Ytukyg

Let $T$ be a linear endomorphism on the vector space $F^3$, where $F$ is the finite field with $p$ elements, whose minimal polynomial is $x^2$.

How many $1$-dimensional and $2$-dimensional $T$-invariant subspaces are there? And how many of these $1$-dimensional and $2$-dimensional invariant subspaces are direct summands with a $T$-invariant subspace?

For the second question, I think that the number of $1$-dimensional $T$-invariant and the number of $2$-dimensional $T$-invariant subspaces should be the same.

I am not sure if it helps, but since the minimal polynomial is $x^2$ and the dimension of the vector space is $3$, the characteristic polynomial is $x^3$.

I am looking for the intuition/understanding more so than the actual answer (although the actual answer might help a lot).

Let's consider the kernel and image of $T$. The kernel of $T$ must be two dimensional, since it cannot be three dimensional since $Tne 0$, and if it were one dimensional, then we would have that the rank of $T$ would be two by rank-nullity, so we couldn't have $T^2=0$. Hence the dimension of the image of $T$ must be one dimensional. Also it is clear that the image of $T$ is a one dimensional subspace of $ker T$.

Now, if $Vsubseteq F^3$ is $T$-invariant, then $T(V)subseteq V$. Since the image of $T$ is one dimensional, either $V$ contains the image of $T$, or $V$ lies in the kernel of $T$.

If $V$ is two dimensional, then if it lies in the kernel of $T$, then it is the kernel of $T$, and hence contains the image of $T$. Thus the two dimensional invariant subspaces are precisely those containing the image of $T$. To count the number of planes containing a line in $F^3$, fix a complement of the line. Then planes containing the line in $F^3$ correspond to lines in that complement, of which there are $(p^2-1)/(p-1)=p+1$.

Conversely if $V$ is one dimensional, then if it contains the image of $T$, it is the image of $T$, so it lies in the kernel. Thus the one dimensional invariant subspaces for $T$ are the one dimensional subspaces of the kernel of $T$. We just counted the number of one dimensional subspaces of a two dimensional $F$ vector space. There are $p+1$ one-dimensional invariant subspaces.

I'm not sure exactly what the second question is asking us to count at all, but if you can clarify that it'd be cool. Might be because it's late, since I'm having trouble even formulating guesses as to what it might be supposed to mean, which suggests my brain is about to cease functioning.

Intuition?

Hard to say exactly what the intuition is, but I guess the idea is that since $T$ has nontrivial kernel, that is easily invariant (as is any subspace contained in it), so let's start looking into that. Similarly, since the kernel is nontrivial, $T$ is not surjective, and the image of $T$ will also be invariant (and any subspace containing it), so it's worth looking into that.