Decomposition of vector space using direct sums

Question

Let $V$ be the vector space, and let $V = U_1 \oplus \cdots \oplus U_n$. This is the direct sum.

1. Let $\beta_i$ be the basis of $U_i$. I want to show that $\beta_1 \cup \cdots \cup \beta_n$ is the basis for $V$.
2. Let $W \subset V$, then show that $W = (W \cap U_1) \oplus \cdots \oplus (W \cap U_m)$.

Answer 1

1. Let $\mathbf v\in V$. Then there are $\mathbf v_i\in U_i$ such that $$\mathbf v=\sum_{i=1}^n\mathbf v_i.$$ Also, there are $a_{ij}\in\mathbb F$ such that $\mathbf v_i=\sum_{j=1}^{m_i}a_{ij}\mathbf e_{ij}$ where $\mathbf e_{ij}\in \beta_i$ since $\beta_i$ is a basis for $U_i$. Then $$\mathbf v=\sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}\mathbf e_{ij}.$$ Hence, $\beta_1\cup\cdots\cup\beta_n$ is a spanning set of $V$.

Suppose that $$\sum_{i=1}^n\sum_{j=1}^{m_i}a_{ij}\mathbf e_{ij}=\mathbf 0$$ Then, for each $i$, $\sum_{j=1}^{m_i}a_{ij}\mathbf e_{ij}\in U_i$ but since $V=U_1\oplus\cdots\oplus U_n$, $U_{i_1}\cap U_{i_2}=\{\mathbf0\}$ for all $i_1,i_2$. Hence, $\sum_{j=1}^{m_i}a_{ij}\mathbf e_{ij}=\mathbf 0$ and by the linear independence of each $\beta_i$, we get $a_{ij}=0$ for all $i,j$ i.e. $\beta_1\cup\cdots\cup\beta_n$ is linearly independent.

2. As Berci said in the comment, the statement is not true. Consider $V=\mathbb R^2$, $U_1=\operatorname{span}(\mathbf e_1)$, $U_2=\operatorname{span}(\mathbf e_2)$ and $W=\operatorname{span}(\mathbf e_1+\mathbf e_2)$. Then $W\cap U_1=W\cap U_2=\{\mathbf 0\}$ while $W\neq\{\mathbf 0\}$.