Is this always true about direct sums? $V$ is a vector space over a field $V$, and $V=V_1\oplus \dots \oplus V_k$, and $B_i$ is a basis for each $V_i$. Then $B=B_1 \cup \dots B_k$ is a basis for $V$?
 A: Yes this is true. The direct sum of vector spaces $V_1,...,V_k$ is characterized through the property that
$$\sum \limits_{i=1,...,k} v_i = 0 \Leftrightarrow \forall i\in\{1,...,k\}: v_i = 0$$
This together with the defining property of a basis lets you deduce the statement of your question.
Another perspective (I don’t know what facts you know about direct sums) is the following. Recall that the field $K$ can be seen as a vector space over itself, it is the prototypical 1-dimensional vector space over $K$. Now given a vector space of dimension $d$, to give a basis $\{b_1,...,b_k\}$ of $V$ amounts to giving an isomorphism $K^{ d} \cong V$. Indeed by the extension theorem / the universal property of a basis there are unique linear maps
$$\begin{array}{rcl}
\varphi: K^{ d} & \rightarrow & V\\
e_i & \mapsto & b_i
\end{array}$$
and
$$\begin{array}{rcl}
\psi: V & \rightarrow & K^{ d}\\
b_i & \mapsto & e_i
\end{array}$$
which are mutually inverse to each other.
So having a basis $B_i=\{b_{i1},...,b_{id_i}\}$ for each vector space $V_i$ is the same as having an isomorphism $K^{ d_i} \cong V_i$ for every $i$, where $d_i = \dim_K V_i$. The universal property of $\oplus$ induces unique linear maps
$$\begin{array}{rcl}
K^{d_1+...+d_n} = K^d_1 \oplus ... \oplus K^d_k &\rightarrow& V_1 \oplus ... \oplus V_k\\
\sum \limits_{i=1}^k \sum \limits_{j=1}^{d_i} \lambda_{ij}e_{ij} & \mapsto & \sum_i \varphi_i(\sum\limits_{j=1}^{d_i}\lambda_{ij}e_{ij}) = \sum \limits_{i=1}^k \sum \limits_{j=1}^{d_i} \lambda_{ij}b_{ij}
\end{array}$$
and
$$\begin{array}{rcl}
V_1 \oplus ... \oplus V_k & \rightarrow & K^{d_1 + ... + d_k}\\
\sum \limits_{i=1}^k v_i & \mapsto & \sum \limits_{i=1}^k \psi_i(v_i)
\end{array}$$
which are again mutually inverse to each other. Hence $B_1 \cup ... \cup B_k$ does indeed form a basis of $V_1 \oplus ... \oplus V_k$.
