Suppose $V$ is a linear space over $\mathbb{R}$, $V_1$, $V_2$, $V_3$ are the subspaces of $V$.
If $V_1 \cap V_2 = \{0\}$, $V_2 \cap V_3 = \{0\}$, $V_3 \cap V_1 = \{0\}$, then is $V_1 + V_2 + V_3$ a direct sum?
If $V_1 \cap V_2 = \{0\}$, $V_3 \cap (V_1 + V_2) = \{0\}$, then is $V_1 + V_2 + V_3$ a direct sum?
Many thanks for any hints.
Best Answer
The first one is not direct sum.
Counter-example: Take $V_1$ to be span of $e_1$, $V_2$ to be span of $e_2$, $V_3$ to be span of $e_1+e_2$
The second is direct sum. ( Which I assume without proof that $V+W$ is direct sum if and only if the union of their basis is linearly independent.)
Proof: Suppose $B_1$ is the basis that span $V_1$, $B_2$ is the basis that span $V_2$,$B_3$ is the basis that span $V_3$.
Since $V_1+V_2$ is direct sum, $B_1\cup B_2$ span $V_1+V_2$.
Since $V_3 \cap (V_1 + V_2) = \{0\}$, $B_1\cup B_2$ is linearly independent with $B_3$, hence $(B_1 \cup B_2 )\cup B_3$ is a basis. Hence is direct sum.