Let $V=\mathbb Q^7$, $\phi: V \rightarrow V$ the $\mathbb Q$-linear map given by the matrix:
$A=\begin{align*} \begin{pmatrix} 1&0&0&0&1&0&2 \\0&2&1&0&0&-1&0\\0&0&2&0&0&0&0\\0&0&0&2&0&0&0\\1&0&0&0&0&-2&-2 \\0&0&0&0&0&1&0 \\0&0&0&0&0&1&1 \end{pmatrix} \end{align*}$
The characteristic polynomial is:
$P^c_{\phi}=X^7-9X^6+32X^5-55X^4+41X^3+2X^2-20X+8= (X-2)^3(X-1)^2(X^2-X-1)$
Determine $\operatorname {dim} [\operatorname{Ker}(\phi -id_V)] $ and $\operatorname {dim} [\operatorname{Ker}(\phi -2id_V)] $. Write $(V, \phi )$, viewed as $Q[X]$-module, in standard form. [Hint: Use the dimensions and the corollary below] and determine the minimum polynomial of $\phi$
Note: Introducing a category Endom($k$). The objects of this category are pairs $(V, \phi) $ consisting of a $k$-vector space $V$ and an endomorphism $\phi$. We have that $k[X]-$module is isomorphic to Endom($k$)
I determined that the rref of $A-I$ and $A-2I$ are
$A-I \sim \begin{align*} \begin{pmatrix} 1&0&0&0&0&0&0 \\0&1&0&0&0&0&0\\0&0&1&0&0&0&0\\0&0&0&1&0&0&0\\0&0&0&0&1&0&2 \\0&0&0&0&0&1&0 \\0&0&0&0&0&0&0 \end{pmatrix} \end{align*}$ $A-2I \sim \begin{align*} \begin{pmatrix} 1&0&0&0&0&0&0 \\0&0&1&0&0&0&0\\0&0&0&0&1&0&0\\0&0&0&0&0&1&0\\0&0&0&0&0&0&1 \\0&0&0&0&0&0&0 \\0&0&0&0&0&0&0 \end{pmatrix} \end{align*}$
so that
$\operatorname {dim} [\operatorname{Ker}(\phi -id_V)]= 1$
$\operatorname {dim} [\operatorname{Ker}(\phi -2id_V)]= 2$
but I have no idea how to write the standard form and find the minimal polynomial. Any help is appreciated.
EDIT:
My (partial) solution:
I determined the the rref of $A-I$ and $A-2I$ and from them
$\operatorname {dim} [\operatorname{Ker}(\phi -id_V)]= 1$ and $\operatorname {dim} [\operatorname{Ker}(\phi -2id_V)]= 2$
Now the standard form given by (1.4.2.1)consists in decomposing my vector space $V$ like this:
$V \cong Q[X]/(f_1) \oplus ...\oplus Q[X]/(f_t)$ , where the polynomials must be ordered such that $f_1|f_2...|f_t$. To do this I use corollary 1.4.7 so:
$P^c_{\phi}=f_1f_2...f_t$ where $f_t= P^{\operatorname min}_{\phi}$ is the minimum polynomial
So:$$(X-2)^3(X-1)^2(X^2-X-1)=f_1f_2...f_t$$
So some posibilities that satisfy $f_1|f_2...|f_t$ are
i) $f_1=(X-2), f_2=(X-2)(X-1), f_3=(X-2)(X-1)(X^2-X-1)$
ii) $ f_1=(X-1)(X-2),f_2=(X-1)(X-2)^2(X^2-X-1)$
iii)$f_1=(X-2)^3(X-1)^2(X^2-X-1)$
maybe there are others ways
I must decide which is the correct normal form. I am sure this is the procedure until this point.
Now to identify the correct normal form, my guess is that I have to look at the dimensions of the eigenspaces is option i), because since I have eigenspaces of dimension 1 and 2 it's the only one where i have something of dimension 1 and something of dimension 2 and that respects the condition $f_1|f_2...|f_t$, meaning $\mathbb Q/(X-2) $ of dim = 1,$\mathbb Q/(X-2) (X-1) $ of dim =2 and $\mathbb Q/(X-2)(X-1)(X^2-X-1)$. However I am not sure if this is correct and more importantly why? I also don't know if I should be considering generalized eigenspaces instead. Moreover I am not sure if the fact that the characteristic polynomial the eigenvalue 2 has geometric multiplicity 2 while in the decomposition $(X-2)$ appears with exponent 1 in $\mathbb Q/(X-2) (X-1) $ of dim=2 and in $\mathbb Q/(X-2)(X-1)(X^2-X-1)$ of dim =4 should be a sign that it's wrong. I would have expected $\mathbb Q/(X-2)^2 $ as one of the factors to have a correspondence with the eigenspace of dim 2 of the eigenvalue 2, but then I wouldn't have the decomposition in normal form. That's again because I don't know the relation between algebraic and geometric multiplicities with the spaces in the decompositions.
Note: Example 1.1.6 below explains the isomorphism between a $k$-vector space together with an endomorphism, i.e. $(V,\phi)$ and $ K[X]$-modules.
You can find here
docdroid.net/rpCR8ku/cathomalg-pdf
the lecture notes where this theorems come from
On $k[X]/(f(X))$, multiplication by $X-\alpha$ has kernel of dimension $0$ if $f(\alpha) \not = 0$ and dimension $1$ otherwise.
