Cyclic and noncyclic T-invariant subspaces Let $p$ be a prime integer and let $F=\Bbb Z/p\Bbb Z$ be the field with $p$ elements. Let $V$ be a vector space over $F$ and $T:V\to V$ a linear operator. Assume that T has characteristic polynomial $x^4$ and minimal polynomial $x^3$.
a) How many $3$-dimensional cyclic $T$-invariant subspaces does $V$ have?
b) How many of the $3$-dimensional cyclic T-invariant subspaces of $V$ are direct summands of $V$?
c) How many non-cyclic $3$-dimensional $T$-invariant subspaces does $V$ have?
d) How many of the $3$-dimensional non-cyclic T-invariant subspaces of $V$ are direct summands of $V$?
I have been thinking over this problem for a long time, but I am not able to solve parts b,c,d. I can show that the answer for part a is $p$. But I need help on the other parts. Can somebody give me a good explanation of what is expected to do in this problem. Thank you.
 A: From the hypotheses, there is a cyclic submodule (i.e., $T$-invariant subspace) of dimension$~3$, but the whole space has dimension$~4$ and is not cyclic. By the structure theorem, the whole space can then be decomposed (non-uniquely) as direct sum of cyclic subspaces of dimensions $3$ and $1$; choose such a decomposition and call the factors $W_3$ and $W_1$.
The factor $W_1$ is not contained in any cyclic $3$-dimensional submodule$~W$, since the generator of$~W_1$ is not in the image of$~T$, whereas all elements of the unique $1$-dimensional submodule of$~W$ are in the image of$~T$. It follows that $W$ is injectively (and therefore bijectively) projected onto $W_3$ parallel to $W_1$. Choosing some generator$~g$ of$~W_3$, one can reconstruct $W$ from the (unique) pre-image of$~g$ under this projection, as that pre-image generates$~W$. Also one easily sees that every element of$~V$ that projects to$~g$ generates a submodule of dimension$~3$, and these submodules are all different. The answer to (a) is then the number of such elements, which is $|W_1|=p$.
This argument also shows that every $3$-dimensional cyclic submodule $W$ is a summand: $V=W\oplus W_1$ in all cases.
By the same reasoning, any $3$-dimensional submodule that does not contain $W_1$ must be cyclic, so any non-cyclic $3$-dimensional submodule contains$~W_1$. Moreover its projection onto $W_3$ is a submodule of dimension$~2$, and $W_3$ being cyclic there is exactly one such submodule (the image of$~T$). So there is exactly one non-cyclic $3$-dimensional submodule of$~V$. This submodule is the direct sum of $W_1$ and a $2$-dimensional cyclic submodule (inside$~W_3$) so it cannot itself be a direct summand of $V$ (which has no $2$-dimensional cyclic direct summands, by the structure theorem).
