# Proving that $V = U_1 \oplus U_2 \oplus \ldots \oplus U_k$.

Problem: Let $V$ be a vectorspace and $\beta$ a basis for $V$. Now make a partition of $\beta$ in a disjoint union of subsets $\beta_1, \ldots, \beta_k$ and let $U_i = \text{span}(\beta_i)$ for every $i = 1, \ldots, k$. Prove then that $V = U_1 \oplus U_2 \oplus \ldots \oplus U_k$.

Proof: Take an arbitrary vector $v \in V$. Then we have to show that there are unique vectors $u_1 \in U_1, u_2 \in U_2, \ldots, u_k \in U_k$ such that \begin{align*} v = u_1 + u_2 + \ldots + u_k. \end{align*} Suppose there are two such ways, i.e. that \begin{align*} v= u_1' + u_2' + \ldots + u_k'. \end{align*} also holds. Then we have \begin{align*} \sum_{i=1}^k u_i = \sum_{i=1}^k u_i', \end{align*} or \begin{align*} (u_1 - u_1') + (u_2 - u_2') + \ldots + (u_k - u_k') = 0 \end{align*}

Now I'm not sure how to proceed. I want to show that $u_i = u_i'$. How can I do that?

As $U_i = \def\sp{\mathop{\rm span}}\sp(\beta_i)$, we may write $$u_i = \sum_{v\in\beta_i} \lambda_v v, \quad u_i' = \sum_{v \in \beta_i} \lambda'_v v$$ for some $\lambda_v, \lambda'_v \in k$ (the scalar field). We have, as the $\beta_i$ are disjoint and $\bigcup \beta_i = \beta$, that $$\sum_{v \in \beta} \lambda_v v = \sum_i u_i = \sum_i u'_i = \sum_{v \in \beta} \lambda'_v v$$ As $\beta$ is a basis, $\beta$ is linear independent, giving $\lambda'_v = \lambda_v$ for all $v$. Hence $u_i = u'_i$ for all $i$.
• I don't see on what basis you can conclude that $\sum_{v \in \beta} \lambda_v v = \sum_{i} u_i$. In the earlier step above you said that one such vector, i.e. only $u_i = \sum_{v \in \beta_i} \lambda_v v$. But now we are speaking of the sums of such vectors. How does this follow? – Kamil Jul 2 '15 at 13:45