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Algebra  2014

# The Matrix Equation over Fields or Rings

Abstract:

Let and let be an algebraically closed field with characteristic 0 or greater than . We show that if and satisfy , then are simultaneously triangularizable. Let be a reduced ring such that is not a zero divisor and let be a generic matrix over ; we show that is the sole solution of . Let be a commutative ring with unity; let be similar to such that, for every is not a zero divisor. If is a nilpotent solution of where , then . 1. Introduction Let be an integer at least . In the first part, we consider matrices with entries in , a field such that its characteristic is or greater than , and denotes its algebraic closure. In the second part, the entries of the matrices are elements of , a commutative ring with unity ; then, is the ring of endomorphisms of the free -module . Let and let be a polynomial with coefficients in the complex field such that . In [1, 2], the matrix is given and the authors study the matrix equations in the unknown : In the present paper, we extend some results obtained in [1, 2] when we replace with or . Two matrices , are said to be simultaneously triangularizable (abbreviated to ) over if there exists such that and are upper triangular matrices. The following result is a slight improvement of [1, Theorem 1] or of [3, Theorem 11′]. Theorem 1. One assumes that or is and that is a polynomial with coefficients in : Then, , are over . Remark 2. Note that the previous result fails when . We now consider (1) and (2). Let be a commutative ring with unity and let be a matrix where the are commuting indeterminates. Let be the ring of the polynomials in the indeterminates and with coefficients in . Thus, the algebra generated by is included in . In particular, there are no polynomial relations, with coefficients in , linking the . We say that is a generic matrix over . When is reduced (for every , implies ), we obtain a precise result. Proposition 3. Let , and let be a reduced ring such that is not a zero divisor. Let be a generic matrix. Then, is the sole solution of (1). Otherwise, we only obtain a partial result. Proposition 4. Let be a commutative ring with unity, , and let be a polynomial with coefficients in such that . Let be a nilpotent solution of (2). Then, all elements of the two-sided ideal, in , generated by are nilpotent. Yet, if is diagonalizable and its spectrum is “good,” then we obtain a complete solution. Theorem 5. Let be similar over to with such that, for every , is not a zero divisor. Let be a nilpotent solution of (2). (i)Then, there is such that where, for every , and .(ii)If moreover is a unit, then . 2. Equation

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