X j = X j X i, for all 1 i, j s (ii) Exercise: Find the 1-dimensional solutions to the equations in (29) (iii) Let X 1, X 2,, X s be a triangularizable n n solution to S, and fix 1 i n Check that the iith entries of the X 1, X 2,, X s form a 1-dimensional solution to S (iv) Exercise: Show that E, F, H = 0 is the only 1-dimensional solution to the system of equations in (30) Using this conclusion and (iii), prove that the system in (30) has no nonzero triangularizable solutions Irreducible Solutions We will use M n to denote the set of all n n matrices 37 (i) Consider the solution in (29i), and note that X + Ỹ 2 [ = 1 ], X Ỹ 2 [ = 1 ], X ( X + Ỹ 2 ) [ = 1 ] ( ) [ ] X, X2 X + Ỹ 1 = 2 Consequently, every matrix in M 2 can be written as a linear combination of products of powers of X and Ỹ (ii) Exercise: For the solution in (29ii), can every matrix in M 2 be written as a linear combination of products of powers of X and Ỹ? (iii) Can you show that every matrix in M r+1 can be written as a linear combination of products of powers of the solution Ẽ, F, H in (30)? 38 Let S be a system of noncommutative polynomial equations f 1 (X 1,, X s ) = 0, f 2 (X 1,, X s ) = 0,, f t (X 1,, X s ) = 0,

20 NONCOMMUTATIVE POLYNOMIAL EQUATIONS and let X = { X 1,, X s } be an n n matrix solution to S We will say that X is irreducible provided every n n matrix in M n can be written as a linear combination of products of powers of matrices from X The irreducible solutions are minimal in a sense that will be justified (in part) later 39 (i) Exercise: An upper triangularizable solution is irreducible if and only if it is 1- dimensional Consequently, the irreducible solutions to XY Y X = Y (see (35)) are all 1-dimensional (ii) Exercise: Let N be an n n matrix commuting with every matrix in M n Then N is a scalar matrix 40 Aside: (i) Recall that n n matrices can be viewed as linear operators on C n Exercise: Show that (0) and C n are the only M n -invariant subspaces of C n (ie, subspaces of C n simultaneously invariant under all of the matrices in M n ; see (24)) (ii) Use (i) to deduce the following: Let X be an irreducible n n solution to the system S, as in (38) Then there is no subspace of C n that is X-invariant (This conclusion helps to explain why we think of irreducible solutions as being minimal ) In particular, there are no common eigenvectors for the matrices in X (iii) Molien s Theorem (1893): Let A be a set of n n matrices, and suppose that (0) and C n are the only A-invariant subspaces of C n Then every matrix in M n can be written as a linear combination of products of powers of matrices in A (We will omit the proof, only noting that it requires the Fundamental Theorem of Algebra) Noncommutative Algebra in Two Variables 41 (i) We can add and multiply polynomials (and Laurent polynomials) in noncommuting variables For example, (1/2)X + X = (3/2)X, (X Y )(X + Y ) = X 2 + XY Y X + Y 2

MATH 4096: SPRING 2009 21 Note that we cannot simplify the last expression to X 2 Y 2 However, scalars always commute with the variables (and with other scalars): (X + 2) 2 = (X + 2)(X + 2) = X 2 + 2X + X2 + 2 2 = X 2 + 4X + 4 (ii) Any polynomial always commutes with itself (iii) It is sometimes useful to think of multiplication in this context as juxtaposition (iv) If f(x, Y ) is a noncommutative polynomial then we can define a scalar product (cf (2)), λf(x, Y ) = λf(x, Y ) In the left hand side λ is an element of C, and in the right hand side λ is a scalar polynomial 42 (i) When no additional conditions are placed on the variables X and Y, we denote the set of all noncommutative polynomials in them by C{X, Y } (ii) It is not hard to verify that C{X, Y } is a complex vector space under the addition and scalar multiplication described in (41) The following is a basis for C{X, Y }: Xi 1 Y j 1 X i 2 Y j2 X j l Y j l l = 1, 2, 3,, i 1, j 1, i 2, j 2,, i l, j l are non-negative integers 43 (i) When we write C{X, Y }/ XY = Y X, we mean the set of noncommutative polynomials in X and Y, with multiplication as in (41), but with the substitution rule XY = Y X For example, in C{X, Y }/ XY = Y X, X 2 Y = Y X 2, (X + Y )X = X 2 + Y X = X 2 XY (ii) By C{X, Y }/ XY = Y X we mean the set of polynomials in X and Y subject to XY = Y X Therefore, C{X, Y }/ XY Y X can be viewed as the commutative polynomial ring in X and Y

22 NONCOMMUTATIVE POLYNOMIAL EQUATIONS (iii) Note that both C{X, Y }/ XY + Y X and C{X, Y }/ XY Y X are vector spaces under polynomial addition and scalar multiplication In each of these two cases, { X i Y j i, j are non-negative integers } is a basis (Exercise: Try to prove this last assertion) (iv) In general, if f 1 (X, Y ), f 2 (X, Y ), f t (X, Y ) are polynomials in X and Y, then C{X, Y }/ f 1 (X, Y ) = 0, f 2 (X, Y ) = 0,, f t (X, Y ) = 0 is the vector space comprised of polynomials in X and Y subject to the substitution rules 0 = f 1 (X, Y ), 0 = f 2 (X, Y ),, 0 = f t (X, Y ) It is often difficult to compute a basis for these abstractly defined vector spaces 44 The polynomial equations f 1 (X, Y ) = 0,, f t (X, Y ) = 0 in (43) are termed relations 45 Aside: Vector spaces equipped with some type of multiplication are usually referred to as algebras Noncommutative Algebra in Several Variables 46 (i) We can add and multiply polynomials in the noncommuting variables X 1, X 2,, X t using the procedure described in (41) For example, (X 1 X5 3 )(3X 2 4X 1 ) = 3X 1 X5 3 X 2 4X 1 X5 3 X 1, where X 3 5 = X 5 X 5 X 5 (ii) The set of all polynomials in the noncommuting variables X 1, X 2,, X s, when not subjected to additional restrictions, is denoted C{X 1, X 2,, X t } (iii) Following (42), C{X 1, X 2,, X t } is a C-vector space with basis l = 1, 2, 3,, X b 1 a 1 X b 2 a 2 X b 3 a 3 X b l a l a 1, a 2,, a l {1, 2,, t} b 1, b 2,, b l are non-negative integers

MATH 4096: SPRING 2009 23 47 (i) In view of (43 45), when we are given a system S of equations f 1 (X 1,, X s ) = 0, f 2 (X 1,, X s ) = 0,, f t (X 1,, X s ) = 0, we can consider the polynomials in the noncommuting variables X 1, X 2,, X t subject to the equations in S The set of all such noncommutative polynomials is denoted C{X 1, X 2,, X t }/ f 1 (X 1,, X s ) = 0, f 2 (X 1,, X s ) = 0,, f t (X 1,, X s ) = 0, and has a vector space structure similar to that described in (43) As in (44), the equations in S are referred to as relations (ii) Exercise: Consider the noncommutative polynomials, in variables E, F, H, subject to the relations in (30) Show that HF E = F EH c 2006 2009 E S Letzter