Question about algebraic multiplicity Let $A$ be a complex matrix with eigenvalue $c$. Suppose $c$ is an eigenvalue with algebraic multiplicity $n$, then is it true that $\dim \ker(A - cI)^{n} = n$? If so, why?
 A: Let $A$ be a complex $m \times m$ matrix and $c$ one of its eigenvalues, whose algebraic multiplicity is $n$. 
Using the Jordan canonical decomposition, we write
$$
(A - cI)^n = (PJP^{-1} - cPP^{-1})^n = P(J - cI)^nP^{-1},
$$
where $J$ is the Jordan matrix of $A$ and $P$ a full-rank (thus invertible) eigenvectors matrix. 
Expanding $(J - cI)^n$  we have
$$
\begin{bmatrix}
(S^{n\times n})^n & \mathbf{0}^{n \times (m-n)} \\ 
\mathbf{0}^{(m-n) \times n} & (J'^{(m-n)\times(m-n)})^n
\end{bmatrix},
$$
where $S$ is composed by diagonal blocks of shift matrices and $J'$ is, by construction, a full-rank Jordan matrix.
Every shift matrix is nilpotent, and its degree equals its dimention. The highest possible nilpotent degree of $S$ is, therefore $n$, when it is composed by one single block. This occurs when the geometric multiplicity of $c$ equals 1. Thus we conclude $(S^{n\times n})^n = 0^{n \times n}$ regardless of the geometric multiplicity of $c$.
It follows that $(J - cI)^n$ is formed by a null diagonal block of dimension $n$ and a full-rank block of dimension $m-n$, therefore, using the fact $P$ has full rank, we deduce
$$
\dim\ker(A - cI)^n = 
\dim\ker (P(J - cI)^nP^{-1}) =
\dim\ker (J - cI)^n = n
$$
A: Algebraic multiplicity $n$ means that
$$A = S \begin{bmatrix} J_1 \\ & J_2 \end{bmatrix} S^{-1}$$
for some nonsingular $S$ and jordan matrices $J_1$ and $J_2$, where $J_1$ is of order $n$ and has the following form:
$$J_1 = \begin{bmatrix} c & d_1 \\ & c & d_2 \\ & & \ddots & \ddots \\ & & & c & d_{n-1} \\ & & & & c \end{bmatrix}, \quad d_i \in \{0,1\}.$$
So,
$$(A - c{\rm I})^n = S \begin{bmatrix} (J_1 - c{\rm I})^n \\ & (J_2 - c{\rm I})^n \end{bmatrix} S^{-1},$$
where the diagonal elements in $J_2 - c{\rm I}$ are all nonzero, so it is a nonsingular matrix. On the other hand,  $J_1 - c{\rm I}$ has the following form:
$$J_1 - c{\rm I} = \begin{bmatrix} 0 & d_1 \\ & 0 & d_2 \\ & & \ddots & \ddots \\ & & & 0 & d_{n-1} \\ & & & & 0 \end{bmatrix}, \quad d_i \in \{0,1\},$$
which is nilpotent and, clearly, $(J_1 - c{\rm I})^n = 0$, so
$$(A - c{\rm I})^n = S \begin{bmatrix} 0 \\ & (J_2 - c{\rm I})^n \end{bmatrix} S^{-1}.$$
The statement now follows trivially from the form of $(A - c{\rm I})^n$ and nonsingularity of $(J_2 - c{\rm I})^n$.
