algebraic multiplicity of an eigenvalue and rank of a matrix

Question

Let $A \in GL_n(\mathbb{C}) $ and let $M_{\lambda}$ denote the algebraic multiplicity of the eigenvalue $\lambda$ (this is zero whenever $\lambda$ is not an eigenvalue of $A$) . prove that $$rank(A - Id) \ge n - M_1$$ I tried considering the characteristic polynomial of $A$, but I have no idea what to do next.

Answer 1

It is known that if $\lambda$ is an eigenvalue of $A$, then its multiplicity $M_{\lambda}$ is greater than or equal to the dimension of its eigenspace i.e $$\dim Ker\left( A-\lambda Id \right)\leq M_{\lambda} $$

If $\lambda$ is not an eigenvalue of $A$ then $ Ker\left( A-\lambda Id \right)= \{0\} $, thus the above inequality is true for all $\lambda \in \mathbb{C}.$

From rank–nullity theorem $$\dim Ker\left( A-\lambda Id \right)=n-rank\left( A-\lambda Id \right)$$ By substituting this into the above inequality, we get $$ rank\left( A-\lambda Id \right) \geq n-M_{\lambda}$$

For $\lambda=1$ we get the result.