# The intersection assumption in the definition of direct sums

Let $$W_1,W_2,\ldots,W_k$$ be subspaces of a vector space $$V$$. We define the sum of these spaces to be the set $$\{v_1+v_2+\cdots+v_k:v_i\in W_i\text{ for }1\leq i\leq k\},$$ which we denote by $$W_1+W_2+\cdots+W_k$$ or simply by $$\sum_{i=1}^k W_i.$$ If, in addition, $$W_j\cap\sum_{i\not=j}W_i=\{\mathbf 0\}$$ for each $$j$$, then we call $$V$$ the direct sum of $$W_1,W_2,\ldots,W_k$$ and write $$V=W_1\oplus W_2\oplus\cdots\oplus W_k$$. What I'm concerned about is whether we can prove that the intersection of all subspaces in the direct sum is the zero space, that is, $$\bigcap_{i=1}^k W_i=\{\mathbf 0\}.$$ I think this is weaker than the statement that each subspace intersects the sum of the other subspaces only in the zero vector but don't know how to justify it. Please give me some help. Thank you.

• Try simple example like $\mathbb{R}^3$ with $W_1=\langle e_1 \rangle$, $W_2=\langle e_2 \rangle$ and $W_3=\langle e_1+e_2 \rangle$ . Certainly $\cap_i W_i = \{0\}$ but $W_3 \cap (W_1+W_2) \ni e_1+e_2$. So we cannot form direct sum with $W_1,W_2,W_3$ but $\cap_iW_i = \{0\}$. Commented Jul 28, 2020 at 4:01
• If $v \in \cap_i W_i \subseteq W_1\cap W_2 = W_1 \cap(W_2 + 0 \cdots +0) \subseteq W_1 \cap (W_2 + W_3 + \cdots + W_n) = \{0\}$. Then $v= 0$. Commented Jul 28, 2020 at 4:18

$$\{\mathbf 0 \}\subseteq W_j\cap W_l\subseteq W_j\cap\sum_{i\not=j}W_i=\{\mathbf 0\}$$ for all $$l\not=j$$. Thus also $$\bigcap_{i=1}^k W_i=\{\mathbf 0\}$$. Moreover, let $$V=\mathbb{R}^2$$, $$W_1=\mathbb{R}(1,0)$$, $$W_2=\mathbb{R}(0,1)$$, $$W_3=\mathbb{R}(1,1)$$. The $$W_1\cap W_2\cap W_3=\{(0,0)\}$$, but $$W_3\cap (W_1+W_2)=W_3\not=\{(0,0)\}$$.

It suffices to see that for $$i\neq j$$, we have $$W_i \cap W_j=\{0\}$$ (if we add more sets to the intersection, it can only get smaller, but since all the $$W_k$$ contain $$0$$, the intersection remains the zero space).

The inclusion "$$\supseteq$$" is easy, as both $$W_i$$ and $$W_j$$ contain $$0$$.

The other inclusion "$$\subseteq$$" follows from the direct sum: $$W_i\cap W_j\subseteq W_i \cap \sum_{k\neq i} W_k = \{0\}.$$

Then from this we can directly deduce that $$\bigcap_{i=1}^k W_i=\{0\}.$$
Whenever, "orthogonal complement" can't be usable, Then, clearly, it's the obvious fact that $$\{0\} \subseteq \bigcap_{i=1}^k W_i$$ , as each $$W_i$$ is a subspace.
Now, if possible let there be some non zero vectors (e.g. vector $$a$$) other than $$0$$ in $$\bigcap_{i=1}^k W_i$$ , so, in each $$W_i$$ .
Then, $$W_j\cap\sum_{i\not=j}W_i$$ should contain non zero vectors other than $$0$$ , as , $$W_j\cap\{0, a+0+0+......\text { 0's k-2 times } \} ⊂ W_j\cap\sum_{i\not=j}W_i$$,(since each $$W_i$$ contains vector $$a$$) But this is a contradiction to $$W_j\cap\sum_{i\not=j}W_i = \{0\}$$. So, $$\bigcap_{i=1}^k W_i=\{0\}.$$