Let $K$ be an algebraically closed Field, $V$ a $K$-vector space with $\dim V < ∞ $ and $f ∈ End(V)$. For an eigenvalue $λ$, let $r(f, λ)$ denote k the multiplicity of $λ$ in the minimal polynomial. Then it holds that $r(f, λ)$ is the size of the largest Jordan block for the eigenvalue $λ$. Show that the geometric multiplicity $\text{geom$(f,\lambda)$}$ is equal to the number of Jordan Blocks.
I'm trying to prove this statement and I don't know if what I've done so far is any good and I don't know how to proceed. I started by assuming that $A$ is the matrix that represent $f$ then $geom(f,\lambda)=\dim(\ker(A-\lambda I_n))$. Furthermore I assume that A is in Jordan Normalform with indexed Jordan Blocks as: $$A= \begin{pmatrix} J_1&0&0&\dots&0 \\0&J_2&0&\dots&0\\ 0&0&J_3&\dots&0 \\ \vdots & \vdots & \vdots &\dots &J_n \end {pmatrix} $$ Since we are given that the algebraic multiplicity of $\lambda$ in the minimal polynomial is the size of the largest Jordan Block we can assume that $r(f,\lambda)=k$ and the: $$J(\lambda,k)=\begin{pmatrix} \lambda_k&1&0&\dots&0 \\ 0 &\lambda_k & 0 &\dots&0\\ \vdots&\vdots&\vdots &\dots&\vdots \\0 &0 &0&0&\lambda_k \end{pmatrix} \rightarrow geom(J,\lambda)=\dim(\ker(J-\lambda I))=1$$ Therefore it follows that $$geom(A,\lambda_i)=\dim(\ker(A-\lambda_iI))\rightarrow geom(A,\lambda)= \sum_{i=1}^{n} geom(J_i,\lambda_i)$$ I gave it a lot of effort and I genuinely don't know whether or not all I did was restate the theorem.