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.
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.
Consider $$ \begin {pmatrix} 0&1\\0&0\\ \end{pmatrix}$$.
This matrix has both Eigen values zero but its rank is one.
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