$V_1$,$V_2$,$V_3$ 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 $-u_1+u_1'$?

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.