Algebraic multiplicity = geometric multiplicity?

Sure, I can give a simple example:

$$ A=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}. $$

The characteristic polynomial is $(\lambda - 1)^2$, so the algebraic multiplicity os $2$, however, geometric multiplicity is $1$, indeed $dim Ker(A-I)=1$.

If $M$ is diagonalizable, then, $$V=E_1\oplus...\oplus E_k.$$ If it's not diagonalisable, then $$V\supset E_1\oplus...\oplus E_k$$ but $$V\neq E_1\oplus...\oplus E_k.$$ Therefore, $$m_1+...+m_k=n:=\dim(V)>\dim(E_1)+...+\dim(E_k),$$ and thus $$(m_1-\dim(E_1))+...+(m_k-\dim (E_k))>0.$$

You can easily show by induction that $m_k\geq \dim(E_k)$ for all $k$. But you can't say more.