The context of my question is the following problem
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 corollary 1.4.2 below] and determine the minimum polynomial of $\phi$
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
