Sequence of subspaces $Ker(\lambda I-A)^k$ stationary at algebraic multiplicity?

Question

Let $A$ be a square $n\times n$ matrix, and $\lambda$ be an eigenvalue of $A$. Then the nullspace or subspace $Ker(\lambda I-A)$ is different from $\{0\}$. And we can obtain an increasing sequence of subspaces $Ker(\lambda I-A)^k$ that is $$Ker(\lambda I-A)\subset Ker(\lambda I-A)^2 \subset\dots$$ Since the dimension of $\mathbb{R}^n$ is finite and equals $n$, then the increasing sequence of subspaces $Ker(\lambda I-A)^k$ will at some point be stationary at some integer $k$. Let $k_0\geq1$ be the smallest integer that verifies this property.

Since the geometric multiplicity $m_g=dimKer(\lambda I-A)$, then it is clear that $m_g\leq dimKer(\lambda I-A)^{k_0}$.

My question is: is there a relationship with such integer $k_0$ with the algebraic multipilicty $m_a$?

Answer 1

From your definition, $k_{0}$ is the degree of the factor $(x-\lambda)$ in the minimal polynomial of $A$, while $m_{a}$ is the degree of $(x-\lambda)$ in the characteristic polynomial of $A$. Thus we know that $k_{0}\leq m_{a}$. But it seems that $m_{g}$ has nothing to do with $k_{0}$ (That is, I don't think any of them is consistently no smaller than the other). Maybe you can try showing that $m_{g}+k_{0}\leq m_{a}+1$, if you'd like to (which I think is a pretty good exercise).

Answer 2

For $k\geq 0$, let $Z_k=\ker (\lambda I - A)^k$ (with $Z_0=\{0\}$) and set $d_k=\dim Z_k$. By assumptions, we have $d_0=0 < d_1 < ... < d_{k_0}= d_{k_0+1}=...$

The map $\lambda I -A$ induces an injective map: $$ \lambda I - A : Z_{k+1}/Z_k \to Z_k/Z_{k-1},$$

so we must have: $d_{k+1}-d_k \leq d_k - d_{k-1}$. In other words the sequence $d_k$ is increasing and concave with $d_0=0$. Therefore, $d_k \leq k d_1$ for all $k\geq 0$. In particular you have the relation $$ m_a \leq k_0 m_g$$ where equality may be obtained by a map with equal size Jordan blocks. So this inequality is the best you can obtain for an upper bound on $m_a$ without further assumptions. The lower bound given in the answer of Aby Coathin may also be shown using the above-mentioned concavity and is also optimal.