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The direct sum of the three subspaces $V_1, V_2, V_3$is defined as $V_1 \oplus V_2 \oplus V_3$ iff in their intersection lies only a zero vector. I want to find the subspace in $\Bbb R^3$ for which the direct sum exists and the subspace where the direct sum isn't defined.

1) Direct sum exists: For example $\{v_1,v_2,v_3\}=\{(1,-1,1), (-1,-1,1),(1,1,1)\}$ (the set is linearly independent)

2) Direct sum doesn't exist: For example $\{v_1,v_2,v_3\}=\{(1,-1,1), (2,-2,2), (1,1,1)\}$ (the set is linearly dependent)

PS: $v_i \in V_i$ and these vectors individually form the subspaces because they are closed under a multiplication by a scalar and under an addition.

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  • $\begingroup$ That is not a correct definition of the triple direct sum. $\endgroup$ – Qiaochu Yuan Mar 8 '17 at 19:22
  • $\begingroup$ Then I don't know. $\endgroup$ – Leif Mar 8 '17 at 19:30
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The direct sum is constructed iteratively. First you construct $V_1 \oplus V_2$, then you construct $(V_1 \oplus V_2) \oplus V_3$. So you have to require that $$V_1 \cap V_2 = \{0\}\,,$$ but also that $$(V_1 \oplus V_2) \cap V_3 = \{0\}\,.$$ Note that this is more restrictive than just requiring $V_1 \cap V_2 \cap V_3 = \{0\}$.

For example, let $e_1, e_2, e_3$ be the standard basis in $\mathbb R^3$ and $[e_1]$ the span of $e_1$. Then the direct sum $$ [e_1] \oplus [e_2] \oplus [e_3]$$ exists, but the direct sum $$ [e_1] \oplus [e_2] \oplus [e_1+e_2]$$ does not exist, even though $$ [e_1] \cap [e_2] \cap [e_1+e_2] = \{0\}\,.$$

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  • $\begingroup$ It doesn't exist, because $e_1$ and $e_2$ can't span whole \Bbb R^3, right? $\endgroup$ – Leif Mar 8 '17 at 21:17
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    $\begingroup$ No, it doesn't exist, because $e_1+e_2 \in [e_1] \oplus [e_2]$. $\endgroup$ – Martins Bruveris Mar 8 '17 at 21:46
  • $\begingroup$ Ok, thank you very much. $\endgroup$ – Leif Mar 8 '17 at 21:50

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