This question is related to these two questions.
Difference between sum and direct sum
Examples of sum and direct sum of vector subspaces
I am going through a book and I managed to prove two necessary and sufficient conditions for a sum of vector subspaces
$$V_1 + V_2 + … V_s$$ to be direct sum $V_1 \bigoplus V_2 \bigoplus … \bigoplus V_s \tag{*}$
The first one is this:
(1) For every $i$:
$$V_i \cap (V_1 + V_2 + V_{i-1} + … V_{i+1} + … V_s) = \{0\}$$
The second one is this:
(2) The zero vector can be represented as sum of vectors from $V_i$ in a unique way
(namely $0\ =\ 0\ +\ 0 +\ \dots +\ 0$).
These two were given as problems in my book.
Now… I wonder if the following one is also a necessary and sufficient condition for the same thing
(3) $V_i \cap (\cup_{j \ne i} V_j) = \{0\}$
I tried but I cannot prove that from the third one it follows $(*)$.
Maybe this third condition is just not enough to prove $(*)$?
Or am I just not seeing how to use (3) to prove (*)?
This third condition is not in my book but I somehow felt it's also a necessary and sufficient condition for $(*)$ so I tried to prove it. But I could not do so.
Best Answer
In $\Bbb R^2$, take$$V_1=\{(x,0)\mid x\in\Bbb R\},\ V_2=\{(0,x)\mid x\in\Bbb R\}\text{ and }V_3=\{(x,x)\mid x\in\Bbb R\}.$$Then $V_1+V_2+V_3$ is not a direct sum (for instance, $(0,0)=(1,0)+(0,1)-(1,1)$), but, for each $i\in\{1,2,3\}$,$$V_i\cap\left(\bigcup_{j\ne i}V_j\right)=\{0\}.$$