Application of the Theorem of Primary decomposition of a vector space
Hi everyone. It's my first topic, so I hope it is right. I would like to share with you this statement, which 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 $i=1,...,r$. 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) $T_i\ne 0$;
(2) $\sum_{i=1}^{r}T_i$ is the identity on $V$;
(3) $T_i\circ T_j=T_j\circ T_i=0$ for all $i\ne j$;
(4) $Im(T_i)=ker(p_i(T)^{e_i})$ for all $i$;
(5) $V=Im(T_1)\oplus...\oplus Im(T_r)$.
I add some notations and terminologies.
- $m^T(t)$ 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
$$ V\ =\ _T\ =\ \{p(T)(u):p(t) \in \mathbb{K}[t]\}$$
- 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. $V= ker(p_1(T)^{e_1})\oplus...\oplus ker(p_r(T)^{e_r})$
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. $V= U_1\oplus...\oplus U_h$, 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.
Attempt: We consider $h_i(t)=\prod_{j\ne i}p_j(t)^{e_j}$, for all $i=1,...,r$. Since $g.c.d.(h_1(t),...,h_r(t))=1$, for the Bezout's identity we have that there exist $g_1(t),...,g_r(t)\in \mathbb{K}[t]: g_1(t)h_1(t)+...+g_r(t)h_r(t)=1$. 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 contraddiction 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 $$ g_1(t)h_1(t)+\dots+p_i(t)^{e_i}q_i(t)h_i(t)+g_r(t)r_r(t)=1$$
where $q_i(t)\in \mathbb{K}[t]:g_i(t)=q_i(t)p_i(t)^{e_i}$. Therefore
$$ p_i(t)^{e_i}[g_1(t)h_1(t)+\dots+q_i(t)h_i(t)+g_r(t)r_r(t)]=1$$
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 contraddiction with the irreducibility of $p_i(t)$.
(2) and (3) follow immediatily from previous arguments. For the other claims it is sufficient to use the Theorem of Primary Decomposition of a vector space and of a $T$-cyclic vector space.
If there are some mistake, you can post it please. Thank you very much, best regards!
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 $i=1,...,r$. 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) $T_i\ne 0$;
(2) $\sum_{i=1}^{r}T_i$ is the identity on $V$;
(3) $T_i\circ T_j=T_j\circ T_i=0$ for all $i\ne j$;
(4) $Im(T_i)=ker(p_i(T)^{e_i})$ for all $i$;
(5) $V=Im(T_1)\oplus...\oplus Im(T_r)$.
I add some notations and terminologies.
- $m^T(t)$ 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
$$ V\ =\ _T\ =\ \{p(T)(u):p(t) \in \mathbb{K}[t]\}$$
- 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. $V= ker(p_1(T)^{e_1})\oplus...\oplus ker(p_r(T)^{e_r})$
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. $V= U_1\oplus...\oplus U_h$, 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.
Attempt: We consider $h_i(t)=\prod_{j\ne i}p_j(t)^{e_j}$, for all $i=1,...,r$. Since $g.c.d.(h_1(t),...,h_r(t))=1$, for the Bezout's identity we have that there exist $g_1(t),...,g_r(t)\in \mathbb{K}[t]: g_1(t)h_1(t)+...+g_r(t)h_r(t)=1$. 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 contraddiction 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 $$ g_1(t)h_1(t)+\dots+p_i(t)^{e_i}q_i(t)h_i(t)+g_r(t)r_r(t)=1$$
where $q_i(t)\in \mathbb{K}[t]:g_i(t)=q_i(t)p_i(t)^{e_i}$. Therefore
$$ p_i(t)^{e_i}[g_1(t)h_1(t)+\dots+q_i(t)h_i(t)+g_r(t)r_r(t)]=1$$
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 contraddiction with the irreducibility of $p_i(t)$.
(2) and (3) follow immediatily from previous arguments. For the other claims it is sufficient to use the Theorem of Primary Decomposition of a vector space and of a $T$-cyclic vector space.
If there are some mistake, you can post it please. Thank you very much, best regards!
Risposte
Hi pandistelle007: welcome in this Italian forum on Mathematics and company.
Why don't you write in Italian? So you have more chance to receive answers to your questions.
Another question: what do you mean by \(m^T\)? The minimal polynomial associated to linear map \(T\)?
Why don't you write in Italian? So you have more chance to receive answers to your questions.
Another question: what do you mean by \(m^T\)? The minimal polynomial associated to linear map \(T\)?
"j18eos":
Hi pandistelle007: welcome in this Italian forum on Mathematics and company.
Why don't you write in Italian? So you have more chance to receive answers to your questions.
Another question: what do you mean by \(m^T\)? The minimal polynomial associated to linear map \(T\)?
Hi j18eos. I prefered writing in English because a few days ago I wrote this exercise in this way in LaTex so I copied it here without changing the language. If it'isnt a problem for you, I leave the exercise in English.
Anyway I understand Italian so, if you want, you can write in Italian.
I have just added some notation in the previous text, I hope it is better now. Thank you, warm regards!
Don't worry... however: I'm waiting for an answer to my question. 
Yes, $m^T$ is the minimal polynomial of the endomorphism $T$. I have already added it in the first post Lol
Lol
Ok, I see it now! 
I traslated the text from English to Italian... I hope in a positive feedback! Thank you very much
Does someone support my arguments? 
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