Exercise 5.2.23

It’s enough to check whether $K_1$ has a common nonzero element with the summation of others or not. Thus we can do similarly for the other cases. Let $V_1$ be the summation of $K_2,K_3,\dots ,K_p$. Now, if there is some nonzero vector $x\in K_1(V_1\cap W_2)$, we may assume that $x=u+v$ with $u\in V_1$ and $v\in W_2$. Since $W_1$ is the direct sum of all $K_i$’s, we know $K_1\cap V_1=\{0\}$. So $x-u=v\ne 0$ is an element in both $W_1$ and $W_2$, a contradiction. Thus we’ve completed the proof.