Limitations on the size of semigroups of matrices (Q2480763)

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Limitations on the size of semigroups of matrices
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    Limitations on the size of semigroups of matrices (English)
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    3 April 2008
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    Let \(S\) be a multiplicative semigroup of the algebra \(M_{n}(\mathbb{F)}\) of \(n\times n\) matrices over an algebraically closed field \(\mathbb{F}\). The authors prove a number of theorems which bound the size of \(S\) in the case that \(S\) is irreducible. Suppose that \(S\) is irreducible and \(\varphi\) is a nonzero linear functional on \(M_{n}(\mathbb{F)}\). If \(m:=\left| \varphi(S)\right| \) is finite, then \(\left| S\right| \leq m^{n^{2}}\), whilst if \(m\) is countable, then \(S\) is also countable. Moreover, if \(\mathbb{F=C}\) and \(\varphi(S)\) is bounded, then \(S\) is also bounded. More generally, if \(S\) is irreducible and \(T:M_{n}(\mathbb{C)}\rightarrow M_{m}(\mathbb{C)}\) is a nonzero linear transformation, then \(S\) is finite, countable or bounded whenever \(T(S)\) has the corresponding property. They also prove a dual property. If \(\mathbb{F=C}\) and \(S\) is irreducible, then \(S\) is finite, countable or bounded whenever some nonzero ideal \(J\) of \(S\) has the corresponding property. For example (assuming \(0\in S\)), if \(k\) is the minimal nonzero rank of the matrices in \(S\), then we can take \(J=\left\{ u\in S \mid \text{rank}(u)=k\right\} \cup\left\{ 0\right\} \). Generally these results are no longer true if \(S\) is not assumed to be irreducible.
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    semigroup
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    irreducible
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    linear functional
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    trace
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    finiteness
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    boundedness
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    ideals
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    matrix algebra
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