Find $a,b$ such that $\mathbb R^3 = V \oplus W$ I have some doubts with this task: 

Let $a,b \in \mathbb R$. In linear space $\mathbb R^3$ we have:
  $$V = span([1,1+a,-2]^T,[2,6,-2-a]^T)= span(v_1,v_2) $$ and $$W = span([0,3,-1-b]^T,[2,2+b,-2]^T) = span(w_1,w_2)$$
  Find $a,b$ such that $\mathbb R^3 = V \oplus W$ 


If $\mathbb R^3 = V \oplus W$ then $V \cap W = \left\{\vec{0}\right\}$ 
I think that I should consider these cases: 
$1.$ $(v_1,v_2,w_1)$ are linear independent and $w_2 = \alpha w_1 $ 
$2.$ $(v_1,v_2,w_2)$ are linear independent and $w_1 = \alpha w_2 $ (but it is equivalent to $1$ so I don't have to check that)
$3.$ $(w_1,w_2,v_1)$ are linear independent and $v_2 = \alpha v_1 $ 
$4.$ $(w_1,w_2,v_2)$ are linear independent and $v_1 = \alpha v_2 $ (but it is equivalent to $3$ so I don't have to check that)
but there are a lot of calculus and I am not sure if (i) this is correct way (ii) how to do this quickly 
Can somebody help me with this problem? Thanks for your time.

Observe that both $V$ and $W$ can be at most two dimensional each. If both are two dimensional then their intersection cannot be just $\{0\}$. If one of them is $0$ dimensional OR both are $1$ dimensional then $\Bbb{R}^3$ cannot be their direct sum. So the only two possibilities are:


*

*$V$ is $1-$dim and W is $2-$dim.

*$V$ is $2-$dim and W is $1-$dim.


In the first case $v_1$ and $v_2$ must be dependent vectors and $v_1,w_1,w_2$ must be independent vectors. That means
$$k\begin{bmatrix}1\\1+a\\-2\end{bmatrix}=\begin{bmatrix}2\\6\\-2-a\end{bmatrix} \implies k=2 \text{ and } a=2$$
For the independence of $\{v_1,w_1,w_2\}$, the following matrix should have full rank.
$$\begin{bmatrix}1&3&-2\\0&3&-1-b\\2&2+b&-2\end{bmatrix} \longrightarrow \begin{bmatrix}1&3&-2\\0&3&-1-b\\0&b-4&2\end{bmatrix} \longrightarrow \begin{bmatrix}1&3&-2\\0&3&-1-b\\0&0&\frac{(b-2)(b-1)}{3}\end{bmatrix}.$$
Thus for full rank, we want $b \neq 1,2$. 
Thus with $a=2$ and $b \neq 1,2$, $\Bbb{R}^3=V \oplus W$.
Similarly you can do the other case.
