Determining the dimensions of the eigenspaces of a matrix given its characteristic equation? Let's say I have a $6$x$6$ matrix with the characteristic equation 
$$\lambda^2 (\lambda - 1)(\lambda - 2)^3 = 0$$
How can I use this to find all possible dimensions for eigenspaces of the matrix, without assigning values?
 A: The geometric multiplicity $\gamma(\lambda) = \dim \operatorname{eigenspace}(\lambda)$ associated to an eigenvalue $\lambda$ can be anything between $1$ and the algebraic multiplicity of $\lambda$. Here the possibilities are: $\gamma(0) = 1$ or $\gamma(0) = 2$; $\gamma(1) = 1$; $\gamma(2) = 1$ or $\gamma(2) = 2$ or $\gamma(2) = 3$. To actually realize these as matrices, you may use Jordan normal forms and that fact that $\gamma(\lambda)$ equals the number of Jordan blocks corresponding to $\lambda$. For example, the case $\gamma(0) = 1$, $\gamma(1) = 1$ and $\gamma(2)=2$ can be realized as $$\begin{pmatrix}0 &1 & & & & \\ &0 & & & & \\ & &1 & & & \\ & & &2 & 1& \\ & & & &2 & \\ & & & & & 2\end{pmatrix}.$$
(Blank entries are zero.)
Another example: the following three matrices have characteristic polynomial $\lambda^3$ and have respective geometric multiplicities $\gamma(0) = 1$, $\gamma(0) = 2$, $\gamma(0) = 3$:
$$\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0& 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0& 0 & 0\end{pmatrix}, \begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0& 0 & 0\end{pmatrix}.$$
A: You know that the dimension of each eigenspace is at most the algebraic multiplicity of the corresponding eigenvalue, so 
1) The eigenspace for $\lambda=1$ has dimension 1
2) The eigenspace for $\lambda=0$ has dimension 1 or 2
3) The eigenspace for $\lambda=2$ has dimension 1, 2, or 3.
