Jordan blocks and the characteristic polynomial For $A \in \mathbb{C}^{n,n}$ and $\{ \lambda_1, \dots , \lambda_r\}$ are the eigenvalues of $A$.
Then my lecture notes say that the characteristic polynomial of $A$ is $$(-1)^n\prod_{i=1}^r(x-\lambda_i)^{a_i}$$
where $a_i$ is the sum of the degrees of the Jordan blocks of $A$ of eigenvalue $\lambda_i$. 
But I thought that there was only one Jordan block for a particular eigenvalue? I could understand it if it said that $a_i$ was just the degree of each block for each eigenvalue, because then for the JCF of an $n\times n$ matrix surely the characteristic equation would be raised to power $n$? If anyone could explain this it would be great! 

Consider 
$$
\begin{bmatrix}
2&1 &0&0&0\\
0&2&\mathbf{0}&0&0\\
0&0&2&1&0\\
0&0&0&2&1\\
0&0&0&0&2
\end{bmatrix}
$$
That matrix has only $2$ as an eigenvalue, but two Jordan blocks, one of size $2$ and the other of size $3$. 
