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The connection between the rank of a matrix and its zero-mode eigenvectors

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The connection between the rank of a matrix and its zero-mode eigenvectors

Hi, I would be most thankful if you could help me prove that if an arbitrary n by n matrix has rank m ¡ n, then the matrix has (n-m) eigenvectors corresponding to the eigenvalue zero. Thank you!


By a change of basis, you can put the matrix in Jordan canonical form. Once that is done, figuring out rank and null eigenvectors becomes trivial.

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