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

Question

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.

Answer 1

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)