Properties of an endomorphism and its minimal polynomial

Question

Recently, I have found this statement and I try to solve as exercise.

Consider a vector space $V$ with finte dimension $n\geq 1$ and $T\in End(V)$ with $m^T(t)=\prod_{i=1}^{r}p_i(t)^{e_i}$, where $p_i(t)$ is a monic and irreducible polynomial in $\mathbb{K}[t]$ and $e_i\geq 1$ for all $i=1,\dots,r$. Then you have to find $r$ polynomials $f_i(t) \in \mathbb{K}[t]$ such that $T_i=f_i(T)\in End(V)$ satisfies the following properties:

(1) $T_i$ isn't the null endomorphism;
(2) $\sum_{i=1}^{r}T_i=id_V$;
(3) $T_i\circ T_j=T_j\circ T_i=0$ for all $i\neq j$;
(4) $Im(T_i)=ker(p_i(T)^{e_i})$ for all $i$;
(5) $V=Im(T_1)\oplus\dots\oplus Im(T_r)$

Attempt: We consider $h_i(t)=\prod_{j\neq i}p_j(t)^{e_j}$. Since $g.c.d.(h_1(t),\dots,h_r(t))=1$, for the Bezout's identity we have that there exist $g_1(t),\dots,g_r(t)\in \mathbb{K}[t]: g_1(t)h_1(t)+\dots+g_r(t)h_r(t)=1$. We set $f_i(t)=g_i(t)h_i(t)$, so $f_i$ should be good for the thesis. What do you think?

For instance, I try to summarize the first claim. We suppose by contradiction that $T_i$ is the null endomorphism; it follows that $m^T(t)|h_i(t)g_i(t)$, so $p_i(t)^{e_i}|g_i(t)$. Hence $$ g_1(t)h_1(t)+\dots+p_i(t)^{e_i}q_i(t)h_i(t)+g_r(t)r_r(t)=1$$ so $$ p_i(t)^{e_i}[g_1(t)h_1(t)+\dots+q_i(t)h_i(t)+g_r(t)r_r(t)]=1$$ Then $p_i(t)^{e_i}$ is invertible in $\mathbb{K}[t]$, so $p_i(t)^{e_i} \in \mathbb{K}^{*}$. Since $\mathbb{K}[t]$ is a domain, $\deg p_i(t)=0$, so $p_i(t)\in \mathbb{K}^*$, a contradiction with the irreducibility of $p_i(t)$.
(2) and (3) follow immediately. For (4) and (5) it is useful to use the Theorem of Primary Decomposition of a vector space and of a $T$-cyclic vector space.

If there are some mistakes, you can post it please.Thank you very much, best regards!

Answer 1

I add some terminologies and notations, related to the text.

• $m^T(t)$ is the minimal polinomial of $T$.
• A vector space is $T$-invariant if $T(V)\subseteq V$.
• A vector space is $T$-cyclic if there exists a $u\in V$ such that $$ V\ =\ <u>_T\ =\ \{p(T)(u):p(t) \in \mathbb{K}[t]\}$$
• We say that a $T$-invariant vector space $V$ is $T$-indecomposable if there do not exist non-trivial $T$-invariant subspaces $U$ and $W$ such that $V=U\oplus W$.

Theorem of primary decomposition of a vector space

Let $V$ be a vector space with $dim_{\mathbb{K}}(V)=n \in \mathbb{N}$ and $T\in End(V)$. We suppose that $m^T(t)=\prod_{k=1}^rp_k(t)^{e_k}$. Then:

1. $V= ker(p_1(T)^{e_1})\oplus\dots\oplus ker(p_r(T)^{e_r})$
2. $C^T(t)=\prod_{j=1}^rp_j(t)^{m_j}$, where $m_j\geq e_j$
3. $ker(p_j(T)^{e_j})=ker(p_1(T)^{m_j})$
4. $\dim ker(p_j(T)^{e_j})=m_j\deg p_j$
5. Set $T_j$ the restriction of $T$ on $ker(p_j(T)^{e_j})$, then $m^{T_j}(t)=p_j(t)^{e_j}$ and $C^{T_j}(t)=p_j(t)^{m_j}$.

Theorem of primary decomposition of a $T$-cyclic vector space

Let $V$ be a vector space with $dim_{\mathbb{K}}(V)=n \in \mathbb{N}$ and $T\in End(V)$. We suppose that $m^T(t)=p(t)^{e}$. Then:

1. $V= U_1\oplus\dots\oplus U_h$, where $U_i$ is $T$-cyclic subspace of $V$.
2. Set $T_i$ the restriction of $T$ on $U_i$, then $m^{T_i}(t)=p(t)^{e_i}$, where $e=e_1\geq e_2\geq \dots\geq e_p\geq 1$.
3. $\dim U_i=e_i\deg q(t)$.
4. This decomposition is unique, unless the order of subspaces.