# If $V_j = \ker(A-\lambda I)^j$, Why does $\dim(V_j)=r_i$ if and only if $V_j = V_{j+1}$?

If we have a matrix $$A$$ and the characterisic polynomial is $$(x-\lambda_1)^{r_1} \cdot...\cdot(x-\lambda_m)^{r_m}$$ and we define for each $$\lambda_i$$, $$V_{j} = \ker(A-\lambda I)^j$$. Why does there exits $$k$$ such that $$\dim V_k = r_i$$, and why does it hold if and only if $$V_k = V_{k+1}$$? Also, is it possible to have $$\dim V_j > r_i$$? I saw this on algorithm to find Jordan form, and I'm trying to understand why the algorithm works.

Let's denote the vector space by $$W$$ with $$\dim W=n$$, in which $$n$$ is the size of $$A$$. I will break down your question into three statements:

Claim 1: $$\{0\}=V_0\subseteq V_1\subseteq \dots \subseteq V_n \subseteq V_{n+1}\subseteq \dots$$

Proof: You can verify it yourself.

Claim 2: There is an index $$j$$ such that $$V_j=V_{j+1}$$. Then $$V_j=V_{j+1}=\dots$$, i.e. the increasing sequence above is stationary.

Proof: The increasing sequence in Claim 1 can't be strict forever, for $$W$$ is finite dimensional. Thus the existence of an index $$j$$ is clear. If $$V_j=V_{j+1}$$, take $$v\in V_{j+2}$$. Then $$(A-\lambda I)^{j+2}v=0\Rightarrow (A-\lambda I)v\in V_{j+1}=V_j \Rightarrow (A-\lambda I)^{j+1}v=0\Rightarrow v\in V_{j+1}$$. So $$V_{j+2}\subseteq V_{j+1}$$, hence $$V_{j+2}=V_{j+1}$$.

Claim 3: Let $$V_{\infty}=\cup_{i\geq 0} V_i$$, then $$V_{\infty}$$ is a vector subspace of $$W$$ and $$\dim V_{\infty}=r_i$$

Proof: This is a basic result in the theory of Jordan Canonical Form. For example, look at Jordan Canonical Form, Theory and Practice- Steven Weintraub, Corollary 2.3.5 (page 30)