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Is the rank of a matrix equal to the # of non-zero eigenvalues of a diagonalizable matrix?

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    Yes. For more information (https://math.stackexchange.com/questions/1349907/what-is-the-relation-between-rank-of-a-matrix-its-eigenvalues-and-eigenvectors) – Yadati Kiran Nov 21 '18 at 01:11
  • Yes if you count with multiplicity. $T\colon V\to V$ is diagonalisable iff $V=\bigoplus_\lambda\ker(T-\lambda I)$. – user10354138 Nov 21 '18 at 01:11

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