Finding the dimensions of subspaces of a Vector space and S-cyclic subspaces using minimal poynomials I've been staring at a chapter in Bill Cooperstein's Advanced Linear ALgebra for some time now and one section is giving me trouble.  It is about elementary divisors and invariant factors.  My question is this:
Let $S$ be an operator on a finite dimensional real vector space and assume that $U=[S,\mathbf{u}_1] \oplus [S,\mathbf{u}_2] \oplus\cdots\oplus [S,\mathbf{u}_6]$ where $[S,\mathbf{u}_i]$ is the $S$-cyclic subspace generated by $\mathbf{u}_i$, where $\mu_{S,\mathbf{u}_i}(x)$ is the minimal polynomial of each $S$-cyclic subspace
and
\begin{align*}
\mu_{S,\mathbf{u}_1}(x) &= \mu_{S,\mathbf{u}_2}(x) = (x^2+1)^5,\\
\mu_{S,\mathbf{u}_3}(x) &= (x^2+1)^4,\\
\mu_{S,\mathbf{u}_4}(x) &= \mu_{S,\mathbf{u}_5}(x) = (x^2+1)^2,\\
\mu_{S,\mathbf{u}_6}(x) &= x^2+1.
\end{align*}
Set $U_i = \{\mathbf{u}\in U \mid (S^2+I_U)^i(\mathbf{u}) = 0\}$ for $i = 1,\ldots,6$.
Determine the dimension of each $U_i$.
We know each minimal polynomial is irreducible but they are not distinct from each other.  What does this tell us about each $S$-cyclic subspace for vector $\mathbf{u}_i$?
 A: Here is a summary of the answer to this perfectly standard exercise.
Feel free to ask more questions.
 Fundamental fact  Let $C$ be a cyclic subspace, and let
$P$ be the minimal polynomial of $S$ on $C$. Then the dimensionality
of $C$ is exactly the degree of $P$.
Indeed, if we denote this degree by $d$ and if the cylic subspace
$C$ is generated by a vector $u$, then $(u,Su,S^2u, \ldots ,S^{d-1}u)$
is a basis of $C$.
So if you denote by $P_i$ the minimal polynomial of $S$ on $[S,u_i]$, then the dimensionality of $[S,u_i]$ is the degree of $P_i$.
Next, for each $i$ we have $U_i=(U_i\cap[S,\mathbf{u}_1]) \oplus 
(U_i\cap[S,\mathbf{u}_2]) \oplus\cdots\oplus (U_i\cap[S,\mathbf{u}_6])$ and for each $j$
$(U_i\cap[S,\mathbf{u}_j])$ is again a cyclic subspace, so you can use the fundamental fact above again. Finally :
$$
\begin{array}{lccl}
{\sf dim}(U_0) &=& 0+0+0+0+0+0 &=& 0 \\
{\sf dim}(U_1) &=& 2+2+2+2+2+2 &=& 12 \\
{\sf dim}(U_2) &=& 4+4+4+4+4+2 &=& 22 \\
{\sf dim}(U_3) &=& 6+6+6+4+4+2 &=& 28 \\
{\sf dim}(U_4) &=& 8+8+8+4+4+2 &=& 34 \\
{\sf dim}(U_5) &=& 10+10+8+4+4+2 &=& 38 \\
{\sf dim}(U_6) &=& 10+10+8+4+4+2 &=& 38 \\
\end{array}
$$
 UPDATE : Explanation on my second claim 
Let $V_j=U_i \cap [S,u_j] $. Let us show that $U_i =\oplus_{j} V_j$.
First, as $V_j \subseteq [S,u_j]$ we see that the sum $\sum_{j} V_j$ is direct,
and as $V_j \subseteq U_i$ we see that the sum is contained in $U_i$.
Conversely, let $w\in U_i$. Because of  $U=[S,\mathbf{u}_1] \oplus [S,\mathbf{u}_2] \oplus\cdots\oplus [S,\mathbf{u}_6]$, there are polynomials $Q_1,Q_2, \ldots ,Q_6$ such that
$$
w=\sum_{j=1}^6 w_j, \ \text{with} \ w_j=Q_j(S)u_j \in [S,u_j]
$$
Then
$$
O=(I+S^2)^i w= \sum_{j=1}^6 ((I+S^2)^i)w_j
$$
Note that in the sum above, the summand number $j$ is in $[S,u_j]$. By the direct sum property, each of those summands must be zero. In other words, $w_j \in V_j$ as wished.
