If two matrices similar to one another are diagonalizable, then certainly this is the case, since the algebraic multiplicity of any eigenvalue they share must be equal (since they are similar), and since they are diagonalizable, those algebraic multiplicities must equal the geometric multiplicities of those eigenvalues. But what if the matrix is not diagonalizable?(adsbygoogle = window.adsbygoogle || []).push({});

BiP

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# Is the geometric multiplicity of an eigenvalue a similar invariant?

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