If $V=W_1+\cdots+W_k$ and $\dim(V) =\sum_{i=1}^{k}\dim(W_i)$, prove that $V = W_1 \oplus\cdots\oplus W_k$ 
Let $V$ be a vector space and let $W_1,\ldots,W_k$ be subspaces of $V$ such that $V = W_1 +\cdots+ W_k$ and $\dim(V) = \dim(W_1) +\cdots+ \dim(W_k)$. Prove that $V = W_1 \oplus\cdots\oplus W_k $.

My attempt: 
$k=1$ is trivial, for $k=2$, 
$\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1 \cap W_2)$ which gives us $\dim(W_1 \cap W_2)=0$, which implies $W_1 \cap W_2=\{0\}$. Hence $V = W_1 \oplus W_2 $. 
For $k=3$, we get $\dim(W_1+W_2+W_3)=\dim(W_1+W_2)+\dim(W_3)-\dim((W_1+ W_2)\cap W_3)$, which implies $\dim(W_1 \cap W_2)+\dim((W_1+ W_2)\cap W_3)=0$. We get the following: 
$1. \,W_1 \cap W_2 = \{ 0 \}$
$2. \, (W_1+ W_2)\cap W_3 = \{0\} $ , $(W_1+ W_3)\cap W_2 = \{0\} $ and $(W_2+ W_3)\cap W_1 = \{0\}$
Are conditions $1$ and $2$ enough to prove that $V = W_1 \oplus W_2 \oplus W_3  $ ? This method isn't good for solving this problem for general $k$, is there any better way to prove this?

In fact
$$\begin{cases}
W_1 \cap W_2 = \{ 0 \} \\
(W_1+ W_2)\cap W_3 = \{0\}
\end{cases}$$
are sufficient conditions. The second one implies $(W_1+W_2) + W_3= (W_1+W_2) \oplus W_3$ and the first one that $W_1 + W_2 = W_1\oplus W_2$. Therefore
$$V= W_1+W_2+W_3 = W_1\oplus W_2 \oplus W_3.$$
However for the general case, I would rather proceed by induction on the dimension $n = \dim V$:


*

*Easy for $n=2$...

*Suppose that the result is true for $n \ge 2$ and that $V = W_1 +···+ W_k$ with $\dim V = \displaystyle\sum_{i=1}^{k} \dim W_i = n+1$. Without loss of generality, we can suppose that $\dim W_k \ge 1$. Then $V = (W_1 +···+ W_{k-1})+ W_k$  and $\dim V = \dim (W_1 +···+ W_{k-1})+ \dim W_k$. With what you proved in your question, you have $V = W \oplus W_k$ where $W = W_1 +···+ W_{k-1}$. Now $\dim W \le n$. So you can apply induction hypothesis to it, getting $W = W_1 \oplus···\oplus W_{k-1}$ and finally $V = W_1 \oplus···\oplus W_{k-1} \oplus W_k$ as desired.

