If matrices A and B commute, then they have a common eigenvector.
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Replying to @AlgebraFact
Actually, this is false. Take any matrix, J, without eigenvectors, and note that it commutes with -J (for example), which likewise is without eigenvectors.
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Replying to @MathPrinceps @AlgebraFact
I suppose we are talking complex numbers, where every matrix surely has an eigenvector.
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Replying to @tifv @AlgebraFact
This proviso (i.e., that the matrices in question have entries drawn from an algebraically closed field) was not, however, included in the original statement.
8:22 PM - 26 Dec 2018
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