In the books 1 and 2, in Somme directe d’une famille de sous-espaces vectoriels, I am reading the following:
1) let $E,F$ two vector subspaces of $V$, $E+F$ is direct sum, $E+F \doteq E\oplus F$, if $$\forall e \in E, f \in F (e+f=0_V \to e=f=0_V)$$
2) let $E,F$ two vector subspaces of $V$, then $$E+F \doteq E\oplus F \leftrightarrow E \cap F =\{0_V\}$$
3) let $E_1,E_2,…,E_p$ $p$-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 $$\forall e_1 \in E, e_2 \in E_2,…,e_p \in E_P (e_1+e_2+…+e_p=0_V \to e_1=e_2=…=e_p=0_V)$$
I think also:
It is correct? Thanks in advance!
No. This doesn’t work. You have to check that $E_i \cap \left(E_1+E_2+\cdots+E_{i-1}+E_{i+1}+\cdots+E_p\right) = \{0\}$ for each $i$.
Just checking that the subspaces don’t pairwise intersect is not enough.
Consider the example: $U = \{ (0,y) \;|\; y \in \mathbb{R} \}$, $V =\{ (x,0)\;|\; x \in \mathbb{R} \}$, and $W = \{ (x,x) \;|\; x \in \mathbb{R} \}$.
Each pair trivially intersect: $U \cap V = U \cap W = V \cap W = \{ (0,0) \}$. But this definitely is not a direct sum: $(-1,0)+(0,-1)+(1,1)=(0,0)$.
[Notice that $W \cap (U+V) = W \cap \mathbb{R}^2 = W \not= \{(0,0)\}$.]
$~$«$\displaystyle\sum_{i=1}^k E_i$ is direct $\,\Leftrightarrow\,$ the $E_i$s intersect trivially pairwise »$~$ is true for $V$ iff $k<3$ or $\dim V=1$.
Proof exercise: Suffices to consider $k=3$, $\dim V=2$. Show if $V=\langle v,w\rangle$ then $\langle v\rangle,\langle w\rangle,\langle v+w\rangle$ is a counterexample to the claim. Notice this does not depend on the field of scalars for the space $V$.