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1. Algebras of Matrices Let K be an algebraically closed field and V a finite dimensional vector space, n = dim(V ). A subspace W is invariant for a collection C of linear transformations if u(W ) ⊆ W for all u ∈ C. A subspace is called nontrivial if is different from (0) and V . A collection of linear transformations is reducible if it has a nontrivial invariant subspace, and irreducible otherwise. A collection of linear transformations is triangularizable if there is a basis of the vector space V such that all transformations in the collection have upper triangular matrix representations with respect to that basis. It is clear that triangularizability is equivalent to the existence of a chain of invariant subspaces (0) = V0 < V1 < · · · < Vn = V with dim(Vj ) = j for each j. (If the collection is triangularizable with respect to the basis {e1 , . . . , en }, let Vj be the linear span of {e1 , . . . , ej } for each j.) We call such a chain a triangularizing chain for the collection. For a linear transformation A with invariant subspaces N ⊂ M , we define the ˜ quotient linear transformation A˜ on M/N by A([x]) = [Ax]. Since M is invariant for A, Ax is in M . Also, since N is invariant for A, if [x] = [y] then [Ax] = [Ay] so A˜ is well-defined. In particular, the restriction of A to an invariant subspace M , denoted by A|M , is the quotient linear transformation for M and N = (0). If F is a family of linear transformations with invariant subspaces N ⊂ M then the quotient of F with respect to M and N is the family of quotients A˜ with respect to M/N where A is in F. Definition 1.1. If P is a property of linear transformations, we say it is inherited by quotients if, for every family of linear transformations F that satisfies P, if N ⊂ M are invariant subspaces for F, then the quotient of F with respect to M/N also satisfies P. We can now state and prove the Triangularization Lemma. Lemma 1.2. (Triangularization Lemma) Let P be a set of properties, each of which is inherited by quotients. If every family of linear transformations of V (with dim(V ) > 1) that satisfies P is reducible, then every collection of transformations satisfying P is triangularizable. Proof. Let F be a family of linear transformations that satisfies P. Take a maximal chain of invariant subspaces for F, 0 = M0 < M1 < · · · < Mn = V . Denote this chain by C and assume that C is not a triangularizing chain. Then there must be a k such that Mk /Mk−1 has dimension at least 2. Then F|Mk /Mk−1 has property P and is reducible by the hypothesis. Therefore it has an invariant subspace L. Define M = {x ∈ Mk : [x] ∈ L}. Since L is a proper subspace of Mk /Mk−1 , M is properly 1

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between Mk−1 and Mk . Since M is an invariant subspace of a quotient by invariant subspaces of F, it is an invariant subspace of F. This contradicts the maximality of C, so C must be a triangularizing chain. ¤ A simple and useful result of the above lemma is the following theorem. Theorem 1.3. Every commutative family of linear transformations is triangularizable. Proof. Let F be a commutative family of linear transformations. Since commutativity is a property inherited by quotients, we need only show F is reducible by the Triangularization Lemma. If every element of F is a scalar, then every subspace is invariant for F, so it is triangularizable. Otherwise, take a nonscalar A in F. Let λ be an eigenvalue for A and let M be the corresponding eigenspace. Since A isnt a scalar, M is nontrivial. For any B in F and x in M , ABx = BAx = B(λx) = λBx. Thus Bx is in M and M is an invariant subspace for F. Therefore F is reducible and triangularizability follows. ¤ The following well-known result is an easy corollary. Corollary 1.4. (Schur’s Theorem)Every linear transformation is triangularizable. Proof. The family {A} is commutative, so it is triangularizable.

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An algebra of linear transformations is a collection of linear transformations that is closed under addition, multiplication, and multiplication by scalars. If A is an algebra of linear transformations and x is any given vector, then {u(x) : u ∈ A} is easily seen an invariant subspace for A. The question of which algebras of linear transformations have nontrivial invariant subspaces has the following classical answer. Theorem 1.5. (Burnside’s Theorem) The only irreducible algebra of linear transformations on the finite dimensional vector space V of dimension greater than 1 is the algebra of all linear transformations of V . 2. Semigroups of Matrices A collection of matrices or linear transformations is called a semigroup if is closed under multiplication. If S is such a semigroup, then it is clear that the algebra generated by it is simply the linear span of S. It therefore follows from Burside’s Theorem that a semigroup of linear transformations is irreducible if and only if it contains a basis for the algebra of all linear transformations of V . Lemma 2.1. Let S be a semigroup of linear transformations and let f be a linear functional on End(V ). If f is nonzero, but f |S = 0, then S is reducible. Proof. Let A be the algebra generated by S. A consists of linear combinations of members of S so f |A = 0. Assume S is irreducible. Then A is irreducible and, by Burnsides Theorem, A = End(V ). Then f = 0 which is a contradiction, so S is reducible. ¤ A more general lemma will be given bellow.

