It is currently Sun Mar 18, 2018 4:50 am

 All times are UTC [ DST ]

 Page 1 of 1 [ 3 posts ]
 Print view Previous topic | Next topic
Author Message
 Post subject: Idempotent EndomorphismsPosted: Sat Jun 25, 2016 7:08 am
 Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
Let $\mathbb{K}$ be a field, let $\mathcal{E}$ be a $\mathbb{K}$-vector space and consider the ring $\displaystyle End_{ \mathbb{K} }\left(\mathcal{E} \right)$ of endomorphisms of $\mathcal{E}$.

1. If $\displaystyle f \in End_{ \mathbb{K} }\left(\mathcal{E} \right)$ is idempotent, then show that $\mathcal{E} = \mathcal{V}\oplus \mathcal{W}$, where $\mathcal{V} = Ker(f) = Im( Id_{\mathcal{E}} - f ) \; \; \& \; \; \mathcal{W} = Im(f) = Ker( Id_{\mathcal{E}} - f )$
2. Conversely, if there exist subspaces $\mathcal{V}$ and $\mathcal{W}$ of $\mathcal{E}$ such that $\mathcal{E} = \mathcal{V}\oplus \mathcal{W}$, then there exists an idempotent element $\displaystyle f \in End_{ \mathbb{K} }\left(\mathcal{E} \right)$ such that $\mathcal{V} = Ker(f) = Im( Id_{\mathcal{E}} - f ) \; \; \& \; \; \mathcal{W} = Im(f) = Ker( Id_{\mathcal{E}} - f )$
3. If $\displaystyle \dim_{\mathbb{K}}(\mathcal{E}) < \infty$, then show that $\forall \, f \in End_{ \mathbb{K} }\left(\mathcal{E} \right) \smallsetminus \{ 0 \} \; \exists \, g \in End_{ \mathbb{K} }\left(\mathcal{E} \right) \, : \, \left( g\circ f \right)^{2} = g\circ f$
4. Is the ring $\displaystyle End_{ \mathbb{K} }\left(\mathcal{E} \right)$ connected?

Top

 Post subject: Re: Idempotent EndomorphismsPosted: Sat Jun 25, 2016 7:11 am
 Team Member

Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Hi Nickos. This is a really nice exercise.

The set

$\displaystyle{\rm{End_{\mathbb{K}}\,(E)}=\left\{f:E\longrightarrow E: f(x+y)=f(x)+f(y)\,,f\,(k\cdot x)=k\cdot f(x)\,,\forall\,x\,,y\in E\,,\forall\,k\in\mathbb{K}\right\}}$

eqquiped with the usual operation of addition of functions and with the composition as it's multiplication, is an associative ring with zero element

$\displaystyle{\mathbb{O}:E\longrightarrow E\,,x\mapsto \overline{0}}$ and $\displaystyle{Id_{E}}$ as it's unity.

1. Let $\displaystyle{f\in \rm{End_{\mathbb{K}}\,(E)}}$ be an idempotent element, that is $\displaystyle{f^2=f\circ f=f}$ . The subsets

$\displaystyle{V=\rm{Ker}\,(f)=\left\{x\in E: f(x)=\overline{0}\right\}\,,W=\rm{Im}\,(f)=\left\{f(x): x\in E\right\}}$

are subspaces of $\displaystyle{\left(E,+,\cdot\right)}$ .

Obviously, $\displaystyle{\left\{\overline{0}\right\}\subseteq V\cap W}$. Let $\displaystyle{x\in V\cap W}$, that is $\displaystyle{x\in\rm{Ker}(f)}$ and

$\displaystyle{x\in\rm{Im}(f)}$. So, $\displaystyle{f(x)=\overline{0}}$ and $\displaystyle{x=f(y)}$ for some $\displaystyle{y\in E}$. Then,

$\displaystyle{f(x)=f(f(y))\implies \overline{0}=f^2(y)\implies \overline{0}=f(y)\implies \overline{0}=x}$. So,

$\displaystyle{V\cap W\subseteq \left\{0\right\}}$

and thus $\displaystyle{V\cap W=\left\{\overline{0}\right\}\implies V+W=V\oplus W\subseteq E}$ . Let $\displaystyle{x\in E}$.

$\displaystyle{f^2(x)=f(x)\implies f\,(f(x))=f(x)\implies f(f(x)-x)=\overline{0}\implies f(x)-x\in \rm{Ker}(f)}$, so:

$\displaystyle{x=\left(x-f(x)\right)+f(x)}$, where:

$\displaystyle{x-f(x)\in \rm{Ker}(f)=V\,,,f(x)\in W=\rm{Im}(f)}$ .

Finally, $\displaystyle{E=V\oplus W}$ . Also, if $\displaystyle{x\in \rm{Ker}(f)}$, then $\displaystyle{f(x)=0}$ and

$\displaystyle{x=x-\overline{0}=x-f(x)=\left(Id_{E}-f\right)(x)\implies x\in \rm{Im}\,(Id_{E}-f)}$. On the other hand, if

$\displaystyle{x\in \rm{Im}(Id_{E}-f)}$, then $\displaystyle{x=y-f(y)}$ for some $\displaystyle{y\in E}$ and then :

$\displaystyle{f(x)=f(y-f(y))=f(y)-f(f(y))=f(y)-f^2(y)=\overline{0}}$,

which means that $\displaystyle{x\in \rm{Ker}(f)}$ .

Therefore, $\displaystyle{V=\rm{Ker}(f)=\rm{Im}(Id_{E}-f)}$ . Similarly,

$\displaystyle{W=\rm{Im}(f)=\rm{Ker}(Id_{E}-f)}$ .

