I tried to write a proof for this theorem and I would like verification on it. I would also appreciate an alternative (perhaps shorter) proof:
Let V be a K - Vector Space. The following statements are equivalent:
1. There is a decomposition of $V=W_1\oplus...\oplus W_n$
2. There exist projections $e_i:V \to V / Id_V= \sum_{i=1}^n e_i$ , $e_i\circ e_j=e_j\circ e_i= 0$ and $Im(e_i)=W_i$
My proof:
$1 \Rightarrow 2$ //
Since $V=W_1 \oplus ... \oplus W_n$ , $\exists$ n projections $e_i:V \to V / V=Ker(e_i)\oplus Im(e_i)$ , $i=1,...,n$
We can define the projections in such a way that $Im(e_j)=W_j$ and $Ker(e_j)=\oplus _{i\neq j} W_i$ :
Let $v=\sum_{i=1}^n v_i \in V$ , with $v_i \in W_i$, and let's define $e_i(v)=v_i$ , $i=1,...,n$
- Let's prove it is a linear map:
Let $\lambda \in K$ and $u, v \in V$ , where $u=\sum_{i=1}^n u_i$ and $v=\sum_{i=1}^n v_i$ , $u_i, v_i \in W_i$ :
$$e_j(\lambda u+v)=e_j\bigg(\sum_{i=1}^n \lambda u_i+v_i \bigg)=\lambda u_j+v_j=\lambda e_j(u)+e_j(v)$$
- Now let's prove it's a projection:
$$e_i^2(v)=e_i(e_i(v))=e_i(v_i)=v_i =e_i(v)$$
- Now let's verify that $Nu(e_j)=\oplus_{i\neq j} W_i$ and $Im(e_j)=W_j$ :
If $v\in \oplus_{i \neq j} W_i$ , then $v=\sum_{i\neq j}v_i$ and $e_j(v)=0 \Longrightarrow \oplus_{i\neq j} W_i\subset Ker(e_i)$
Let $v \in Ker(e_j) \Rightarrow e_j(v)=0 \forall i=1,...,n$ , but we also have that $v=v_j+\sum_{i\neq j}v_i \Rightarrow e_j(v)=v_j$
Then $v_j=0$ and $v=\sum_{i\neq j} v_i \in \oplus_{i\neq j} W_i \Longrightarrow Ker(e_i) \subset W_i$
$$\therefore Ker(e_j)=\oplus_{i\neq j} W_i \;\; \forall j=1,...,n$$
Let $w \in Im(e_i) \Rightarrow \exists u \in V / e_i(u)=w=v_i \in W_i \Longrightarrow Im(e_i) \subset W_i$
Let $v_i \in W_i \Rightarrow e_i(v_I)=v_i \in Im(e_i) \Longrightarrow W_i \in Im(e_i)$
$$\therefore Im(e_i)=W_i \;\; \forall i=1,..,n$$
- We have just proven that $Im(e_i)=W_i$, so are left to prove that $Id_V= \sum_{i=1}^n e_i$ and $e_i\circ e_j=e_j\circ e_i= 0$ :
Let $v \in V$ , then:
$$\Rightarrow v=\sum_{i=i}^n v_i = \sum_{i=i}^n e_i(v) \Rightarrow Id_V(v)= \sum_{i=i}^n e_i(v) \;\; \forall v \in V$$
$$\therefore Id_V= \sum_{i=1}^n e_i$$
$$\Rightarrow e_i\circ e_j(v)=e_i(v_j)=0 \;\; (e_j \circ e_i \; is \; analogous)$$
$$\therefore e_i\circ e_j=e_j\circ e_i= 0$$
$2 \Rightarrow 1$//
Assume that $\exists$ n projections $e_i:V \to V$ $/Id_v=\sum_{i=1}^n e_i$ , $e_i\circ e_j=e_j\circ e_i= 0$ and $Im(e_i)=W_i$
Let $v\in V$, then $Id_V(v)=\sum_{i=1}^n e_i(v) \in \sum_{i=1}^n W_i$ $\forall v\in V$
$$\therefore V=W_1+...+W_n$$
Assume there is a $v\in \bigcap\limits_{i=1}^n W_i$ , then $v=e_i(v)$ $\forall i=1,...,n$
So $e_i\circ e_j(v)=e_i(v)=v=0$
$$\therefore \bigcap\limits_{i=1}^n W_i = {0} \;\; and \;\; V=\oplus_{i=1}^n W_i$$
$$\blacksquare$$