Let $F$ be a splitting field for the characteristic polynomial of a matrix $A\in F^{n\times n}$. Denote by $S(A)$ the set of eigenvalues of $A$. Denote by $F^n$ denote the set of n-dimensional column vectors.
Let $M(A,\alpha):=\{v\in F^n|(A-\alpha I)^n v=0\}$. There was a result in a certain paper that the dimension of this particular vector space is equal to the algebraic multiplicity of $\alpha$. However, they kinda jumped on the proof of this particular thing. Can someone enlighten me how it happened that way? Thank you very much.
Hope this answer will make things clearer and giving the context:
The geometric multiplicity of an eigenvalue $\alpha$ is the dimension of the corresponding eigenspace $\ker(A-\alpha I)$. Now, in determining the Jordan normal form of a matrix, we're led to consider the sequence of nested subspaces: $\ker(A-\alpha I)^i\subset \ker(A-\alpha I)^{i+1}$ $(i=0,1,\dots,)$: $$\{0\}\subset\ker(A-\alpha I)\subset\ker(A-\alpha I)^2\subset\dotsm $$
The sequence of corresponding dimensions is involved in the number of Jordan blocks of a given size.
It happens this sequence is, for the first values of $i$ a (strictly) increasing sequence, then it stabilises, for dimensions reasons.
The smallest integer $m$ such that $\ker(A-\alpha I)^m= \ker(A-\alpha I)^{m+1}=\cdots$, is equal to the multiplicity of $\alpha$ as a root of the the characteristic polynomial