Prove that $V$ is the direct sum of $W_1, W_2 ,\dots , W_k$ if and only if $\dim(V) = \sum_{i=1}^k \dim W_i$ Let $W_1,\dots,W_k$ be a subspace of a finite dimensional vector space $V$ such that the $\sum_{i=1}^k W_i = V$.

Prove: $V$ is the direct sum of $W_1, W_2 , \dots, W_k$ if and only $\dim(V) = \sum_{i=1}^k \dim W_i$.


Define $\ \ B_i$ a basis of $ \ \ W_i\ \ $  for $\ \ i \in {1,..,k } $. Since sum( $W_i$)$= V$ we know that  $\bigcup_{i=1}^k B_i$ is a spanning list of $V$.
Now we add the condition : sum dim($B_i$)$=$sum dim($W_i$)$=$dim($V$). 
This means that $\bigcup_{i=1}^k B_i$ is a basis. (Spanning list of $V$ with same dimension).
Can you conclude from here the direct sum property?
If we define Basises as before and we know that $W_i$ is a direct sum can you conclude ( knowing that sum( $W_i$)$= V$ , wich is given)  that $\ \ \bigcup_{i=1}^k B_i$ is a Basis of $V$? 
