Direct sum of two vector subspaces, show that $V = V_1 \oplus V_2$ Consider $$V_1 = \{(x,2x,3x)~|~x \in \mathbb{R}\}$$ $$V_2 = \{(0,y,z)~|~y,z \in \mathbb{R}\}$$ of the row vector 3-space $\mathbb{R}^3$.
How do i show that $V_1 + V_2 = \mathbb{R}^3$?
For a start $V_1+V_2 \subseteq \mathbb{R}^3$ because both $V_1,V_2$ are subspaces of $\mathbb{R}$.
Now we try to prove $\mathbb{R}^3 \subseteq V_1+V_2$. Let $a \in \mathbb{R}^3$, that is for each $a \in \mathbb{R}^3$, there exists $v_1 \in V_1,v_2 \in V_2$ such that $a = v_1+v_2$. But i do not know how to continue. 
 A: Hint:
Find the values of $b_1, b_2, b_3$ such that  $$a=(a_1,a_2,a_3) = (a_1,2a_1,3a_1)+ (b_1,b_2,b_3)$$
A: For every $P=(a, b, c) \in \mathbb{R}^3$; we have: 
$$(a, b, c)=(a, 2a, 3a)+(0, b-2a, c-3a);$$ 
notice that $(a, 2a, 3a) \in V_1$ and $(0, b-2a, c-3a) \in V_2$.


Also if we have $(a, b, c)=P_1+P_2$; for some $P_1=(x_1, 2x_1, 3x_1) \in V_1$ and $P_2=(0, x_2, x_3) \in V_2$.
Then we must have: 
$$ 
             \left\{ \begin{array}{lcc}
             a=x_1+0 , \\
             b=2x_1+x_2 , \\
             c=3x_1+x_3 , \\ 
             \end{array}
             \right.
$$ 
so we can conclude that $x_1=a, x_2=b-2a, x_3=c-3a$;
i.e. $P_1=(a, 2a, 3a), P_2=(0, b-2a, c-3a)$;
which implies the uniqueness of $P_1, P_2$.
A: Another approach is that $$V_1 = \operatorname{span}\{(1,2,3)\},\ V_2 = \operatorname{span}\{(0,1,0),(0,0,1)\}$$ which implies that $$V_1+V_2 = \operatorname{span}\{(1,2,3),(0,1,0),(0,0,1)\}.$$
Since $$\det\begin{pmatrix}
1& 2& 3\\
0& 1& 0\\
0& 0& 1
\end{pmatrix} = 1,$$
these three vectors are linearly independent and thus $\dim(V_1+V_2) = 3.$ Finally $V_1+V_2\subseteq \mathbb R^3$ gives $V_1+V_2 = \mathbb R^3$ because dimensions are equal.
