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Is it true to say that matrix rank is equal to the number of eigenvalues different from 0?

If not please give me contradiction example since I didn't find one.

Bernard
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dan
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2 Answers2

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Consider $$ \begin {pmatrix} 0&1\\0&0\\ \end{pmatrix}$$.

This matrix has both Eigen values zero but its rank is one.

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The dimension of eigenspace corresponding to $\lambda=0$ is the nullspace of $A$ therefore, by rank-nullity theorem, the statement is true only when the geometric and algebraic multiplicity are equal for $\lambda=0$, that is when the eigenvalue $\lambda=0$ is not defective.

Refer to the related

user
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