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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.

peter.petrov
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2 Answers2

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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 $i\in\{1,2,3\}$,$$V_i\cap\left(\bigcup_{j\ne i}V_j\right)=\{0\}.$$

José Carlos Santos
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  • Hm... Nice example, thanks. I want to make sure I understand you properly. So basically these are 3 lines (1-dimensional subspaces) through the origin of the 2d space $\Bbb R^2$. Correct? They satisfy (3) but their sum is not direct. Is their sum equal to $\Bbb R^2$ actually? Seems it is, right? – peter.petrov Sep 07 '21 at 21:26
  • Yes to everything. In particular, since $V_1+V_2=\Bbb R^2$, $V_1+V_2+V_3$ is also equal to $\Bbb R^2$. – José Carlos Santos Sep 07 '21 at 21:29
  • I see. Thanks again. – peter.petrov Sep 07 '21 at 21:31
  • ...whereas $\bigcup_i V_i$ is just a star-shaped region in $\mathbb{R}^2$, which is not convex and thus not a vector space. – leftaroundabout Sep 08 '21 at 08:29
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The third condition is not sufficient. Consider $V_1=\{(x,0):x\in\Bbb R\}$, $V_2=\{(x,x):x\in\Bbb R\}$, and $V_2=\{(0,x):x\in\Bbb R\}$. Then $V_1\cap(V_2\cup V_3)=\{(0,0)\}$, but $\Bbb R^2\neq V_1\oplus V_2\oplus V_3$.

pancini
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