It is well known that for a $n \times n$ matrix $A$ ,
the charicteristic polynomial $p(x)$ satisfies
$p(x)=\prod_{\lambda : eigenvector} (x-\lambda)^{a(\lambda)}$
where $a(\lambda )$ is the algebraic multiplicity of $\lambda$
The minimal polynomial of $A$ , $\mu _A (x) $ can also be represented as a form of
$\mu_A (x)=\prod (x-\lambda)^{b(\lambda)}$
The question is: if the geometric multiplicity of $\lambda$ is $g( \lambda )$, is
$b(\lambda ) \le a(\lambda )-g(\lambda )+1$ ?
I think I saw this inequality somewhere, but I'm failing to find how to prove it, and I'm not even sure if it works. Please tell me whether it works, and a proof if it does.
Best Answer
Yes, this is true. Suppose that the Jordan normal form of $A$ has Jordan blocks with eigenvalue $\lambda$ of sizes $k_1,\dots,k_m$. Then $$a(\lambda)=\sum_{i=1}^mk_i,$$ $$g(\lambda)=m,$$ and $$b(\lambda)=\max(k_1,\dots,k_m).$$ Thus $$a(\lambda)-g(\lambda)=\sum_{i=1}^m(k_i-1)\geq b(\lambda)-1$$ since $b(\lambda)$ is one of the $k_i$. (This is assuming $\lambda$ actually is an eigenvalue at all; if not then $a(\lambda)=g(\lambda)=b(\lambda)=0$ and the inequality still holds.)