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 Post subject: Idempotent EndomorphismsPosted: Sat Jun 25, 2016 7:08 am
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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?

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

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

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