1

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:

  • 4) let $E_1,E_2,...,E_p$ $p$-vector subspaces of $V$, then $$E_1+E_2+...+E_p \doteq E_1\oplus E_2 \oplus ... \oplus E_p \leftrightarrow \forall i,j \in \{1,2,...,p\}(i\neq j \to E_i \cap E_j = \{ 0_V \})$$

It is correct? Thanks in advance!

mle
  • 2,287

2 Answers2

2

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 $W \cap (U+V) = W \cap \mathbb{R}^2 = W \not= \{(0,0)\}$.]

Bill Cook
  • 29,244
  • therefore 1),2),3) are correct but 4) is not!? – mle Feb 26 '14 at 02:10
  • 1
    Yes. (1) and (2) are equivalent (they both can serve as the definition of the direct sum of two subspaces). (3) is a generalization of (1) and my answer (checking that each subspace doesn't intersect the sum of the remaining subspaces) is the proper generalization of (2). So if you call my answer (4), then (3) and (4) are equivalent with (1) and (2) as special cases. :) – Bill Cook Feb 26 '14 at 02:18
  • therefore: $$E_1+E_2+...+E_p \doteq E_1\oplus E_2 \oplus ... \oplus E_p \leftrightarrow \forall i \in {1,2,...,p}(E_i \cap \left(E_1+E_2+\cdots+E_{i-1}+E_{i+1}+\cdots+E_p\right) = {0_V})$$ It is correct?! :) Thanks soo much in advance! – mle Feb 26 '14 at 02:31
  • 1
    @ГарнакОлэговытц Yes, that last statement is correct. – anon Feb 26 '14 at 02:47
  • @anon, thans soo much for answer! :) – mle Feb 26 '14 at 02:48
1

$~$«$\displaystyle\sum_{i=1}^k E_i$ 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 $V$.

anon
  • 151,657
  • what it means "$E_i$ s intersect trivially pairwise" ? – mle Feb 26 '14 at 12:07
  • @ГарнакОлэговытц Given your question, what do you think it means? :) If $P(x,y)$ is a property of a pair of things $x$ and $y$, then we say ${x_1,x_2,\cdots}$ is pairwise $P$ if $P(x_i,x_j)$ for all distinct $i,j$. For example, $P(x,y)$ might say "$x$ and $y$ are coprime," in which case we can speak of a list of integers being pairwise coprime. Two subspaces intersect trivially if their intersection is the trivial subspace $0$. Thus, "$E_i$s intersect trivially pairwise" means $E_i\cap E_j=0$ whenever $i\ne j$. – anon Feb 26 '14 at 18:01