KOLCHIN’S THEOREM

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Theorem 2.2. (Levitzki’s Theorem) Every semigroup of nilpotent operators is triangularizable. Proof. Let S be such a semigroup. Since the nilpotency is a property inherited by quotients, it suffices by the Triangularization Lemma to show that S is reducible. For any element A in S, tr(A) = 0 since the only eigenvalues of a nilpotent linear transformation are zero. Therefore tr is a nonzero functional on End(V ) that is zero on S, so S is reducible by Lemma 2.1. ¤ Definition 2.3. Let f be a linear functional on End(V ). We say that f is permutable on a family F of linear transformations in End(V ) if, for any A1 , . . . , An ∈ F and any permutation τ of {1, . . . , n}, we have f (A1 · · · An ) = f (Aτ (1) · · · Aτ (n) ). We say that f is multiplicative on F if f (AB) = f (A)f (B) for all A and B in S. Clearly, if f is multiplicative on F then f is permutable on F. Lemma 2.4. Let f be a linear functional on End(V ). If S is a semigroup of linear transformations, then f is permutable on S if and only if (i) f (AB) = f (BA), and (ii) f (ABC) = f (BAC) for all A, B and C in S. Proof. Properties (i) and (ii) are clearly implied by permutability. For the other direction, let f be a linear functional satisfying (i) and (ii). We prove that f is permutable on S by induction on the number of letters. By using (i), we have that f (ABC) = f (CAB) = f (BCA) and (ii) allows us to rearrange the first two letters. Thus f is permutable on three letters from S. Assume that f is permutable on fewer than n letters of S. We want to show that f (Aτ (1) · · · Aτ (n) ) = f (A1 · · · An ) for all A1 , . . . , An ∈ S and any permutation τ . S is a semigroup, so the products of Ai ’s are still in S. We have f (Aτ (1) · · · Aτ (n) ) = f ((A∗ · · · An )(A∗ · · · A∗ )) = f ((A∗ · · · A∗ )(A∗ · · · An )) = f (A∗ · · · An−1 )(A∗ · · · A∗ )(An )) = f ((A∗ · · · A∗ )(A∗ · · · An−1 )An ), where the first and third equality follow as τ is a permutation, the second follows from (i), and the fourth comes from (ii). Since An−1 An is in S, showing that the last line is equal to f (A1 · · · An−1 An ) reduces to the case on n − 1 elements and the result is proved. ¤ Lemma 2.5. Let F be a family of linear transformations and let f be a nonzero linear functional on End(V ). If f is permutable on F, then F is reducible. Proof. Let S and A be, respectively, the semigroup and the algebra generated by F and note that A = span(S). As f is permutable on F and S consists of products of members of F, f is permutable on S. For P any linear transformations A, B, and Pn Pn n C in A we can write A = i=1 αi Ai , B = j=1 βj Bj , C = k=1 γk Ck with the Ai , Bj , and Ck in S and αi , βj , and γk in K. Then f (ABC) = ... = f (BAC) by using the linearity of f and its permutability on S. Similarly, f (AB) = f (BA). By Lemma 2.4, f is permutable on A. Assume F is irreducible. Then A = End(V ) by Burnside’s Theorem. Therefore we can take A and B in A such that AB −BA 6= 0. Let J 6= 0 be the semigroup ideal of A generated by AB −BA. For any X and Y in A we have that f (X(AB −BA)Y ) =

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f (XABY ) − f (XBAY ) = 0 since f is permutable on A. Therefore f |J = 0 and by Lemma 2.1, J is reducible. (But End(V ) has no nontrivial ideals, and thus f is 0 on End(V ).) Then A is reducible. This is a contradiction, so F is reducible. ¤ In certain situations, this result gives us triangularizability. Theorem 2.6. (Kolchin’s Theorem) If every member of a semigroup S of linear transformations is unipotent, then S is triangularizable. Proof. Let S be in S. Then we can triangularize S and, since it is unipotent, its diagonal consists entirely of 1’s. If n = dim(V ), then tr(S) = n so the trace is constant on S. Constancy is a special case of permutability, so tr is a nonzero linear functional which is permutable on S. By Lemma 2.5, S is reducible. The property of being unipotent is inherited by quotients, so S is triangularizable by the Triangularization Lemma. ¤