For your problem, it means that each $k[X]/(f_i)$ term in the standard form would either give $+1$ or $+0$ towards $\dim \ker(\phi - \alpha\,\text{id})$, depending on whether $f_i(X)$ is divisible by $X - \alpha$. Thus nonzero powers of $X - 1$ only occur in one of the $f_i$ (which one could it be?), and of $X - 2$, in two of them. So only one option to distribute the $f_i$ remains, namely, $i = 2$ and $f_1 = X-2$ (remember that $f_i$ divides $f_{i+1}$, and by by 1.4.7 $f_t$ is the minimal polynomial and $\prod_i f_i$ is the characteristical polynomial).
In general case, computing just the dimensions of $\ker(A-\lambda \,\text{id})$ won't be enough, you'll have to do the same to powers of $A-\lambda \,\text{id}$ until the sequence stabilizes for given $\lambda$. But that is probably known to you from the ODE course.
Edit 1. The quotient algebra $R = \Bbb Q[X]/(f)$ is a $\deg f$-dimensional vector space. For any $q \in \Bbb Q[X]$ the "multiplication by $q(X)$" map $m_q$ is a linear operator $R \to R$. For example, if you fix a basis of $R$, you could represent the operator as a matrix and forget about the polynomials at all. And its kernel $\ker m_q$ is first and foremost a linear subspace in $R$.
Now for its dimension in case $q = X - \alpha$ and $f = qh$. The algebra $R$ consists of cosets of polynomials wrt the ideal $(f)$, but in each of them we can fix exactly one representative whose degree is strictly lower than $f$'s. Let's work with these representatives. Suppose $v \in \ker m_q$. Then, as you said, the polynomial $v$ must be a multiple of $h$, but this implies that $v$ is either zero or has degree at least $\deg h$ — which is also the maximal possible degree among our representatives. So $v$ can only be a scalar multiple of $h$. That's a $1$-dimensional space worth of $v$'s.
The rationale behind the choice is given in second paragraph of the first part of my post. Fix a $\lambda \in \Bbb Q$. When you act on $k[X]/(f_1) \oplus \ldots \oplus k[X]/(f_t)$ with $(A - \lambda I)$ by multiplication, it acts separately on all the direct summands. On some of these summands it will act like an invertible operator (with $0$-dimensional kernel), on some ones it will have (exactly!) $1$-dimensional kernels, and the kernel dimensions on summands add up to rank deficiency of the whole matrix $A - \lambda I$ (the numbers that you originally computed). Thus, as I said,
The factor $X^2-X-1$ will obviously show up only in the minpoly too (as we have only one copy of it), so, as a maximum over the factors, $t$ will be equal to $2$. What's left to arrange is three copies of $X-2$ into two 'boxes' such that both boxes contain at least one copy and the second box can't contain less copies than the first.
Edit 2.
Wikipedia: The rank deficiency of a matrix is the difference between the lesser of the number of rows and columns, and the rank.
You can think of this in terms of the Jordan form over $\Bbb C$. Then the vector space $\Bbb C^7$ that $A$ acts upon splits into a direct sum of generalized eigenspaces, and each g. eigenspace further splits into smaller $A$-cyclic subspaces (cyclic = of form $\langle v, Av, \ldots \rangle$). You compute the $\ker (A - \lambda I)$ to isolate the true eigenvectors w/ e.val. $\lambda$ from the generalized EV's w/ $\lambda$ and unrelated vectors (well, we only need its dimension here, but I'm sure in ODE you also had to calculate the vectors). Each Jordan block w/ $\lambda$ gives exactly one $\lambda$-eigenvector, and thus the total number of Jordan blocks is equal to the dimension. Then you have to compute the kernel of the second power, which would contain the true e.vect. and the generalized e.vect of height 1, and so on.
In the problem you posted the situation is, in a way, more or less the same. The "standard form" (invariant factor) decomposition is more of a minimal-length description that still allows to recover the operator action, while the Jordan decomposition is "divide and conquer" until you get operators that act in a very simple way (the Jordan blocks). They are closely related, see the second bullet point of 1.3.13 and 1.4.9 for some examples.
$A$ acts on $V$ by multiplication, and it is a linear operator. Same with the identity transformation $I$. For any two operators that act on $V$, their linear combination is still a linear operator on $V$, and so is their product. Since we have $I$, $A$, linear combinations and multiplication, we can assign a linear operator $p(A)$ to every polynomial $p(X) \in \Bbb Q[X]$. (In functional analysis, you can define and use $p(A)$ even for more general continuous functions $p$, but that won't work for $\Bbb Q$ and is a digression anyway.)
The universe of polynomials $p(X)$ acting on $\Bbb Q[x]/(f_1) \oplus \ldots \oplus \Bbb Q[x]/(f_t)$ is parallel to the universe of $p(A)$'s and $V$. However, 1.4.2 states that the only things that differ between these two universes are labels on their elements, and the phenomena they exhibit are essentially the same. That's the meaning of the sentence right after 1.4.2.1 (a square commutative diagram, if you wish).
In my first edit, I identified the two universes without even thinking about it — formally and pedantically I was wrong, but from a practical viewpoint it was just an abuse of language. However in a course with 'Categories' in its name you should be really careful with such things :-)