# Direct sum of $3$ subspaces

$V_1$,$V_2$,$V_3$ are subspaces of vector space $V$. How to prove that if $V_1 \cap \left(V_2+V_3\right) = V_2 \cap \left(V_1+V_3\right) = V_3 \cap \left(V_2+V_3\right)=\{0\}$ so $V_1\oplus V_2 \oplus V_3$ ?

I tried to prove this way: let $w$ a vector such that $w= u_1+u_2+u_3$ where $u_i$ is in $V_i$ . Now I assume that there is another way to write this vector (and want to get a contradiction in order to prove direct sum) : $w= u_1'+u_2'+u_3'$ again $u'_i$ is in $V_i$ .

now I subtract and get $0 = u_1-u_1'+u_2-u_2'+u_3-u_3'$ but how to continue in order to show that each one of the elements is zero ? (then i will get $u_i=u_i'$ and this prove direct sum)

From $0=u_1-u_1' +u_2-u_2'+u_3-u_3'$ you know that $-u_1+u_1' = u_2-u_2'+u_3-u_3'$. Note that $-u_1+u_1' \in V_1$, and $u_2-u_2'+u_3-u_3' \in V_2+V_3$. You know that $V_1 \cap (V_2+V_3)=\{0\}$. What can you conclude about $-u_1+u_1'$?