2. It's known that $\displaystyle{V\cap W=\left\{\overline{0}\right\}}$ and $\displaystyle{E=V+W}$.

Let $\displaystyle{x\in E}$. Then, $\displaystyle{x=v+w}$ for some $\displaystyle{v\in V\,,w\in W}$ and if $\displaystyle{x=v'+w'}$, where

$\displaystyle{v'\in V\,,w'\in W}$, then :

$\displaystyle{v+w=v'+w'\iff v-v'=w-w'\implies v-v'\in V\cap W\implies v-v'=\overline{0}\implies v=v'\implies w=w'}$ .

According to this analysis, we define $\displaystyle{f:E\longrightarrow E\,,f(v+w)=w}$ and this function is well defined.

Let $\displaystyle{x\,,y\in E}$ and $\displaystyle{k\in\mathbb{K}}$ . There are $\displaystyle{v_1\,,v_2\in V\,,w_1\,,w_2\in W}$ such that

$\displaystyle{x=v_1+w_1\,,y=v_2+w_2}$ and thus:

$\displaystyle{f(x+y)=f\,((v_1+v_2)+(w_1+w_2))\,,f(k\cdot x)=f\,(k\cdot v_1+k\cdot w_1)}$ .

Due to the fact that $\displaystyle{V\,,W}$ are subspaces of $\displaystyle{\left(E,+,\cdot\right)}$ we have that

$\displaystyle{v_1+v_2\in V\,,w_1+w_2\in W\,,k\cdot v_1\in V\,,k\cdot w_1\in W}$

and then:

$\displaystyle{f(x+y)=w_1+w_2=f(x)+f(y)\,,f(k\cdot x)=k\cdot w_1=k\cdot f(x)}$.

Consequently, the function $\displaystyle{f}$ is $\displaystyle{\mathbb{K}}$ - linear, that is $\displaystyle{f\in \rm{End_{\mathbb{K}}\,(E)}}$ .

Obviously, $\displaystyle{\rm{Im}(f)\subseteq W}$. If $\displaystyle{w\in W}$, then $\displaystyle{x=\overline{0}+w\in E}$,

where $\displaystyle{\overline{0}\in V}$

and $\displaystyle{f(x)=w}$, which means that $\displaystyle{W=\rm{Im}\,(f)}$ .

Let $\displaystyle{x=v+w\in E\,,v\in V\,,w\in W}$ .

If $\displaystyle{x\in \rm{Ker}(f)}$, then $\displaystyle{f(x)=\overline{0}\implies w=\overline{0}\implies x\in V}$. On the other hand, if

$\displaystyle{v\in V}$, then $\displaystyle{v=v+\overline{0}\,,\overline{0}\in W}$ and $\displaystyle{f(v)=\overline{0}\implies v\in \rm{Ker}(f)}$ ,

thus: $\displaystyle{V=\rm{Ker}(f)}$. Futhermore, for each $\displaystyle{x=v+w\in E\,,v\in V\,,w\in W}$, we have that :

$\displaystyle{f^2(x)=f\,(f(x))=f(w)=f(x)}$, cause :

$\displaystyle{f(w)=f(x-v)=f(x)-f(v)=f(x)-\overline{0}=f(x)}$

since $\displaystyle{v\in \rm{Ker}(f)}$ .

3. Without a solution to this.

4. Let $\displaystyle{\dim_{\mathbb{K}}\,E=n<\infty}$. So, $\displaystyle{\left(E,+,\cdot\right)\simeq \left(\mathbb{K}^{n},+,\cdot\right)\implies}$

and let $\displaystyle{f:E\longrightarrow \mathbb{K}^{n}}$ be an isomorphism of left $\displaystyle{\mathbb{K}}$ - modules.

We define $\displaystyle{h:\rm{End_{\mathbb{K}}\,(E)}\longrightarrow \rm{End_{\mathbb{K}}\,(\mathbb{K}^{n})}}$ by

$\displaystyle{g:E\longrightarrow E\mapsto h(g):\mathbb{K}^{n}\longrightarrow \mathbb{K}^{n}\,,h(g)(x)=(f\circ g\circ f^{-1})(x)}$

which is well defined, $\displaystyle{\mathbb{K}}$ - linear, it maintains the multiplication and also is bijection. So,

$\displaystyle{\left(\rm{End_{\mathbb{K}}\,(E)},+,\circ\right)\simeq \left(\rm{End_{\mathbb{K}}\,(\mathbb{K}^{n})},+,\circ\right)\simeq \left(\mathbb{M}_{n}\,(\mathbb{K}),+,\cdot\right)}$ as rings.

Since the field $\displaystyle{\left(\mathbb{K},+,\cdot\right)}$ is connected, so is $\displaystyle{\left(\mathbb{M}_{n}\,(\mathbb{K}),+,\cdot\right)}$

and thus the ring $\displaystyle{\left(\rm{End_{\mathbb{K}}\,(E)},+,\circ\right)}$ is connected.

I am not so sure about the last question. What's your opinion ?

P.S. For the last one check here : Is This Ring Connected?

Top

 Post subject: Re: Idempotent EndomorphismsPosted: Sat Jun 25, 2016 7:12 am
 Team Member

Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
Thank you for your solution, Mr. Papapetros!

I think that your arguments in (4) are valid.

I hope that part (3) will also be resolved soon - nevertheless, i do accept your answer!

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 3 posts ]

 All times are UTC [ DST ]

#### Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta