I'm trying to solve the following excercise and I would like to ask for a hint as I'm confused:
Let $V$ be a finite dimensional vector space over a filed $K$. Let $V_1, V_2, V_3$ and $V_4$ be subspaces of $V$ such that $V=V_1 \bigoplus V_2 \bigoplus V_3 \bigoplus V_4$. If $\beta_i$ is a basis of $V_i$ $(i=1,2,3,4)$, prove that:
$$ \bigcup_{i=1}^4 \beta_i $$
is a basis of $V$
My attempt:
Let: $\dim V_1 = k_1, \dim V_2 =k_2, \dim V_3 = k_3, \dim V_4=k_4 $, it is clear that $\dim V = k_1+k_2+k_3+k_4$
Let:
$\beta_1 = \{ r_1,...,r_{k_1}\}$, $\beta_2 = \{ s_1,...,s_{k_2}\}$, $\beta_3 = \{ t_1,...,t_{k_3}\}$, $\beta_4 = \{ u_1,...,u_{k_4}\}$ bases of $V_1, V_2, V_3$ and $V_4$ We claim that they are linearly independent. Let $\alpha_1,...,\alpha_{k_1}, \beta_1, ..., \beta_{k_2}, \gamma_1,..., \gamma_{k_3}, \delta_1,...,\delta_{k_4}$ scalars such that:
$$ \sum_{i=1}^{k_1} \alpha_i r_i + \sum_{i=1}^{k_2} \beta_i s_i + \sum_{i=1}^{k_3} \gamma_i t_i + \sum_{i=1}^{k_4} \delta_i u_i = 0$$
We see that: $\sum_{i=1}^{k_1} \alpha_i r_i \in V_1$, $\sum_{i=1}^{k_2} \beta_i s_i \in V_2$, $\sum_{i=1}^{k_3} \gamma_i t_i \in V_3$ and $\sum_{i=1}^{k_4} \delta_i u_i \in V_4$. Let $w_1=\sum_{i=1}^{k_1} \alpha_i r_i$, $w_2=\sum_{i=1}^{k_2} \beta_i s_i$, $w_3 = \sum_{i=1}^{k_3} \gamma_i t_i$ and $w_4 = \sum_{i=1}^{k_4} \delta_i u_i$. Then:
$$ w_1+w_2+w_3+w_4=0 \Rightarrow w_1=-w_2-w_3-w_4 $$
So that $w_1 \in V_1 \cap (V_2+V_3+V_4)$...
And I actually don't know how to continue or if this is the best way to approach the proof. I know that, in order for $\beta$ to be a basis of $V$, their vectors must be linearly independent, and that is it! since $\dim V = k_1+k_2+k_3+k_4$ and $\beta$ has $k_1+k_2+k_3+k_4$ vectors