Exercise 2.C.15

Proof. Let $u_1,\dots ,u_n$ be a basis of $V$. Define $U_j$ by

for $j=1,\dots ,n$. Each $U_j$ is clearly $1$-dimensional. Moreover, $V=U_1+\cdots +U_n$, because this sum contains all the basis vectors. Thus, for every $v\in V$, we can write it uniquely in the form

for some scalars $a_1,\dots ,a_n\in 𝔽$ (by 2.29). Since each $u_j$ is in $U_j$, by the definition of direct sum, $U_1+\cdots +U_m$ is a direct sum. □