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 Post subject: Idempotent Endomorphisms
PostPosted: 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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PostPosted: Sat Jun 25, 2016 7:11 am 
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Joined: Mon Nov 09, 2015 1:52 pm
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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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PostPosted: Sat Jun 25, 2016 7:12 am 
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Joined: Tue Nov 10, 2015 8:25 pm
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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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