I am using the following definition:

  • Def.: let $E_1,...,E_p$ $p$-vector subspaces of $V$, $E_1+E_2+...+E_p$ is direct sum, $E_1+E_2+...+E_p \doteq E_1\oplus E_2 \oplus ... \oplus E_p$, if $$\forall e_1 \in E, e_2 \in E_2,...,e_p \in E_P (e_1+e_2+...+e_p=0_V \to e_1=e_2=...=e_p=0_V)$$

and I must to proof:

  • Prop.: let $E_1,...,E_p$ $p$-vector subspaces of $V$, then: $$\sum_{i=1}^pE_i \doteq \bigoplus_{i=1}^p E_i \leftrightarrow \forall i\in \{1,...,p\}(E_i \cap \sum_{t \in \{1,...,p\}-\{i\}}E_t=\{0\})$$

  • Proof: I thought (by induction), for $p=2$ I have $E_1,E_2$ $2$-vector subspaces of $V$, and I must to proof $$ E_1+E_2 \doteq E_1 \oplus E_2 \leftrightarrow \forall i \in \{1,2\}(E_i \cap \sum_{t \in \{1,2\}-\{i\}}E_t=\{0\})$$ but $\forall i \in \{1,2\}(E_i \cap \sum_{t \in \{1,2\}-\{i\}}E_t=\{0\})$ means $E_1 \cap E_2=\{0\} \wedge E_2 \cap E_1=\{0\}$ and it is true by CLIC and because $\cap$ is commutative; I must to proof ($p\to p+1$) $$[\sum_{i=1}^pE_i \doteq \bigoplus_{i=1}^p E_i \leftrightarrow \forall i\in \{1,...,p\}(E_i \cap \sum_{t \in \{1,...,p\}-\{i\}}E_t=\{0\})]\to [\sum_{i=1}^{p+1}E_i \doteq \bigoplus_{i=1}^{p+1} E_i \leftrightarrow \forall i\in \{1,...,p+1\}(E_i \cap \sum_{t \in \{1,...,p+1\}-\{i\}}E_t=\{0\})]$$ But I don't know to continue, I am confused...mmmm How can I do?

  • 1
    $\begingroup$ Unfortunately, induction won't work here. $\endgroup$ – DiffeoR Feb 26 '14 at 14:30
  • $\begingroup$ @DiffeoR, ah ok.. by contradiction? $\endgroup$ – mle Feb 26 '14 at 14:32
  • 1
    $\begingroup$ Yes ! that's the way to do. All the best. $\endgroup$ – DiffeoR Feb 26 '14 at 14:34
  • Proof of $\bf\leftarrow$

Let $e_i\in E_i$ such that $$e_1+e_2+\cdots+e_n=0$$ so forall $i$ we have $$e_i=-\sum_{j\ne i} e_j\in E_i \cap \sum_{j \in \{1,...,n\}-\{i\}}E_j=\{0\}$$

  • Proof of $\bf\rightarrow$ (by contraposition)

If there's $i$ such that $$E_i \cap \sum_{j \in \{1,...,n\}-\{i\}}E_j\ne\{0\}$$ so let $x\ne0$ in this intersection hence $$x=e_i=\sum_{j\ne i} e_j$$ so $$e_i-\sum_{j\ne i} e_j=0$$ but $$e_i\ne0$$ and this means that $E_1+E_2+...+E_p$ isn't a direct sum.QED.

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  • $\begingroup$ Two proofs in one! Nice! $\endgroup$ – amWhy Feb 27 '14 at 13:10

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