Show that $V = V_1 \oplus \dots \oplus V_m \Leftrightarrow B = B_1 \cup \dots \cup B_m$ is a basis of $V$. Let $V$ be finite-dimensional and suppose $V_1,...,V_m$ are subspaces of $V$. Furthermore, suppose that $B_i$ is a basis of $V_i$ for each $i$. Show that $V = V_1 \oplus \dots \oplus V_m \Leftrightarrow B = B_1 \cup \dots \cup B_m$ is a basis of $V$.
How do I go about proving this?

Update:
So I think I've proved the $\Rightarrow$ side
Assume $V = V_1\oplus \dots \oplus V_m$ 
This means that $V = V_1 + \dots + V_m$ and $V_1 \cap \dots \cap V_m =$ {$0$}
So $V_1,...,V_m$ spans $V$ and their intersection is trivial
Suppose that $V = V_1\oplus \dots \oplus V_m$ has a basis $B = B_1\cup \dots \cup B_k$ where $k < m$
Since there are less bases than subspaces, this implies that some $V_i$ shares the same basis with some other $V_i$ for $i = (1,2,...,m)$ meaning there is linear dependence in the subspaces. 
But, this contradicts the statement $V = V_1 + \dots + V_m$ ($V_1,...,V_m$ spans $V$) since subspaces cannot span a vector space if there is linear dependence. 
Thus, there must be a unique basis $B_i$ for each $V_i$ meaning the basis for $V$ must be $B = B_1 \cup \dots \cup B_m$
Any corrections on that side of the proof or suggestions for the other side are greatly appreciated! 
 A: Let's start with the $\Rightarrow$ side.

Assume $V = V1 \oplus \cdots \oplus V_m$
This means that $ V = V_1 + \cdots +V_m$ and $V_1 \cap \cdots \cap V_m= \{0\}$
So $V_1,...,V_m$ spans $V$ and their intersection is trivial.

Alright so far, but saying that $V = V_1 \oplus \cdots \oplus V_m$ implies that $V_i \cap V_j = \{ 0 \}$ for any $i \neq j \in \lbrace 1, \ldots, m \rbrace$. Note that this is stronger than $V_1 \cap \cdots \cap V_m= \{0\}$
After that, you've lost me. It looks like you want to prove the statement by contradiction: you assume that $B_1 \cup \ldots \cup B_k$ is a basis for some $k < m$ and want to conclude that this cannot be true. However, this does not imply that $B_1 \cup \ldots \cup B_m$ is a basis. Proof by contradiction only works if you assume the exact opposite of what you want to prove. "$B_1 \cup \ldots \cup B_k$ is a basis for some $k < m$" is not the oppositie of "$B_1 \cup \ldots \cup B_m$ is basis", as there is still a third option covered by neither of those: $B_1 \cup \ldots \cup B_k$ is not a basis for any $k \leq m$.
Let's try a different approach. Remember that, to show that $B$ is a basis for $V$, we need two things:


*

*$B$ spans $V$

*the elements of $B$ are linearly independent


Start with the first one. Take an arbitrary vector $v \in V$. Using that $V = V_1 + \ldots + V_m$, there exist some $v_1 \in V_1, \ldots, v_m \in V_m$ such that $ v = v_1 + \ldots + v_m$. Each $v_i$ can be written as a linear combination of elements of $B_i$ (because $B_i$ is a basis for $V_i$!), so $v$ itself can be written als a linear combination of elements in $B = B_1 \cup \ldots \cup B_m$. This shows that $B$ spans $V$.
Try the linear independence yourself: start by assuming that there exist vectors $v_1, \ldots, v_N \in B$ and non-zero numbers $r_1, \ldots, r_N$ with $r_1v_1 + \ldots + r_Nv_n = 0$. This time, try to reach a contradiction with the fact that $V_i \cap V_j = \lbrace 0 \rbrace$ for $i \neq j$ and that $B_i$ is a linearly independent set for each $i$. Let me know what you get.
