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Is it true that, for any subalgebra $\cal S$ of the algebra of linear operators in a finite-dimensional vector space $V$ over a field, $$ \bigcap_{A\in\cal S}\ker A=\{0\}\hbox{ and } \bigcup_{A\in\cal S}A(V)=V \quad\hbox{implies that} \quad\hbox{some $A\in\cal S$ is non-singular? } $$

(A more naive version of this question was answered by Benjamin Steinberg.)

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  • $\begingroup$ My updated answer to your original question shows the answer is no. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 18, 2021 at 11:48
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    $\begingroup$ Why? ${\cal S}V\ne V$. $\endgroup$
    – Anton Klyachko
    Commented Jul 18, 2021 at 11:54
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    $\begingroup$ Sorry you are right. No answering questions before coffee. $\endgroup$
    – Benjamin Steinberg
    Commented Jul 18, 2021 at 12:05
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    $\begingroup$ Consider the set of 4 x 4 matrices, whose lower left 3 x 3 block is zero. $\endgroup$
    – user130903
    Commented Jul 18, 2021 at 14:44
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    $\begingroup$ @Zero, why not 3x3 matrices with 2x2 zero block? Maybe, you wish to post an answer (to make the question ``officially closed'')? Thanks! $\endgroup$
    – Anton Klyachko
    Commented Jul 18, 2021 at 18:48

1 Answer 1

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Let $S$ b be the set of 3 x 3 matrices whose lower left 2 x 2 block equals zero. Then $S$ is an algebra satisfying the conditions, but containing no invertible matrix.

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