I explored quite a few online resources and math StackExchange questions too. I know that in general, rank is not equal to the number of non-zero eigenvalues. But I was confused by seeing this Solution to Problem 21.1 a. Here they seem to conclude rank based on the theorem "Eigenvectors with distinct eigenvalues are always linearly independent" Also, the column space of the matrix contains all the non-zero eigen vectors.
So, I was wondering if I can make a statement as
Rank >= number of non-zero(counted without repetitions) eigenvalues
or should/can it be
Rank >= number of distinct non-zero eigen values
I could not create any counterexample which disproves this statement. Nor could I analytically prove it is correct. Thank you for the help!
If $A \in \mathbb{C}^{n \times n}$, $$\dim R(A) = n - \dim N(A) \geq n - \dim GE(A, 0),$$ where $$GE(A, \lambda) = \{x \in \mathbb{C}^n : (A - \lambda I)^k x = 0 \text{ for some } k \geq 0\}.$$ It is well known that $\dim GE(A, \lambda)$ is the multiplicity of $\lambda$ in the characteristic polynomial $K_A(\lambda) = \det(\lambda I - A)$ of $A$ (see chapter 2 of https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/linalg.pdf).
Thus Rank $\geq$ number of nonzero eigenvalues counted with multiplicities as roots of $K_A$.
Edit: $K_A(\lambda) = \det(\lambda I - A)$ is a complex polynomial of degree $n$. By the Fundamental theorem of algebra, $K_A$ can be factored as $$K_A = (\lambda - \lambda_1)^{d_1}(\lambda - \lambda_2)^{d_2}\dots(\lambda - \lambda_K)^{d_K}.$$ $d_j$ is called the multiplicity of the root $\lambda_j$.